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Quantization

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<p><strong>CHAPTER &ndash; 20</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>Quantization of Energy</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>Atomic models</strong></p> <p>Initially, atoms are supposed to be indivisible and the smallest particles to be found. However the study of electric discharge through gases led all to believe that atom is not indivisible but is composed of negative and positive charges. After which different scientists had suggested different model of atomic structure.</p> <p><strong>Thomson's atomic model</strong></p> <p>Sir J. J. Thomson was first scientist to purpose structure of an atom. On the basis of his experiment on electric discharge through gases, he concluded that an atom consists negatively charged electrons and positively charged protons. According to him, an atom is a positively charged spherical mass having radius around 10<sup>&ndash;10</sup> m in which negatively charged electrons are distributed like seeds in water melon or like plums in pudding. Because of this reason, Thomson's model is also known as plum pudding model. The total positive charge distributed uniformly over spherical mass is equal to the total charge carried by electrons embedded and hence an atom is electrically neutral.</p> <table> <tbody> <tr> <td width="179">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p>&nbsp;</p> <p><strong>Drawbacks of Thomson's atomic model:</strong></p> <p>Thomson's atomic model had soon to be discarded for the following reasons,</p> <ol> <li>i) Spectral series in the hydrogen atom could not be explained by Thomson's model</li> <li>ii) Thomson's model could not explain scattering of a - particles through large angles in Rutherford's experiment of a- scattering.</li> </ol> <p>&nbsp;</p> <p><strong>Rutherford's Atomic Model:</strong></p> <p><strong>a</strong><strong> - Scattering experiment:</strong></p> <p>In order to find out the structure of an atom and to test the Thomson's atomic model, Ernest Rutherford and his coworkers (Geiger and Mardsen) performed series of experiments in 1911, known as a - scattering experiment. This experiment gave some useful information about the atomic structure.</p> <table> <tbody> <tr> <td width="31">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>Experimental arrangement for a- scattering experiment is as shown in figure. a- particles emitted by a radioactive element (e.g. Radium) kept inside a lead cavity are confined in a narrow beam by using a lead slit. The collimated beam of a - particles is then allowed to fall on a thin gold foil (~ 4 &times; 10<sup>&ndash;7</sup> m thick). A zinc sulphide (ZnS) coated screen is placed on the other side to observe flash of light produced by a- particles. The whole setting is kept inside a vacuum chamber to prevent scattering of a - particle from air molecules.</p> <table> <tbody> <tr> <td width="126">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p><strong>&nbsp;</strong></p> <p><strong>Fig:</strong> &nbsp;Scattering of a - particles</p> <p>&nbsp;</p> <p>On the basis of Thomson's atomic model, it was expected that, they could not be deflected as electrons were light particles. However, they observed quite different observations. On these experiments they observed,</p> <ol> <li>Most of the a- particles were found to cross the gold foil without deviation.</li> <li>Some of the a - particles were deflected by small angles. However, some a- particles were suffered fairly large angle deflection. &nbsp;</li> </ol> <p>iii.&nbsp;&nbsp;&nbsp; A very few number of a - particles were found to suffer a deflection of very large angle of around 180&deg;.&nbsp;</p> <p>Thomson's atomic model could not explain the scattering of&nbsp; a - particles through large angles like 180&deg; because a - particles are heavy particles moving with high velocities and electrons are light particles so cannot cause the deflection of 180&deg;. Only deflection through small angles could be understood according to Thomson's atomic model. From those experimental observation Rutherford with his coworkers concluded following suggestion for structure of atom,</p> <ol> <li>i) As most of a - particles pass through the atom, the atom is mostly empty or contains very light particles (electrons) which may not have significant effect on deflection.</li> <li>ii) Large angle deflection of 180&deg; of positively charged a- particles could happen only if whole positive charge inside an atom is concentrated in a small region and deflection is due to the force of repulsion between +ve charge of atom and a- particle.</li> </ol> <p>iii)&nbsp;&nbsp; Large angle deflection also suggests that almost all mass of an atom is concentrated in a small region.</p> <p><strong>&nbsp;</strong></p> <p><strong>Rutherford atomic model:</strong></p> <p>On the basis of observations observed in a- scattering experiment, Rutherford (and Coworkers) suggested the structure of an atom as follows in 1911;</p> <ol> <li>i) An atom can be supposed as a sphere of size about 10<sup>&ndash;10</sup> m however all the positive charge and almost whole mass of atom is concentrated in a small region at centre named as nucleus.</li> <li>ii) Electrons are spread over the outer parts of nucleus in an atom thereby leaving lot of empty space in the atom.</li> </ol> <p>iii)&nbsp;&nbsp; The total positive charge concentrated in nucleus is equal to the total charge carried by electrons around the nucleus and hence an atom is electrically neutral.</p> <ol> <li>iv) Electrons outside the nucleus would be pulled toward nucleus due to electrostatic force of attraction. Thus, Rutherford suggested that the electrons are revolving around the nucleus in stationary orbit and the necessary centripetal force is provided by force of attraction between positively charged nucleus and negatively charged electrons.</li> </ol> <p>&nbsp;</p> <p><strong>Drawbacks of Rutherford's atomic model:</strong></p> <p>Rutherford had tried to overcome the failure of Thomson's atomic model but at the end Rutherford atomic model also has some drawbacks which can be listed as follows,</p> <ol> <li>i) According to modern theory of electromagnetism, a charged particle in accelerated motion should emit or radiate energy. Therefore, in an atom where electrons are in accelerated motion should also emit energy continuously which should result continuous decrease in radius of orbit and finally electron must get into the nucleus spirally, which is against the stability of an atom.</li> <li>ii) As Rutherford suggested, electrons revolve around the nucleus in all possible orbits and hence must emit radiations of all possible wavelength. i.e. must emit a continuous spectra. But experimental observations suggest that the atom like hydrogen emits a line or discrete spectrum. Thus, Rutherford assumption is against the experimental result.</li> </ol> <p><strong>&nbsp;</strong></p> <p><strong>Bohr's atomic model:</strong></p> <p>As Rutherford's atomic model failed to account the stability of atom and line spectra of hydrogen atom, Neil Bohr tried to overcome the drawbacks of Rutherford's model of an atom. For this he spent several months in Rutherford's laboratory in 1912 and at the end he suggested an atomic model in 1913 for which he was awarded Nobel Prize in physics in 1922. After long study in Rutherford's laboratory Bohr purposed an atomic model and he gave his postulate as,</p> <ol> <li>i) An atom consists a positively charged heavy mass at its centre and carry almost all mass of atom.</li> <li>ii) Electrons can revolve round the nucleus only on certain permitted circular orbit called as stationary orbits (or energy levels) of different radius.</li> </ol> <p>iii)&nbsp;&nbsp; Electrons revolve round the nucleus in permissible orbits and are those in which angular momentum of the electron equals to integral multiple of h/2p, where h is a constant called as Planck's constant.</p> <p>&nbsp;</p> <p>If m and v be the mass and velocity of an electron revolving in an orbit of radius r, then according to Bohr's Postulate,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; L = ,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; n = 1, 2, 3, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;. is principal quantum number.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; mvr&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = n &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;. (1)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; This is also known as Bohr's quantization condition. &nbsp;</p> <ol> <li>iv) All the stationary orbits correspond to a certain energy level and an electron moving round in a stationary orbit does not radiate or absorb energy as expected by modern theory of electromagnetism. An electron radiates or absorbs energy only when it jumps from one stationary orbit to other. Electron emits energy when it jumps from higher energy level to lower energy level and it absorbs energy while jumping from lower energy level to higher energy level.</li> </ol> <p>If En<sub>1</sub> and En<sub>2</sub> are energies associated with the orbits of principal quantum number n<sub>1</sub> and n<sub>2</sub> respectively, then energy of emitted radiation while electron jumps from n<sub>2</sub> to n<sub>1</sub> is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf&nbsp; = En<sub>2</sub>&nbsp; &ndash; En<sub>1</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;. (2)</p> <p>Where, f is frequency of emitted radiation.</p> <p>&nbsp;</p> <p><strong>Bohr's theory of Hydrogen atom</strong></p> <p>The radii of permissible orbits as well as energy and frequency of electron revolving round the nucleus can be calculated by using Bohr's postulates.</p> <p>Hydrogen atom consist a nucleus having positive charge (charge carried by a proton) and an electron having</p> <p>negative charge is revolving round the nucleus in a stationary orbit.</p> <p>&nbsp;</p> <p>Consider m and e be the mass and charge of an electron revolving around the nucleus in a stationary orbit having radius r. Also, consider v be the velocity of electron. Then force of attraction between positively charged nucleus and negatively charged electron can be given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; F<sub>e</sub><sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>= K</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; F<sub>e</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip; &nbsp;(1)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where e<sub>o</sub> is permittivity of free space and</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1 in CGS system</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp; = 9 &times; 10<sup>9</sup> Nm<sup>2</sup> C<sup>&ndash;2 </sup>&nbsp;in SI.</p> <p>As electron revolves round the nucleus in an orbit of radius r with velocity v, the centripetal force required can&nbsp; be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; F<sub>c</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip; (2) &nbsp;</p> <p>Here, the necessary centripetal force for the motion of electron in circular orbit is provided by the electrostatic force of attraction between positively charged nucleus and negatively charged electron. Therefore, we can have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; F<sub>e</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = F<sub>c</sub></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>or,&nbsp; &nbsp; &nbsp; = <sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>....................&nbsp; (3)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; We have form Bohr's quantization condition,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; mvr&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; v<sup>2</sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;..&nbsp; (4)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Substituting this value in equation (3) we get &nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &times;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;. (5)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; And in general,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; r<sub>n</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;..&nbsp; (6)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; n = 1, 2, 3&hellip;&hellip; (Principal quantum number)</p> <p>This equation gives the radius of n<sup>th</sup> orbit of hydrogen atom and from the equation it is clear that the permissible radii are proportional to n<sup>2</sup>.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Substituting all the values in equation (6)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; e<sub>o</sub> &nbsp;&nbsp; = 8.85 &times; 10<sup>&ndash;12</sup> C<sup>2</sup> N<sup>&ndash;1</sup> m<sup>&ndash;2</sup> &nbsp; ,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; h = 6.62 &times; 10<sup>&ndash;34</sup> Js</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; m&nbsp;&nbsp; = 9.1 &times; 10<sup>&ndash;31</sup> kg &nbsp;&nbsp; ,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; e = 1.6 &times; 10<sup>&ndash;19</sup> C</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; r<sub>n</sub> &nbsp;&nbsp; = 5.3 &times; 10<sup>&ndash;11</sup> m &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;= 0.53 &times; 10<sup>&ndash;10</sup> m</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \&nbsp;&nbsp; r<sub>1</sub> = 0.53</p> <p>This value of radius is called as Bohr radius and is denoted by a<sub>o</sub> in general.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; i.e. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Bohr radius,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; a<sub>o</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.53</p> <p>With this value and equation (6), we can give the radius of other orbits in hydrogen atom as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; r<sub>n</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = n<sup>2</sup> a<sub>o</sub></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>\&nbsp; r<sub>2</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 2<sup>2</sup> a<sub>o</sub> &nbsp;&nbsp; = 4a<sub>o</sub></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>r<sub>3</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 3<sup>2</sup> a<sub>o</sub> &nbsp;&nbsp; = 9a<sub>o</sub></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;&hellip;&hellip;</p> <p><strong>&nbsp;</strong></p> <p><strong>Velocity and frequency of electron </strong></p> <p><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </strong>We have from the Bohr's quantization condition</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; mvr&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip; (1)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Again, radius of n<sup>th</sup> orbit can be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; r&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; with this value</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp; &nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \ v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;..&nbsp; (2)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; In general, v<sub>n</sub> = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;&nbsp; (3)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; We have frequency,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; f&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (∵ v = r w)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; f&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = .</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;= &nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \&nbsp;&nbsp; f &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;. (3)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; This is the expression for frequency of orbital electron.</p> <p><strong>&nbsp;</strong></p> <p><strong>Energy of electron in n<sup>th</sup> orbit of Bohr </strong></p> <p>As electron is revolving around the nucleus with certain velocity, it certainly has kinetic energy. Also, the electron will be in force of attraction towards positively charged nucleus, it has potential energy as well. Thus, the total energy of an electron revolving round the nucleus in certain orbit is sum of its kinetic and potential energy.&nbsp;</p> <p>Consider an electron is revolving around the nucleus in n<sup>th</sup> stationary orbit of radius r with velocity v, then kinetic energy of electron is,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K.E.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; mv<sup>2</sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; m = mass of electron</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; again we have &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &THORN;&nbsp;&nbsp; K.E. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = m &nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \ Kinetic energy of electron (K. E.) &nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip; (i)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Potential energy of electron in n<sup>th</sup> orbit can be given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P.E.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = (electrostatic potential) &times; (Charge of an electron)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; (&ndash;e)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; P.E.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Substituting value of r</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P.E.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \ Potential energy of an electron in n<sup>th </sup>orbit</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P.E.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;. (2)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Now, the total energy of an electron in n<sup>th</sup> orbit is</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = K. E + P.E.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; &ndash;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \&nbsp;&nbsp; E&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; In general,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>n</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (3)</p> <p>Hence, the total energy is negative which means that the electron is bound to the nucleus. Which also suggest that the outer orbits have more energy than the inner one.&nbsp;&nbsp;</p> <p><strong>&nbsp;</strong></p> <p><strong>Hydrogen Spectrum</strong></p> <p>When an electron in hydrogen atom jumps from higher energy state to lower energy state then there will be emission of radiation of certain energy, If we consider an electron jumps from higher state n<sub>2</sub> to lower energy state n<sub>1</sub>, then energy of emitted photon is given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf &nbsp; = En<sub>2</sub> &ndash; En<sub>1</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (1)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; En<sub>2</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = Energy of electron in n<sub>2</sub></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; En<sub>1</sub><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>= Energy of electron in n<sub>1</sub></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; from above calculation, we have</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; En<sub>2</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;.. (2)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; En<sub>1</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;.. (3)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \ hf &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = En<sub>2</sub> &ndash; En<sub>1</sub></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; &ndash;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; &ndash;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp; f &nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (4)</p> <p>Wave number of a wave is counted as the number of complete wave in unit length and is defined and given mathematically as reciprocal of wave length. It is denoted by .</p> <p>\&nbsp;&nbsp;&nbsp;&nbsp; Wave number &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (5)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = R &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;. (6)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; R = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;is a constant called as Rydberg's constant.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; If we substitute the values for m, e, e<sub>o</sub>, c &amp; h, then value of R will be</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; R &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1.097 &times; 10<sup>7</sup> m<sup>&ndash;1</sup></p> <p><strong>&nbsp;</strong></p> <p><strong>Energy levels in hydrogen atom</strong></p> <p>As calculated above, the energy of an electron revolving round the nucleus in n<sup>th</sup> stationary orbit in hydrogen atom is given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>n</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;. (i)</p> <p>We can represent the energy of different energy level according to the number of orbit i.e. principal quantum number in a diagram which is called as energy level diagram.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Substituting values of m, e, e<sub>o</sub> and h in equation (i) we obtain,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>n</sub><sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>=</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; J</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \E<sub>n</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash; eV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (∵ 1 eV&nbsp; = 1.6 &times; 10<sup>&ndash;19</sup> J)</p> <p>This expression represents the energy of an electron in n<sup>th</sup> orbit of hydrogen and negative sign signifies that the electron is bound with nucleus.</p> <p>For&nbsp; n =1, the energy of electron in ground state can be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>1</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = eV &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash;13.6 eV</p> <p>Similarly for n= 2, i.e. &nbsp;first excited state, the energy of electron will be</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>2</sub><sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>= &nbsp; eV &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash;3.4 eV</p> <p>On the same way, energy of second, third excited state can be given respectively as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>3</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = eV &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash; 1.5 eV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>4</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp; eV &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash; 0.85 eV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;&hellip;&hellip;.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E <sub>&yen;</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>All these values can be represented in a single diagram as shown in figure below. It is clear that the difference in energy levels goes on decreasing for increasing principal quantum number.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>Hydrogen Spectra</strong></p> <p>Whenever an electron jumps from higher energy state to a lower energy state, the difference of energies of two states will be radiated as radiation of certain frequency and wave length called as spectral line. Wavelength of spectral line depends on the two energy states between which transition of electron occurs. Thus we can have different spectral lines, which can be divided into different group according to their wavelength called as spectral series. There are different types of spectral series in hydrogen atom as follows;</p> <table> <tbody> <tr> <td width="7">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <ol> <li><strong>I) Lyman Series: </strong></li> </ol> <p>The spectral series in hydrogen which is obtained corresponding to the trasition of electron from higher energy state to the lowest energy sate i.e. ground state is known as Lyman series. For Lyman series the wave number of the spectral lines can be give as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; = R</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; R&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;&hellip;.. (i)</p> <p>Where, n<sub>2</sub> = 2, 3, 4 &hellip;&hellip;. wave length of spectral lines of this series are 1215 , 1025 etc. This series lies in ultraviolet region.&nbsp;</p> <ol> <li><strong>II) Balmer Series: </strong></li> </ol> <p>A spectral series in hydrogen atom which corresponds to the transition of electron from higher energy state to first excited state (i.e. n = 2) is known as Balmer series. For this series wave number can be given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;= R</p> <p>&THORN; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; R &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;&hellip;&hellip;. (ii)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where n<sub>2</sub> = 3, 4, 5 &hellip;&hellip;&hellip;. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>Wavelength of spectral lines of this series are 6561, 4860 etc this series lies in the visible region.</p> <p><strong>III)&nbsp;&nbsp;&nbsp; Paschen Series </strong></p> <p>It is the spectral series in which spectral lines correspond to the transition of electron from higher energy state to the second excited sate (i.e. n = 3). Wave number can be given as,</p> <p>&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;= R</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; R &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;. (iii)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where, n<sub>2</sub> = 4, 5, 6 &hellip;&hellip;&hellip;</p> <p>Wavelength of spectral lines in this series are 1875nm, 1281nm etc and these lines fall on the infrared region.</p> <p>&nbsp;</p> <ol> <li><strong>IV) Brackett Series:</strong></li> </ol> <p>This spectral series is the result of electronic transition into third excited sate (i.e. n = 4) from higher energy states (i.e.&nbsp; n = 5, 6, 7 &hellip;.. ). The wave number is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;= R</p> <p>&THORN; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; R &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (iv)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where, n<sub>2</sub> = 5, 6, 7 &hellip;&hellip;&hellip;</p> <p>Spectral lines in this series have wavelength 4051nm, 2625 nm etc. These lines belong to far infrared region.</p> <p><strong>&nbsp;</strong></p> <ol> <li><strong>v) P-fund series </strong></li> </ol> <p>The spectral series in hydrogen atom corresponding to the transition of electron from higher energy state to the fourth excited state (i.e. n = 5) is known as P-fund series. Wave number of spectral lines in this series is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; R</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; R &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;. (v)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where, n<sub>2</sub> &nbsp;&nbsp; = 6, 7, 8 &hellip;&hellip;&hellip;..</p> <p>Members of this series have wavelength 7458 nm, 4652nm etc. which lie in far infrared region.</p> <p><strong>&nbsp;</strong></p> <p><strong>Limitation of Bohr's theory of atomic model </strong></p> <p>Bohr's theory of atomic model was successful in many aspects like introducing the quantum concept in atomic study, explaining stability of atom, explaining spectral series of hydrogen atom etc. However, it suffered or experienced some failures which can be mentioned as follows,</p> <ol> <li>i) It can explain only hydrogen and hydrogen like atoms and cannot explain the structure of atom having many electrons.</li> <li>ii) It explains only about the circular orbits and does not discuss why there are only circular orbits and not elliptical.</li> </ol> <p>iii)&nbsp;&nbsp; It cannot explain about the intensity of spectral lines.</p> <ol> <li>iv) This theory also fails to explain the wave nature of electron.</li> <li>v) Fine structure of atom cannot be explained on the basis of this theory.</li> </ol> <p>&nbsp;</p> <p><strong>Excitation and Excitation potentials </strong></p> <p>In normal condition, the electron in hydrogen atom lies in the first orbit or normal state i.e. ground level, therefore known as ground state level. However, if some (sufficient) energy is supplied to this electron by some means (bombard by slow moving particles like electron, neutron etc) the electron may jump into the higher energy state (i.e. n = 2, 3 &hellip;&hellip;) which are known as excited state of atom. In such a condition the atom is said to be excited and the process is known as Excitation.</p> <p>The amount of energy required to excite an atom i.e. to move electron from its ground state to higher energy state is known as excitation energy and the corresponding potential is known as excitation potential.</p> <p>The amount of energy required to excite an electron form ground state (i.e. n = 1) to its first excitation state (i.e. n =2) is called first excitation energy. The energy required to excite an electron from ground state to first excite state in hydrogen atom can be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = E<sub>2</sub> &ndash; E<sub>1</sub></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash;3.4 &ndash; (&ndash; 13.6)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 10.2 eV.</p> <p>Similarly, energy required to excite electron to second excited state in hydrogen atom is,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = E<sub>3</sub> &ndash; E<sub>1</sub></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &ndash;1.51 &ndash; (&ndash;13.6)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 12.09 eV &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Second Excitation energy)</p> <p>and so on.</p> <p>An electron volt (eV) is the amount of energy gained by an electron while moving through a potential difference of 1V. Thus, form above calculation we can easily say that the first excitation potential in case of hydrogen atom is 10.2V, second excitation potential is 12.09V and so on.</p> <p><strong>Ionization and Ionization Potential</strong></p> <p>As we discussed earlier when an electron gets sufficient energy it jumps into higher energy state from lower (ground) energy state. On receiving sufficiently large amount of energy, the electron may jump out of the atom. In such condition, the atom is said to be ionized and the process is known as ionization. The minimum amount of energy required to ionize an atom is called as ionization energy and the corresponding potential is known as ionization potential.</p> <p>&nbsp;</p> <p>As the electron must jump from the ground state to out of the atom, this we can say from n = 1 to n = &yen;, the ionization energy in case hydrogen atom is,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = E<sub>&yen;</sub> &ndash; E<sub>1</sub></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0 &ndash; (&ndash; 13.6)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 13.6 eV</p> <p>Thus the corresponding potential is 13.6V for hydrogen atom.</p> <p>&nbsp;</p> <p><strong>Emission and absorption Spectra:</strong></p> <p>When an electron in excited state jumps to lower energy state then there will be emission of radiations having different wavelength, according to the difference in energy levels. On the basis of nature of emitted spectral lines, spectra are classified into three types,</p> <p>&nbsp;</p> <ol> <li>i) Line spectra: A spectra in which discrete spectral lines having particular wavelength or frequency can be observed is called line spectra. This type of discontinuous spectra is &nbsp;&nbsp; produced by excited atoms or ions. Hydrogen, Mercury sodium spectrum are some &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; examples of line spectra.</li> <li>ii) Band spectra: The emission spectrum which consist a separate group of spectral lines is known as band spectrum. In general band spectra are produced by gases which contain more than one atom. The familiar example of band spectra is spectra produced by carbon dioxide. Each atom in the molecule emits its own line spectra but due to overlapping of the line spectrum of different atoms of the molecules we get a group of lines called as band. Lines within a band are closer at one side than the other end.</li> </ol> <p>iii)&nbsp;&nbsp; Continuous spectra: The spectra which consist of unbroken or continuous range of wavelength are called continuous spectra. Continuous spectra are produced by hot solids, liquid and highly compressed gases. In these states of substance, the atoms and molecules are so closely packed that change in energy takes place in such an extent that radiation of all possible wavelength are produced. For example, spectrum from sun.</p> <p>Line spectra and band spectra are characteristic of atoms producing them where as the continuous spectra are not.&nbsp; &nbsp;</p> <p>Whenever electron is to be taken from lower energy state to the higher energy sate, one must supply some energy to the electron. For this atom/electron absorb radiation called as absorption spectra. In practices, when a light from a source having continuous spectrum is passed through a medium in gaseous state, it is observed that the resulting spectrum appears as a continuous spectrum in which some characteristic wavelengths are absent. The absent wavelengths are seen as dark lines in the spectrum, such a spectrum is called as absorption spectra.</p> <p>A substance which emits lights of certain wavelength as emission spectra at certain temperature would also absorb light of same wavelength at the same temperature. On absorbing the light of the wavelength, atom will be excited into higher energy state from lower energy state. Therefore, some radiations are absent in the spectrum. This is the reason due to which there are some dark lines in the solar spectrum.</p> <p>Absorption spectra are also of three types line, band and continuous spectra in the similar way of emission spectra.</p> <table> <tbody> <tr> <td width="109">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p>&nbsp;</p> <p><strong>Dual Nature of Light </strong></p> <p>Phenomenon of light like interference diffraction, polarization show the wave nature of light where as phenomenon like photoelectric effect, Compton Effect show the particle nature of light. From these different types of phenomenon we can say that light has both characteristic of particle as well as wave. This characteristic of light is explained as Dual nature of light.</p> <p><strong>&nbsp;</strong></p> <p><strong>de - Broglie's theory</strong></p> <p>In the universe, all energies are in the form of matter and radiations, and can be transformed from one to other. As nature is symmetric in many respect, if light radiation show dual characteristic, matter should also exhibit dual characteristics. On the basis of these facts, de - Broglie thought that, if light can show the dual characteristics of wave and matter then matter also posses dual nature of matter and wave and gave his postulate in 1924 in his Ph .D thesis (later he was awarded Nobel prize for this work) as, wave is always associated with a moving body or particle where wave length of the associated wave is given by the relation,&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;&hellip;&hellip;&hellip; (i)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; h&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = Planck's constant</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; m&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = mass of moving body or particle</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = Velocity of particle</p> <p>The wave associated with a body in motion is called as matter wave or de- Broglie's wave and the wavelength is known as de- Broglie's wavelength.</p> <p>&nbsp;</p> <p><strong>de- Broglie's wavelength </strong></p> <p>Planck explained quantum nature of radiation in his quantum theory. According to him, the energy of photon having frequency f is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =hf &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (1)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where, h is planck's constant.</p> <p>On the basis of de - Broglie's&nbsp; theory if we consider that the radiation as a particle having mass m, then using Einstein's mass energy relation energy associated with the radiation can be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = mc<sup>2</sup> &nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (2)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where, c is velocity of radiation (light)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Comparing equation (1) and (2), we get,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = mc<sup>2</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; h. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = mc<sup>2</sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (∵ c= fl)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (3)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Here, mc &nbsp;&nbsp;&nbsp;&nbsp; = P, momentum of photon</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (4)</p> <p>When a body of mass m is moving with velocity v, then above equations can be arranged as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &THORN;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip; (5)</p> <p><strong>&nbsp;</strong></p> <p><strong>de-Broglie's wavelength of an electron </strong></p> <p>Whenever an electron is accelerated by a certain potential difference from rest, it will gain kinetic energy equal to potential energy provided by potential difference.</p> <p>Consider an electron having mass 'm' and charge 'e' be accelerated by a potential difference of V volts. If v be the velocity gained by the electron, then kinetic energy gained by the electron can be given as the work done (P.E.) in moving electron, i.e.&nbsp;&nbsp;&nbsp;</p> <ol> <li>E. = Work done</li> </ol> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &THORN;&nbsp;&nbsp; mv<sup>2</sup> &nbsp;&nbsp;&nbsp; = Charge &times; Potential difference</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; mv<sup>2</sup> &nbsp;&nbsp;&nbsp; = eV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>Then, de - Broglie's wavelength associated with the electron in motion can be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;= &nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p><strong>&nbsp;</strong></p> <p><strong>Wave theory of electron</strong></p> <p>Consider an electron having mass m is revolving around the nucleus in an orbit of radius r with velocity v.</p> <p>It forms a standing wave around the orbit, so that the path length of electron must be equal to integral multiple of wavelength of de Broglie's wave formed by the electron.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; i.e. 2pr &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = nl &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip; (i)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; n = 1, 2, &hellip;&hellip;&hellip;.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; We have for de- Broglie's wave &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \&nbsp;&nbsp; 2 p r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &THORN;&nbsp;&nbsp; mvr&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;</p> <p>Which is in accordance with Bohr's quantization condition.</p> <p>&nbsp;</p> <p><strong>Heisenberg's Uncertainty Principle</strong></p> <p>In classical Physics, we deal with the property of particles of large size. During such study, physical quantities like position and momentum can be measured precisely at the same time. In similar way, energy and time can also be measured accurately at a single time. However, when we check about study of micro particles (i.e. quantum particles) there arises uncertainty when we try to measure such quantities simultaneously. This theme is formulated by Germsm physicist Werner Heisenberg, which is known Heisenberg's uncertainly principle. This principles states that, "it is impossible to measure both momentum and position of a micro particle at a single time accurately and the product of uncertainty in momentum and position is always equal or greater than ".</p> <p>If DP and DX be uncertainties in momentum and position respectively, then according to Heisenberg's uncertainty principle,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D P &times; DX&nbsp; &sup3;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Similarly for energy and time</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; DE &times; Dt&nbsp; &sup3;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; DE = Uncertainty in energy</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Dt = Uncertainty in time.</p> <p>The pair of physical quantities involved in uncertainty relation are called conjugate quantities.</p> <p>&nbsp;</p> <p><strong>LASER</strong></p> <p>The word LASER stands for Light Amplification by Stimulated Emission of Radiation. It is a highly coherent, highly intense and parallel beam of monochromatic light which is produced by stimulated emission. Before discussing the laser process we shall discuss few related terms:</p> <p>&nbsp;</p> <p><strong>Stimulated absorption and meta-stable state</strong></p> <p>In general, an atom always remains in normal state. However, if the atom interacts with a photon having energy hf = E<sub>2</sub> &ndash; E<sub>1</sub> then an electron in ground state may jump into higher energy state by absorbing the photon. This process of absorption of photon is known as induced or stimulated absorption. Whenever an electron jumps from lower energy state to higher energy state i.e. atom gets excited, it cannot remain for long time in this state, the life time in excited state of atom is around 10<sup>&ndash;8</sup> sec. However, in certain energy levels, atoms remain in excited state for comparatively long time (10<sup>&ndash;3</sup> sec). These energy levels are called as meta-stable state.</p> <p>&nbsp;</p> <p><strong>Spontaneous emission of radiation</strong></p> <p>When an electron in ground state gets sufficient energy by some means it will jump into higher energy state i.e. excited state. However, it cannot stay for long in that excited state and so jumps down to lower energy state emitting radiation. This process happens itself and also emission of radiations are in random direction and uncoordinated, that is why the process is known as spontaneous emission of radiation. If electrons jumps from higher energy state having energy E<sub>2</sub> to lower energy state having energy E<sub>1</sub> then energy of emitted radiation (photon) is given as,</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = E<sub>2</sub> &ndash; E<sub>1</sub></p> <table> <tbody> <tr> <td width="45">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;</p> <p>&nbsp;</p> <p><strong>&nbsp;</strong></p> <p><strong>Stimulated emission of radiation</strong></p> <p>Whenever an electron gets excited, it remains in that state for a very short time. In the mean time if another photon having energy exactly equal to energy difference between two energy levels is introduced into the atom then the injected photon stimulates the excited electron to jump to lower energy state with emission of photon. In such an emission all characteristics of emitted photon will be same as that of introduced photon which means wavelength, frequency, phase, direction etc are same for both emitted and injected photon. If E<sub>1</sub> and E<sub>o</sub> be energies of two different energy levels, the energy of photon to be introduced will be,</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf &nbsp; = E<sub>1</sub> &ndash; E<sub>0</sub></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup></p> <p><sup>&nbsp;</sup></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup></p> <p><sup>&nbsp;</sup></p> <p><strong>&nbsp;</strong></p> <p><strong>Population inversion</strong></p> <p>In normal condition, almost all atoms are in ground or normal state and only very few atoms are in excited state. Thus the number of atoms in ground state is much greater than the number of atoms in excited state. As we explained earlier, if these atoms get interact with photon having suitable energy will be excited. In such process we may reach a condition where number of atoms in excited state becomes greater than the number of atoms in ground state. This condition of having excess number of atoms in excited state than normal state is known as population inversion. The process of achieving population inversion is known as optical pumping.</p> <table> <tbody> <tr> <td width="66">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p><strong>Optical Pumping</strong></p> <p>During the process of optical pumping, the atom in ground state i.e. E<sub>1</sub> is first pumped to second excited state i.e. E<sub>3</sub> by using a photon of suitable energy. Preventing the decay of atom from E<sub>3</sub> to E<sub>1</sub> the atom is imposed to decay from second excited state to first excited state E<sub>2</sub>, which is a meta stable state. The decayed atoms remain for quite long time in meta-stable state. Thus there will be a situation where number of atoms in excited state becomes greater than number of atoms in normal state and population inversion takes place.</p> <p>&nbsp;</p> <p><strong>Stimulation</strong></p> <p>When a photon having suitable energy (hf = E<sub>2</sub> &ndash; E<sub>1</sub>) is made to incident on atoms in excited state, it interact with electron of excited atom and stimulate it to jump to lower energy state. As the emitted photon will be same to incident photon in all aspects, there will be two photons after that process. If we proceed this in a crystal having population inversion, those two photons will stimulate another two atoms to go to normal state from excited state there by making number of photons four. This process continues and hence number of photons increases geometrically and within very short period of time there will be an intense beam of photons which are all in same direction. This process is known as stimulation and is the principle of production of LASER.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Figure:</strong> Stimulated emission</p> <p>According to the lasing materials used there are different types of laser such as solid laser, gas laser, chemical laser, semiconductor laser, dye laser. Most commonly used leaser are solid and gas lasers.</p> <p>&nbsp;</p> <p><strong>Ruby Laser:</strong></p> <p>&nbsp;Simple construction of a Ruby Laser is as shown in fig. below. It consists of a ruby rod which has optically flat ends, one end of which is partially silvered and the other is perfectly silvered. The rod is kept inside a glass tube surrounded by a helical Xenon flash tube which works as a optical pumping system. Liquid nitrogen is circulated inside, so as to prevent excess heating of rod due to non radiative transition of election from higher energy state to lower energy state.</p> <table> <tbody> <tr> <td width="80">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Fig:</strong> Ruby laser</p> <table> <tbody> <tr> <td width="100">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>Ruby rod is a crystal of Aluminum oxide (Al<sub>2</sub>O<sub>3</sub>) containing 0.05% chromium oxide (Cr<sub>2</sub>O<sub>3</sub>). As it is a mixture, some aluminum atoms in crystal lattice are replaced by chromium ions (Cr<sup>3+</sup>). The energy level diagram of the system is shown in given figure. Chromium atoms in ground state are excited from ground state (E<sub>1</sub>) to second excited state E<sub>3</sub> (Energy 2.25eV) directly by absorbing light of wavelength 550 nm produced by Xenon flash tube. Chromium atoms in second excited state immediately come down to first excited state E<sub>2</sub> (Energy 1.79eV) with non radiative transition. As first excited state in chromium atom is meta-stable it is slightly long lived. Hence population inversion could be reached easily. Some of these atoms undergo spontaneous emission from first excited state (E<sub>2</sub>) to ground state (E<sub>1</sub>) radiating a photon having energy hf = E<sub>2</sub> &ndash; E<sub>1</sub> = 1.79 eV and wavelength 694.3 nm. Photons thus emitted move back and forth between two parallel mirrors at two ends of rod and hence stimulate other excited atoms to come down to ground state. As a result a highly intense beam of photon having wavelength 694.3nm will be formed and comes out from the partially silvered end.</p> <p>&nbsp;</p> <p><strong>Helium - Neon Laser (He &ndash; Ne laser)</strong></p> <p>A general construction of Helium - Neon laser is as shown in figure which consist of a glass discharge tube having length around 50 cm and diameter around 5mm. A mixture of Helium and Neon in the ratio 9 :1 is kept inside the tube and there are two electrodes to apply high potential in the mixture. Pressure inside the tube is maintained at very low of around 1 torr. The two ends of the tube are cut at Brewster's angle and two plane mirrors are kept parallel at two end of tube. Among two mirrors used, one is perfectly silvered where as the other is partially silvered.</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Fig:</strong> He-Ne Laser</p> <p>&nbsp;</p> <p>The energy level diagram for the system is as shown in figure. When a high potential is applied between electrodes the gases atoms are ionized. Electrons thus produced attain high velocity and collide with Helium atom and excite He- atom from its ground state 1S level to 2S level where energy of electron is 20.61 eV. He - atom remain in this for long time. These excited He atoms interact with the Neon atoms and excite Neon atoms from their ground state 2P to 5S level as the energy of 5S level of Neon atom is 20.66 eV nearly equal to that of He - atom in 2S level. As a result of such interaction there will be population inversion of Neon atoms. Now, excited Neon atoms come down to its ground state in three different actions. Fist, electron jumps from 5S level to 3P level of Ne atom where energy is 18.70 eV emitting a photon of energy hf = (20.66 &ndash; 18.70)= 1.96 eV&nbsp; and corresponding wavelength of 632.8nm. After which the electron jumps to 3S levels with spontaneous emission or as diffusion to the walls and finally comes down to 2P level with non radiative transitions. The photons emitted in transition of electron from 5S level to 3P level of Neon move back and forth between two parallel mirrors and hence an intense beam will be formed in very short time and comes out from the partially silvered mirror in the form of laser beam.</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;</p> <p><strong>Properties of laser</strong></p> <p>The main features of the laser beam are:</p> <ol> <li>It is highly monochromatic and contains light of almost single wavelength.</li> <li>The beam is highly intense and contains photon having same direction. So it diverges hardly.</li> <li>It is coherent and all the photons are in phase or have constant phase difference.</li> </ol> <p>&nbsp;</p> <p><strong>Uses of Lasers</strong></p> <p>Since it was invented in 1960, laser has contributed and helped human life as it has various uses. Some of them are listed below.</p> <ol> <li>i) Medical use: Because of its unidirectional and coherent nature, a laser beam has different uses in medical field. For examples, blood less micro &ndash; surgery, erase tumors, treatment of soft tissue like eye etc.</li> <li>ii) Industrial use: Leaser beam can be used in industrial area for different purpose like drilling, cutting, welding as well as in the production of microelectronic component.</li> </ol> <p>iii)&nbsp;&nbsp; Laser beams are used to locate distant object and also to find the distance of far objects, depth of sea etc.</p> <ol> <li>iv) Laser beams are used in production of hologram sticker or holography and also used in photography.</li> <li>v) Laser beams are used to transfer communication signals.</li> <li>vi) Laser beams are used in various scientific researches.</li> </ol> <p>&nbsp;</p> <p><strong>X-rays</strong></p> <p>While working in discharge tube in 1895, German physicist Wilhelm Roentgen observed various effects produced by some unknown radiations. He experienced that a photographic plate completely wrapped by a black sheet to be affected by the discharge. He also observed the fluorescent screen covered by a thick black paper placed near to discharge tube become luminous. He continued observations in different setting, placing different type of objects between discharge tube and photographic plate. From all these experiments he concluded that there must be some radiations having high penetrating power are produced from the discharge tube at low pressure. As other characteristics of such radiations were unknown at that time, he named these rays as X- rays (As we suppose unknown variable as x).&nbsp;</p> <p>In the further experiments after its discovery it was observed that these radiations were not deflected in electric and magnetic field suggesting the radiations as electromagnetic. As cathode ray is a beam of electrons the radiations could not be supposed as cathode rays. Actually, these radiations are produced when fast moving electron beam (cathode rays) strikes a target metal of high atomic number (tungsten, molybdenum etc.)</p> <p>Thus X- ray can be produced when fast moving electrons are made to strike a target metal of high atomic number and melting point.&nbsp;</p> <p>&nbsp;</p> <p><strong>Coolidge X- rays tube</strong></p> <table> <tbody> <tr> <td width="50">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p>&nbsp;</p> <p><strong>Fig.: </strong>Coolidge X-ray Tube</p> <p><strong>&nbsp;</strong></p> <p>Coolidge in 1913, devised an instrument to produce X- rays known as Coolidge X- rays tube. The construction of Coolidge X- ray tube is as shown in figure. It consists of a highly evacuated (about 10<sup>&ndash;5</sup> mm of Hg) glass tube having a filament F at one end connected with a low tension. At the other end, a target metal (of high atomic number and high melting point) cut at 45&deg; is placed. Target metal and the filament are connected to a high potential such that filament and target becomes cathode and anode respectively. A metal reflector C is used to focus the electrons on the target metal. A cooling system is arranged to cool the target which gets heated during the production of X - rays, water is generally used for this purpose.</p> <p>When a small current is passed though filament by using low potential, it will be heated and electrons are produced from it. These electrons are accelerated by applying a very high potential between filament and target. The electrons, focused by the metal reflector are then allowed to strike the target with high speed. When fast moving electrons strike the target metal X- rays are procured, however only around 2% energy of electrons is used to produce X- rays and the rest 98% of energy produces heat on target metal. Because of this reason a metal of high atomic number and melting point is used in the production of X- rays. In general Tungsten is used in Coolidge X- rays tube. Water is circulated on target to cool it down.</p> <p>&nbsp;</p> <p><strong>Intensity Control</strong></p> <p>Intensity of X- rays actually symbolizes the number of X- rays present in a beam. An electron can produce only one X- ray. Hence, the intensity of X- rays depends on the number of electrons striking the target metal. Number of electrons depends on the current passing through the filament. So, controlling the current passing through filament by adjusting the low tension/ potential applied we can control the intensity of X- rays.</p> <p>&nbsp;</p> <p><strong>Quality control</strong></p> <p>Frequency of X- rays resembles the quality of X -rays. Thus quality of X- rays means energy of X - rays. As we discussed earlier, energy of electrons is used (converted) to produce X- rays, energy of X- rays depends on the energy of electrons striking the target. Kinetic energy of electrons accelerated by a potential difference of V volts is given as mv<sup>2</sup> = eV. Where v is velocity of electron, m is mass of electron and e is electronic charge. Thus energy of electron depends on the applied high potential. Therefore, energy or in other words quality of X- rays can be controlled by adjusting the high potential applied. Low energy X- rays are called as soft X- rays and high energy X- rays are called as hard X - rays</p> <p>When all energy of electrons is converted into energy of X- rays, then the X- rays having maximum frequency are produced. Therefore the maximum energy of X- rays produced can be given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Max<sup>m</sup> energy of X - rays = mv<sup>2</sup> = eV&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf<sub>max</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = mv<sup>2</sup> = eV</p> <p>&THORN; &nbsp;&nbsp;&nbsp;&nbsp; f<sub>max</sub> &nbsp;&nbsp;&nbsp; =</p> <p>Thus, frequency and hence quality of X- rays depends on the potential difference applied between anode (target) and cathode (filament).</p> <p>If l <sub>min </sub>&nbsp;be the minimum possible wavelength of X- ray produced and c be the velocity light in air then the minimum wave length can be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l<sub>min</sub> &nbsp;&nbsp;&nbsp; = = &nbsp;</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l<sub>min</sub>&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p><strong>Properties of X - rays</strong></p> <p>X-rays were named as X- rays because their properties were unknown at that instant. But, now we know almost all properties of X- rays. These are highly energetic electromagnetic radiations having very short wavelength. Nature and properties of X - rays can be listed as follows,</p> <ol> <li>i) X- rays are electromagnetic radiations having very short wavelength of the range of 1 to 100 .</li> <li>ii) X- rays travel with velocity of light.</li> </ol> <p>iii)&nbsp;&nbsp;&nbsp;&nbsp; They are not deflected in electric as well as magnetic fields.</p> <ol> <li>iv) X- rays affect photographic plates.</li> <li>v) X- rays have ionizing power i.e. they can ionize the gas, atom or matter through which they pass. &nbsp;</li> <li>vi) X- rays can produce photoelectric effect.</li> </ol> <p>vii)&nbsp;&nbsp;&nbsp; X - rays are highly energetic radiations so have penetrating power. They can pass though opaque solids like wood, paper, thin sheet of metals.</p> <p>viii)&nbsp; X- rays show the phenomenons like diffraction, interference, polarization etc. similar to that of ordinary lights.</p> <ol> <li>ix) Excess exposure of X- ray on human body or on body of other living beings may cause harmful effects.</li> <li>x) X- rays produces fluorescence effect in certain material for example zinc sulphide.</li> <li>xi) When X- rays fall on surface of certain materials a part of radiation pass though and the rest is transformed into radiations of some other form called as secondary radiations they are classified as (i) scattered X-rays (ii) characteristic X-rays (iii) scattered b- rays <br /> (iv) characteristic b - rays.</li> </ol> <p>&nbsp;</p> <p><strong>Uses of X- rays</strong></p> <p>X- rays have been found to have great applications on human life and society, as they are highly energetic radiations. We can list some of them as follow,</p> <ol> <li><strong> Radio therapy:</strong> Controlled X- rays exposure is used to destroy malignant tumors and also to cure skin diseases. According to the property of the tumors and diseases soft and hard X- rays are used in this purpose.</li> <li><strong> Diagnosis:</strong> X - rays are used for diagnosis. X- rays are used to find fractures in bone, to find stones, diseased organs, foreign matters etc. in the human body. X- rays are also used to diagnose the disease in lungs, kidneys, intestines etc. of the body.</li> <li><strong> Detective departments:</strong> X- rays are commonly used in detective departments to detect the smuggling of precious metals, explosives and other contraband goods like opium in sealed pearls. They are also used in mints, where coins are made.</li> <li><strong> Engineering:</strong> X- rays are used to detect cracks, flaws in grinders, holes and fractures in prepared goods. Penetrating power of X- rays are used to detect cracks in body of airplanes and cars.</li> <li><strong> Scientific research:</strong> X- rays are also used in scientific research especially in study of atomic structure, crystal structure and behavior of different materials.</li> <li><strong> Industry:</strong> X- rays are also used to find the faults if any in the industrial products like radio valves, tennis balls, tyres etc. They are also used to test the uniformity and quality of paintings.</li> </ol> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>Diffraction of X- rays and Laue's experiment</strong></p> <p>After discovery of X- rays, the properties of X- rays were studied regularly by scientists. In 1900 the properties of X- rays were studied by Haga and Wind and tried to show the wave nature of X- rays similar to that of ordinary lights. They have tried to observe the diffraction of X- rays as we know diffraction of light is possible only when the width of slit equals to wavelength of light they used very fine slits for this purpose. However, in 1913 Laue and his coworkers showed the diffraction of X- rays through an experiment known as Laue's experiment.</p> <table> <tbody> <tr> <td width="7">&nbsp;</td> </tr> <tr> <td>&nbsp;</td> <td>&nbsp;</td> </tr> </tbody> </table> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>&nbsp;</p> <p>The experimental setup for Laue's experiment is as shown in figure. In which, the X - rays produced by a sources S are collimated by using two slits S<sub>1</sub> and S<sub>2</sub>. The collimated beam of X- rays is allowed to pass through a crystal, in the original experiment Laue used zinc sulphide crystal. The beam of X -rays are allowed to fall on a photographic plate after crossing the crystal placed. After sufficient exposure of photographic plate into X- rays, the film is developed.</p> <p>On observing the photographic film, it was observed that there is a central and big spot where the beam strikes directly. However there are many faint and regular spots around the central spot. This regular pattern of spots is then called as Laue Pattern.&nbsp;&nbsp;</p> <p>From this experiment, mainly two different results could be explained. As the spots around the central spot must be the points of constructive interference of X- rays scattered though the crystal on the path of X- rays. Which clearly describe the X- rays as electromagnetic wave as they are diffracted from the crystal lattices. On the other hand, there are small spots around the central spot in regular order, this observation suggests that the atoms in the crystal must be arranged in a regular manner. Since the wavelength of X- rays is very short, it is difficult or some how impossible to construct a diffraction grating to produce diffraction of X- rays, this fact led Laue to believe that three dimensional arrangement of atoms and the space between them might act like diffraction grating.</p> <p><strong>X -ray spectra </strong></p> <p>X -rays are produced from X - ray tube and they are of different wavelength. The intensity of X - rays at a specified accelerating potential are called X - ray spectra. Urey and his co-workers analyzed the intensity and wavelength of X- ray beam, obtained in the X- ray tube, by using different potentials and same target and they plotted a graph between them. The graph is obtained as shown in figure. X- ray spectrum are basically of two types.</p> <ol> <li><strong> Continuous spectra </strong></li> </ol> <p>Continuous X-ray spectra consist of X- rays spectra of all possible wavelength within the range.&nbsp; For the comparatively lower potential of around 30kV, 40kV and upto 50kV the graph is smoothly varying.&nbsp;</p> <p>If we change the target material form molybdenum to copper, the shape and intensity of continuous spectra will change but cut off wavelength (l<sub>min</sub>) will not change. It depends on kinetic energy of fast moving electrons that strike the target and not on the nature of target material.</p> <ol> <li><strong> Characteristics X- rays spectra</strong></li> </ol> <p>When the accelerating potential is increased, it is observed that for certain wavelength of X - rays, the intensity of X- rays become maximum. These are called as characteristic X- rays spectra. When an accelerating potential exceeds a certain limit, characteristics spectra are formed due to superposition of the continuous spectrum as shown in fig.&nbsp;</p> <p>&nbsp;</p> <p><strong>Bragg's Law:</strong></p> <p>William H. Bragg and William L. Bragg studied the diffraction of X- rays in detail. In the process it is supposed that the atoms arranged in a regular manner in parallel planes called as crystal lattice and the space between them behaves as slit. Bragg considered that when monochromatic beam of X- rays incident on the atoms in a crystal lattice each atom acts as a source of scattered radiation of same wavelength. As wavelength of X- rays is of the order of</p> <p>inter atomic distance in the crystal, diffraction occurs from atoms forming an array of parallel planes i.e. crystal lattice. Bragg's law states that the intensity of reflected X- rays will be maximum when path difference between reflected beams from two different planes becomes equal to an integral multiple of wavelength of X- rays. This occurs in practice for certain angle of incidence while at other angles intensity of reflected beam will be minimum. The condition for maximum intensity explained above is called Bragg's diffraction condition and explains the observed Laue pattern.</p> <p>Consider a crystal having inter planner distance d. Consider a set of parallel planes consisting planes XX' YY', ZZ' an so on. Suppose a parallel beam of X- rays incidents on crystal lattice with glancing angle q. Let two rays of X- rays I<sub>1</sub> and I<sub>2</sub> be reflected from atoms A and B lying on two consecutive crystal lattice XX' and YY' and travel along R<sub>1</sub> and R<sub>2</sub> respectively. According to Bragg, the intensity of this reflected beam will be maximum when path difference between two reflected rays R<sub>1</sub> and R<sub>2</sub> equals to integral multiple of wavelength of X - rays. To calculate path difference between two reflected rays let us draw perpendiculars AC and AD from A to I<sub>2</sub> and R<sub>2</sub> respectively.</p> <p>Then clearly path difference between R<sub>1</sub> and R<sub>2</sub> is</p> <p>P.d.&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; CB + BD&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;.. (i)</p> <p>Now from geometry we can have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &ETH; BAC = &ETH;BAD&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = q</p> <p>Again in right angled triangle BAC,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Sin q =</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; CB&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = AB Sin q</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; CB &nbsp;&nbsp;&nbsp;&nbsp; = d sin q &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip; (2)</p> <p>Similarly in D BAD</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Sin q = &nbsp;</p> <p>&THORN; &nbsp;&nbsp;&nbsp;&nbsp; BD&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = d sin q &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip; (3)</p> <p>Now, the path difference between reflected rays is</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P.d. &nbsp;&nbsp; = CB + BD</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = d sin q + sin q</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P.d.&nbsp;&nbsp;&nbsp; = 2d sinq &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;... (4)</p> <p>Now, from Bragg's law, for maximum intensity, path difference must be equal to integral multiple of l.</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2d sin q &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = nl &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;&hellip;&hellip;&hellip;&hellip; (5)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Where, n =1, 2 , 3, 4 &nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.</p> <p>This equation is known as Bragg's diffraction condition or Bragg's equation.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; If n = 1,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2d sin q = l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; This gives the spectrum of first order.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; If n = 2, &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2 d sin q = 2l &nbsp;&nbsp;&nbsp;&nbsp; This gives the spectrum of second order.</p> <p>Applying Bragg's law, we can calculate wavelength of X- rays if other terms are known.</p> <p>&nbsp;</p> <p><strong><br /> </strong></p> <p>Solved Questions</p> <p><strong>&nbsp;</strong></p> <p><strong>Short Answer Questions</strong></p> <p><strong>&nbsp;</strong></p> <ol> <li><strong> What are the major achievements of Bohr's theory of atom?</strong></li> </ol> <p><strong>Ans: </strong>After Bohr's theory of atomic model explained, the world of science got to the following success,</p> <ol> <li>Introduction of quantum mechanics.</li> <li>Mechanical explanation of hydrogen series.</li> </ol> <p>iii.&nbsp;&nbsp;&nbsp;&nbsp; Explanation of stability of atom.</p> <ol start="2"> <li><strong> In hydrogen atom we have only one electron, but its emission spectrum shows many lines. Why? </strong></li> </ol> <p><strong>Ans:</strong> The spectral line is obtained when an electron jumps from higher energy state to lower energy state. Though there is only one electron in H - atom there are several orbits allowed for this electron. Again a sample contains large number of H - atom. Thus there are verity of possibilities of transitions producing spectral lines.</p> <ol start="3"> <li><strong> An electron and a proton are possessing same amount of kinetic energy. Which of two has greater de- Broglie wave length? </strong></li> </ol> <p><strong>Ans:</strong> Let both electron and proton possess same kinetic energy E, then</p> <p>Kinetic energy &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = mv<sup>2</sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;= eV</p> <p>We have&nbsp; de - Broglie's wavelength as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>Now, de - Broglie's wavelength for electron and proton can be given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l<sub>e</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&amp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l<sub>p</sub>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; As M<sub>p</sub> &gt; M<sub>e</sub> de - Broglie's wavelength of electron is greater than that of proton.</p> <ol start="4"> <li><strong> A proton and electron have got same de- Broglie's wave length which has greater energy? </strong></li> </ol> <p><strong>Ans:</strong> We have de - Broglie's wavelength as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp; l<sup>2</sup> &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;</p> <p>or,&nbsp;&nbsp;&nbsp; E&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp;</p> <p>If both electron and proton have same de- Broglie's wavelength, we can have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; a&nbsp;&nbsp;&nbsp;</p> <p>Since the mass of electron is less than that of proton electron has greater energy.</p> <ol start="5"> <li><strong> An electron is in third excited state. How many photon wavelength are possible? </strong></li> </ol> <p><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; [HSEB 2053] </strong></p> <p><strong>Ans: </strong>Suppose E<sub>1</sub>, E<sub>2</sub><sup>, </sup>E<sub>3</sub> and E<sub>4</sub> are respectively ground state, first excited state, second excited state and third excited state of an atom. Then as shown in fig. We can have 6 different form of transition. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;</p> <p>E<sub>4</sub> &nbsp;&nbsp;&nbsp; &reg;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>1</sub></p> <p>E<sub>4</sub> &nbsp;&nbsp;&nbsp; &reg;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>2</sub></p> <p>E<sub>4</sub>&nbsp;&nbsp;&nbsp;&nbsp; &reg;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>3</sub></p> <p>E<sub>3</sub>&nbsp;&nbsp;&nbsp;&nbsp; &reg;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>2</sub></p> <p>E<sub>3</sub><sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>&reg;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>1</sub></p> <p>E<sub>2</sub> &nbsp;&nbsp;&nbsp; &reg;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>1</sub></p> <p>&nbsp;</p> <p>&nbsp;</p> <ol start="6"> <li><strong> Differentiate between excitation and ionization potential. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; [HSEB 2052]</strong></li> </ol> <p><strong>Ans:</strong> The minimum amount of energy required to move an electron from its ground state to any higher energy orbit is called as excitation energy and the corresponding potential is known as excitation potential. For example,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; First excitation energy&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; E<sub>2</sub> &ndash; E<sub>1</sub></p> <p>&amp; &nbsp;&nbsp;&nbsp;&nbsp; First excitation potential &nbsp;&nbsp; =&nbsp;&nbsp; volt</p> <p>On the other hand, the minimum amount of energy required to eject out an electron from the atom completely is called ionization energy and the corresponding potential is known as ionization potential.&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; ionization energy&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; E<sub>&yen;</sub> &ndash; E<sub>1</sub></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>&nbsp;ionization potential&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; &nbsp; Volt.</p> <ol start="7"> <li><strong> Hydrogen atom is stable in ground sate. Why? </strong></li> </ol> <p><strong>Ans: </strong>In hydrogen atom, the electron in ground state means the electron is in the lowest energy state possible i.e. first orbit and therefore there would be no possible transition of electron to the lower orbit. Hence it is stable in ground state.</p> <ol start="8"> <li><strong> What are the evidences for dual nature of light and matter? </strong></li> </ol> <p><strong>Ans: </strong>When some physical entity shows both wave and particle nature then it is said to possess dual characteristics or nature. The interference phenomenon by light and polarization of light are evidences of wave nature of light where as phenomenon like Compton effect and photoelectric effect are evidences of particle nature of light. On the other hand, finite mass of an electron is evidence of particle nature of electron and diffraction of electron and its use in electron microscope is an evidence of its wave nature.</p> <ol start="9"> <li><strong> What is the application of de Broglie's hypothesis?</strong></li> </ol> <p><strong>Ans:</strong> The electron microscope is based on the de- Broglie's hypothesis. According to which, a beam of electron behaves as a wave which can be converged or diverged by applying electric or magnetic field.</p> <ol start="10"> <li><strong> What is nature of matter wave? </strong></li> </ol> <p><strong>Ans: </strong>According to de- Broglie's principle, the wave associated with the moving matter is called a matter wave. The wave length of such wave is given as,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;l&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Actually, such matter waves are not composed of matter. The intensity of wave at a point indicates the probability of finding the particle at that point.</p> <ol start="11"> <li><strong> A hot body emits more light than a cold body. Explain. </strong></li> </ol> <p><strong>Ans:</strong> When a body is heated i.e. when temperature of a body increases, the Kinetic energy of atoms will increase. Due to which they will collide violently and large number of atoms will be excited. When these excited atoms come back to ground state, they will emit photon of certain frequency. As large number of atoms gets excited in hot body than cold body, the hot body emits more light than a cold body.</p> <ol start="12"> <li><strong> What is the meaning of negative energy of an electron in an orbit? </strong></li> </ol> <p><strong>Ans: </strong>The negative energy of an electron in an orbit means that it is bound towards nucleus and if it gains sufficient energy by some means to make its total energy zero, then the electron will become free from atom.</p> <ol start="13"> <li><strong> How many lines can be drawn in the energy level diagram of hydrogen atom? </strong></li> </ol> <p><strong>Ans:</strong> Spectral lines can be drawn or achieved for any transition of electron form any higher energy state to lower energy state. As we can assume number of orbits any positive integer there can be infinite number of spectral lines in energy level diagram of hydrogen atom.</p> <ol start="14"> <li><strong> What is the ratio of energies of orbital electron in 2<sup>nd</sup> and 3<sup>rd</sup> orbits of hydrogen? </strong></li> </ol> <p><strong>Ans: </strong>The energy of an electron in n<sup>th</sup> orbit of hydrogen atom is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>n</sub> &nbsp; =&nbsp;&nbsp;&nbsp; &nbsp; = eV</p> <p>From this, it is clear that energy of election in 2<sup>nd</sup> and 3<sup>rd</sup> orbit can be given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>2&nbsp;&nbsp;&nbsp;&nbsp; </sub>=&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; eV</p> <p>&amp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>3</sub>&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; eV</p> <p>\&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; =&nbsp;&nbsp;&nbsp; /</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>2</sub> : E<sub>3</sub> = 9 : 4</p> <ol start="15"> <li><strong> What is the ratio of kinetic energy to the electrostatic potential energy of an electron in its orbit? </strong></li> </ol> <p><strong>Ans: </strong>We have, knitic energy of an electron in n<sup>th</sup> orbit as,</p> <ol> <li>E. =</li> </ol> <p>and potential energy of electron in n<sup>th</sup> orbit as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P.E.&nbsp;&nbsp;&nbsp; =</p> <p>Now, &nbsp; =</p> <p>\K. E. :&nbsp; P.E.&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1: 2</p> <ol start="16"> <li><strong> Point out importance of de- Broglie's wave? [HSEB 2055]</strong></li> </ol> <p><strong>Ans: </strong>Importance of de - Broglie's wave can be remembered as follows.</p> <ol> <li>i) A matter will have a wave associated with it only when it is in motion. Greater is the momentum shorter will be the wavelength.</li> <li>ii) It helps to justify Bohr's quantization condition of hydrogen atom i.e. the stable states of electron in the atom are governed by integer rules.</li> </ol> <p>iii)&nbsp;&nbsp;&nbsp; De - Broglie waves are independent of charge of particles.</p> <ol start="17"> <li><strong> The de- Broglie wavelength of a particle of kinetic energy E is </strong><strong>l</strong><strong>. What would be the wavelength of the particle if its kinetic energy changed to </strong><strong> ? </strong></li> </ol> <p><strong>Ans:</strong> We have kinetic energy as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K.E.&nbsp; =&nbsp;&nbsp; E = mv<sup>2</sup></p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp; m<sup>2</sup> v<sup>2</sup> = 2mE &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip; (1)</p> <p>Again, l&nbsp; =&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &THORN; mv&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip; (2)</p> <p>From equation (1) &amp; (2)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; 2mE</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp; l<sup>2</sup> &nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;</p> <p>or,&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;</p> <p>When Kinetic energy becomes</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l<sup>'</sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; &nbsp; = 2.</p> <p>\&nbsp;&nbsp;&nbsp;&nbsp; Hence, wavelength would be double.</p> <ol start="18"> <li><strong> How de- Broglie's wave is different from electromagnetic wave? </strong></li> </ol> <p><strong>Ans: </strong>There are two different velocities associated with a particle in motion, one of which refers to the mechanical motion of particles (v) and the other refers to the propagation of the associated wave (u). These two are related as u = . As we know v &lt; c hence velocity of matter wave u &lt; c. Thus, de - Broglie's wave is different from electromagnetic wave.</p> <ol start="19"> <li><strong> Why is the wave nature of particle is not observable in daily life? </strong></li> </ol> <p><strong>Ans:</strong> According to de - Broglie's theory, the wavelength of a wave associated with a body in motion is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l = , where h is Planck's constant and has value 6.62 &times; 10<sup>&ndash;34</sup> Js. In our daily life, the velocity of bodies being small, the wavelength of matter wave would be so small such that they cannot be detected. For example, if a body having mass 1kg is moving with velocity 6.62 m/s, then its wavelength will be 1 &times; 10<sup>&ndash;34</sup>m (very small.)</p> <ol start="20"> <li><strong> Is the de- Broglie wavelength of a Photon of an electromagnetic radiation equal to wavelength of radiation? </strong></li> </ol> <p><strong>Ans: </strong>Wavelength of de - Broglie's wave of a photon is l = .&nbsp;Also, momentum of a photon having frequency f is given as P = =</p> <p>&THORN;&nbsp; P&nbsp; = &nbsp;&nbsp; &THORN;&nbsp; l&nbsp; =</p> <p>Therefore, de - Broglie wavelength of a photon of an electromagnetic radiation equals to the wavelength of radiation.</p> <ol start="21"> <li><strong> Show that electron cannot exist in nucleus.</strong></li> </ol> <p><strong>Ans: </strong>We know that the size of nucleus is about 10<sup>&ndash;14</sup> m. Therefore the uncertainty in position if an electron exists in nucleus would be 10<sup>&ndash;14</sup>m. If we consider DP as uncertainty in momentum, then using uncertainty principle we must have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; D P &times; DX &sup3;</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; m. Dv &times; Dx &sup3; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; [∵&nbsp;&nbsp;&nbsp; P = m v &THORN; Dv]</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; Dv&nbsp; &sup3; &nbsp; &sup3;&nbsp; &nbsp; &sup3; 1.16 &times; 10<sup>10</sup>m/s</p> <p>Which shows velocity of electron greater than velocity of light, which is impossible. Hence electron cannot exist in nucleus.</p> <ol start="22"> <li><strong> Find the longest and shortest wavelength of spectral lines in Balmer series. (R = 1.097 &times; 10<sup>7</sup> m<sup>&ndash;1</sup>) </strong></li> </ol> <p><strong>Ans:</strong> A spectral series which corresponds to the transition of electron from higher energy state is (i.e. n = 3, 4, 5 &hellip;&hellip;) to the lower energy sate (n = 2) is Balmer series. Wave number of which is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = R</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = R ,&nbsp;&nbsp;&nbsp;&nbsp; n<sub>2</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 3, 4, 5 &hellip;&hellip;</p> <p>Longest wavelength is obtained when n<sub>2</sub> = 3</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; i.e. &nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; R</p> <p>&THORN; &nbsp;&nbsp;&nbsp;&nbsp; l<sub>max</sub> &nbsp;&nbsp; = 6.563 &times; 10<sup>&ndash;7</sup> m</p> <p>&THORN; &nbsp;&nbsp;&nbsp;&nbsp; l<sub>max</sub> &nbsp;&nbsp; =&nbsp; 6563</p> <p>Shortest wavelength is obtained when n<sub>2</sub> = &yen;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; i.e. &nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = R</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l<sub>min &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sub>&nbsp;= 3.646 &times; 10<sup>&ndash;7</sup> m</p> <p>&THORN; &nbsp;&nbsp;&nbsp;&nbsp; l<sub>min &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sub>&nbsp;= 3646</p> <ol start="23"> <li><strong> What is the difference between hard and soft X- rays? </strong></li> </ol> <p><strong>Ans:</strong> X- rays which have longer wavelength (greater than 4 ) have low energy and hence low penetrating power. These are called as soft X- rays. However X-rays which have wavelength shorter than 4 are highly energetic and are more penetrating and called as hard X- rays.&nbsp;&nbsp;</p> <ol start="24"> <li><strong> Can aluminum be used as a target in X- ray tube? </strong></li> </ol> <p><strong>Ans: </strong>Aluminum cannot be used as a target in X- ray tube. A metal of high atomic number and melting point can produce energetic X- rays and also holds high temperature during production of X- rays. As aluminum has low melting point and low atomic number and so cannot be used as a target in X- ray production.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Not only aluminum, this problem arises for other soft metals as well)</p> <ol start="25"> <li><strong> How would you increase penetrating power of X- rays?</strong></li> </ol> <p><strong>Ans:</strong> Penetrating power of X- rays depends on its energy. More energy means more penetrating power. As energy of X- rays depends on the energy of electron striking the target metal and which depends on potential difference between filament and target penetrating power of X- rays can be increased by increasing potential difference between filament and target, Penetrating power of X- rays can be increased by increasing the applied potential.</p> <ol start="26"> <li><strong> X- rays are not deflected by the electric and magnetic field? Why?</strong></li> </ol> <p><strong>Ans:</strong> Only charged particles are deflected in electric and magnetic field. Electromagnetic radiations are not deflected in electric and magnetic field. As X - rays are electromagnetic in nature, they are not deflected in electric and magnetic field.</p> <ol start="27"> <li><strong> Production of X- rays is inverse process of photoelectric effect. Do you agree? Justify. </strong></li> </ol> <p><strong>Ans:</strong> In photoelectric effect, energy of a photon is transformed into Kinetic energy of electron. On the other hand, during production of X- ray kinetic energy of electron is converted into a photon/ radiation. Thus production of X- rays can be considered as reverse process of photoelectric effect.</p> <ol start="28"> <li><strong> What are the difference between X- rays and ordinary light? </strong></li> </ol> <p><strong>Ans:</strong> Both X- rays and light are transverse in nature though they have many differences. Some of them are as follows,</p> <table> <tbody> <tr> <td width="259"> <p><strong>X- rays</strong></p> </td> <td width="275"> <p><strong>Light</strong></p> </td> </tr> <tr> <td width="259"> <p>1.&nbsp;&nbsp; X- rays are electromagnetic waves of very short wave length ranging from 1 to 100 .&nbsp;&nbsp;</p> </td> <td width="275"> <p>1.&nbsp;&nbsp; Ordinary lights are electromagnetic waves of comparatively longer wavelength ranging from 4000 to 7000 .</p> </td> </tr> <tr> <td width="259"> <p>2.&nbsp;&nbsp; X- rays are more energetic and have high penetrating power.</p> </td> <td width="275"> <p>2.&nbsp;&nbsp; Lights are less energetic and so do not have such penetrating power.</p> </td> </tr> <tr> <td width="259"> <p>3.&nbsp;&nbsp; X- rays are invisible to human eye.</p> </td> <td width="275"> <p>3.&nbsp;&nbsp; Lights are visible to human eye.</p> </td> </tr> </tbody> </table> <ol start="29"> <li><strong> X- rays are diffracted by Crystals. Why ? </strong></li> </ol> <p><strong>Ans:</strong> Diffraction of wave occurs when the width of slit through which wave passes is in the order of wavelength of wave. In a crystal, atoms are arranged in a regular manner, in a plane known as crystal lattice. The spacing between crystal lattices behaves as slit which is in the order of wavelength of X- rays. Hence, X- rays are diffracted by crystal and are not diffracted in a slit used for diffraction of normal visible light. (This is why X- rays are used in study of crystal structure.)</p> <ol start="30"> <li><strong> X- rays can penetrate through the flesh but not through bone. Why? </strong></li> </ol> <p><strong>Ans:</strong> The penetrating power of X- rays depends on the energy of electron striking the target. X- ray can penetrate through the soft matter but not through the hard one. So, X- rays can penetrate through the flesh (light element on human body) but are unable to penetrate through heavier elements like bone made up of calcium and phosphorus. (This is why X- rays are used to detect factures in bone.)</p> <p><strong>&nbsp;</strong></p> <p><strong>Numerical Examples</strong></p> <ol> <li><strong> The first member of Balmer series of hydrogen atom has a wavelength of 6563</strong>.<strong> Compute the wave length of its second member.</strong></li> </ol> <p><strong>Ans:</strong> We know Balmer series corresponds to the transition of electron from higher orbit to lower orbit (n = 2).&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Wave number of spectral lines in Balmer series is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; R , &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; n<sub>2</sub> = 3, 4, 5 &hellip;&hellip;.</p> <p>For first member,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; n<sub>2</sub>&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 3 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>and l&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 6563 = 6563 &times; 10<sup>&ndash;10</sup> m</p> <p>\ &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; = R &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = R</p> <p>or,&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = R</p> <p>or,&nbsp;&nbsp;&nbsp; R&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>\ &nbsp;&nbsp;&nbsp; R&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1. 097 &times; 10<sup>7</sup> m<sup>&ndash;1</sup></p> <p>For second member, n<sub>2</sub> &nbsp;= 4</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; R</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.097 &times; 10<sup>7</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.097 &times; 10<sup>7</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.097 &times; 10<sup>7</sup> &times;</p> <p>or, &nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 2.056 &times; 10<sup>6</sup></p> <p>\ &nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 486.1 &times; 10<sup>&ndash;9</sup> m = 4861</p> <ol start="2"> <li><strong> Compute the velocity of the electron in the 9<sup>th</sup> orbit of Bohr. How many times does the electron go around the orbit in 1 sec. </strong></li> </ol> <p><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </strong><strong>(e = 1.6 &times; 10<sup>&ndash;19</sup> C, h = 6.6 &times; 10<sup>&ndash;34</sup> JS, m<sub>e</sub> = 9.1 &times; 10<sup>&ndash;31</sup> kg,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K = </strong><strong> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 9 &times; 10<sup>9</sup> Nm<sup>2</sup>C<sup>&ndash;2</sup>)</strong></p> <p><strong>Ans:</strong> Given,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; e&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.6 &times; 10<sup>&ndash;19</sup> C</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; h&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 6.6 &times; 10<sup>&ndash;34</sup> Js</p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>m<sub>e</sub> &nbsp; =&nbsp;&nbsp;&nbsp; 9.1 &times; 10<sup>&ndash;31</sup> Kg</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp; = 9 &times; 10<sup>9</sup> Nm<sup>2</sup> C<sup>&ndash;2</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; n&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 9</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; ?</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; f&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; ?</p> <p>We have velocity of electron as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp;</p> <p>\ &nbsp;&nbsp;&nbsp; v &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 2.437 &times; 10<sup>5</sup> m/s</p> <p>Again we have,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp; mvr&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; <sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp; =&nbsp;</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp; r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 4.3 &times; 10<sup>&ndash;9</sup> m</p> <p>Now f&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp; =&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 9.1 &times; 10<sup>12</sup> Hz</p> <p>(To calculate frequency you can also use the relation f = )</p> <ol start="3"> <li><strong> Find the radius of Li<sup>++</sup> ion in its ground state assuming Bohr's model to be valid? </strong></li> </ol> <p><strong>Ans: </strong>In Li<sup>++</sup> ion, no of electron &nbsp;&nbsp;&nbsp;&nbsp; = 1</p> <p>and no. of proton &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 3</p> <p>Now for an electron in any circular orbit of radius r, we can have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; =&nbsp; &nbsp; .</p> <p>or,&nbsp;&nbsp;&nbsp; mv<sup>2</sup> r &nbsp; =&nbsp; &nbsp; 3e<sup>2</sup> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;. (1)</p> <p>Again,&nbsp; form Bohr's postulate,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; mvr&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &hellip;&hellip;&hellip;.. (2)&nbsp; &nbsp;</p> <p>Dividing equation (1) by (2), we get</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;</p> <p>From equation (2)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp; r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; &nbsp; &times; &nbsp; = &nbsp; &times;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; &nbsp;&nbsp; =&nbsp; &nbsp;</p> <p>\ &nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;r&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; 1.77 &times; 10<sup>&ndash;11</sup> m</p> <ol start="4"> <li><strong> A hydrogen atom initially in the ground level absorbs photon, which excites it to the n = 4 level. Determine the wavelength and frequency of the photon. </strong></li> </ol> <p><strong>Ans:</strong> Energy in ground level of hydrogen</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>1</sub>&nbsp; =&nbsp; &ndash;13.6 eV = &ndash;13.6 &times; 1.6 &times;10<sup>&ndash;19</sup> J</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>1</sub>&nbsp;&nbsp; =&nbsp; &ndash; 2.176 &times; 10<sup>&ndash;18</sup> J</p> <p>Energy in level n = 4 is, E<sub>4</sub>&nbsp; =&nbsp; &ndash;0.85 eV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; &ndash;0.85 &times; 1.6 &times; 10<sup>&ndash;19</sup> J</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; &ndash;1.36 &times; 10<sup>&ndash;19</sup> J</p> <p>Difference in energies in two levels.</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E<sub>4</sub> &nbsp;&ndash; E<sub>1</sub>&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &ndash;1.36 &times; 10<sup>&ndash;19</sup> &ndash; (&ndash;2.176 &times;10<sup>&ndash;18</sup>)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &ndash;1.36 &times; 10<sup>&ndash;19</sup> + 2.176 &times; 10<sup>&ndash;18</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 2.04 &times; 10<sup>&ndash;18</sup> J</p> <p>Hence, energy of incident photon must be,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 2.04 &times; 10<sup>&ndash;18</sup> J</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; f=&nbsp;&nbsp; &nbsp; Hz</p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>=&nbsp;&nbsp;&nbsp; &nbsp; Hz</p> <p>\ &nbsp;&nbsp;&nbsp; f&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 3.09 &times; 10<sup>15</sup> Hz</p> <p>Again we have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; fl</p> <p>&THORN;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; &nbsp;=&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 97. 06 &times; 10<sup>&ndash;9</sup> m</p> <p>\ &nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 97.06 nm</p> <ol start="5"> <li><strong> Light corresponding to transition from n = 5 to n = 2 in hydrogen atom falls on a cesium metal surface whose work function is 1.9eV. Calculate the maximum Kinetic energy of the photo electrons emitted. </strong></li> </ol> <p><strong>Ans:</strong> Energy of photon is given as E = hf = h</p> <p>Again wavelength of photon emitted in transition of electron from n = 5 to n = 2 is given as</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; R</p> <p>or,&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.097 &times; 10<sup>7</sup></p> <p>\ &nbsp;&nbsp;&nbsp; &nbsp;Energy of Photon E&nbsp; =&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.097 &times; 10<sup>7</sup> &times; &times; 6.6 &times; 10<sup>&ndash;34</sup> &times; 3 &times; 10<sup>8</sup></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>=&nbsp;&nbsp;&nbsp; 4.56 &times; 10<sup>&ndash;19</sup> J.</p> <p>Here work function of cesium metal = 1.9eV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.9 &times; 1.6 &times; 10<sup>&ndash;19</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 3.04 &times; 10<sup>&ndash;19</sup> J</p> <p>According to Photoelectric equation</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp; mv<sup>2</sup>&nbsp;&nbsp; =&nbsp;&nbsp; E &ndash; f</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; 4.56 &times; 10<sup>&ndash;19</sup> &ndash; 3.04 &times; 10<sup>&ndash;19</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; 1.52 &times; 10<sup>&ndash;19</sup> J</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp; 0.95 eV.</p> <ol start="6"> <li><strong> The spacing of atomic planes in a crystal is 1.3 </strong><strong> and when a monochromatic X- rays incidents on them at a glancing angle of 5&deg;, a first order image is produced. Calculate the wavelength of X- ray. What is the glancing angle for second order image?</strong></li> </ol> <p><strong>Solution: </strong></p> <p>Given,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Inter planner distance (d)&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; 1.3 &nbsp; = &nbsp;1.3 &times; 10<sup>&ndash;10</sup> m</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Glancing angle (q)&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; 5&deg;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; For first order (n)&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; 1</p> <p>We have, for maximum intensity</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2d sin q&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; n l</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;= &nbsp;</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp; =&nbsp;&nbsp;&nbsp;&nbsp; 2.26 &times; 10<sup>&ndash;11</sup> m</p> <p>For 2<sup>nd</sup> order image, n<sub>2</sub> = 2</p> <p>and</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; 2d sin q<sub>2</sub> = 2l</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;= d sin q<sub>2</sub></p> <p>or, 0.17 &times; 10<sup>&ndash;10</sup> = 1.3 &times; 10<sup>&ndash;10</sup> &times; Sin q<sub>2&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sub></p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; Sin q<sub>2</sub> =</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; q<sub>2</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = Sin<sup>&ndash;1</sup> &nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 7.5&deg;</p> <ol start="7"> <li><strong> An X- ray tube works at a d.c. potential of 50 kV Only 0.4% of energy of the cathode rays is converted into X- radiation and heat is produced in target at a rate of 600 W. Estimate. </strong></li> <li><strong>i) The current passed into the tube </strong></li> <li><strong>ii) The velocity of electrons striking the target, (m = 9.10<sup>&ndash;31</sup> kg, e = 1.6 &times; 10<sup>&ndash;19</sup>C)</strong></li> </ol> <p><strong>Solution: </strong></p> <p>Given,</p> <ol> <li>P.d. = 50 KV = 50 &times; 10<sup>3</sup> V</li> </ol> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; The rate of production of heat (P) = 600W</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; According to question, we have</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (100 &ndash; 0.4) % of energy of electrons being converted into heat energy</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 99.6% of IV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 600&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &times; IV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 6 &times; 10<sup>4</sup> &nbsp; = 99.6 &times; I &times; 50 &times; 10<sup>3</sup></p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; I &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;&nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;I &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.012 A</p> <ol> <li>ii) Velocity of electron (v) = ? &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</li> </ol> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; We have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Kinetic energy of electron &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = Potential energy between electrodes</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; i.e. &nbsp;&nbsp;&nbsp;&nbsp; mv<sup>2</sup> &nbsp;&nbsp;&nbsp; = eV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1.34 &times; 10<sup>8</sup> m/s</p> <ol start="8"> <li><strong> What is the P.d. between the filament and target of an X- ray tube if the tube is to produce X - rays wavelength of 0.05 nm. </strong></li> </ol> <p><strong>Solution:</strong></p> <p>Here,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Wavelength &nbsp;&nbsp; of X-ray to be produced</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.05 nm&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (l = 0.05 nm)</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.05 &times; 10<sup>&ndash;9</sup> m</p> <p>We have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf<sub>max</sub> &nbsp; = eV</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; h = eV</p> <p>or&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; V &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; V&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 24800V&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; V &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 24.8 KV</p> <ol start="9"> <li><strong> The cathode ray tube that generates the picture in early colour televisions were sources of X- rays. If the acceleration voltage in a television tube is 15 KV, what is the shortest wavelength of X- ray produced by the television? </strong></li> </ol> <p><strong>Solution: </strong></p> <p>Potential difference (V) &nbsp;&nbsp;= &nbsp;15 KV&nbsp; = &nbsp;15 &times; 10<sup>3</sup> V</p> <p>Wavelength of X- ray produced (l) = &nbsp;?</p> <p>We have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = eV</p> <p>or, &nbsp;&nbsp;&nbsp; h &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = eV</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.0828 &times; 10<sup>&ndash;9</sup> m</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.8 &times; 10<sup>&ndash;10 </sup>m&nbsp;&nbsp; =&nbsp; &nbsp;0.8</p> <ol start="10"> <li><strong> Protons are accelerated from rest by a potential difference of 4.00 KV and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting x - rays? How does your answer compare to the minimum weave length if 4.00KeV electrons are used instead? </strong></li> </ol> <p><strong>Solution: </strong></p> <p>Here in first case i.e. when proton is used</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; P.d &nbsp;&nbsp;&nbsp; = 4.00 KV &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 4 &times; 10<sup>3</sup> V</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; minimum wavelength (l)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = ?</p> <p>We have</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; &nbsp;hf = &nbsp;eV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;eV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; h &nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;eV</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;&nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;0.31 &times; 10<sup>-9</sup> m</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;0.31 nm</p> <p>In second case when electron is used,</p> <p>Energy of electron (E)&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 4.00 KeV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;4 &times; 10<sup>3</sup> eV</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;4 &times; 10 <sup>3</sup> &times; 1.6 &nbsp;&times; 10<sup>&ndash;19</sup> J</p> <p>Again we have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; E &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;hf</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; hf&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;E</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; h &nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;E</p> <p>or,&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;0.31 &times; 10<sup>&ndash;9</sup> m</p> <p>\ &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;0.31nm &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 0.31 nm</p> <p>Hence, wavelengths of X- rays in both of the cases are equal.</p> <ol start="11"> <li><strong> X -rays are incident on the Zinc sulphide crystal. Its crystal spacing is 3.08 &times;10<sup>&ndash;4</sup> cm and the first order reflection takes places at a glancing angle of 12&deg;. Calculate the wavelength of incident X- rays. </strong><strong>[HSEB 2069]</strong></li> </ol> <p><strong>Solution:</strong></p> <p><strong>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </strong>Here,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Crystal spacing (d) = 3.08 &times; 10<sup>&ndash;4</sup> cm &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 3.08 &times; 10<sup>&ndash;6</sup>m</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; For first order reflection (n) = 1</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Glancing angle (q) = 12&deg;</p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>We have,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2d Sinq = nl</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; nl&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 2 d Sinq</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; or,&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 2 &times; 3.08 &times; 10<sup>&ndash;6</sup> &times; Sin 12&deg;</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; \&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = 1.28 &times; 10<sup>&ndash;10</sup> m.</p> <ol start="12"> <li><strong> Calculate the wavelength of the de-Broglie wave associated with thermal neutrons at temperature 27&deg;C.(Given mass of neutron=1.69&times;</strong><strong>10<sup>&ndash;27 </sup></strong><strong>kg, Boltzmann constant=1.38&times;</strong><strong>10<sup>&ndash;23 </sup></strong><strong>JK</strong><strong><sup>&ndash;1 </sup></strong><strong>)</strong></li> </ol> <p>Given,</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; T&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 273+27=300K</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; K&nbsp;&nbsp;&nbsp;&nbsp; =&nbsp;&nbsp;&nbsp; 1.38&times;10<sup>&ndash;23 </sup>JK<sup>&ndash;1</sup></p> <p><sup>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; </sup>m<sub>e</sub> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =1.67 &times; 10<sup>&ndash;27</sup> Kg</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; l=?</p> <p>we have</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; v &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; =</p> <p>&nbsp;</p> <p>so l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;</p> <p>l &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; = &nbsp;</p> <p>=1.45&times;10<sup>-10 </sup>m</p> <p><strong>&nbsp;</strong></p> <p><strong>EXERCISE</strong></p> <p><strong>&nbsp;</strong></p> <p><strong>Short Answer Questions</strong></p> <ol> <li>Distinguish between fission and fusion reaction.</li> <li>What is meant by chain reaction?</li> <li>In the production of X-ray, how will you control the penetrating power of X-rays?</li> <li>How Paschen series in originated in Hydrogen spectra? Differentiate between excitation potential and ionization potential.</li> <li>What do you meant by De Broglie waves?</li> <li>Which has more energy, a photon in the infrared or one in the ultraviolet? Give reasons.</li> <li>Write relativistic mass variation formula and explain various quantities involved in it.</li> <li>Point out the importance of a de Broglie wave.</li> <li>Why a glowing gas, such as that in a neon tube, gives only certain wavelengths of light?</li> <li>Differentiate between stimulated and spontaneous emission of radiations. [HSEB 2056]</li> <li>What do you mean by uncertainty principle? [HSEB 2056]</li> <li>Can aluminum be used as a target in X-ray tube? [HSEB 2058]</li> <li>Production of x-rays is the reverse phenomenon of photoelectric effect. Justify this statement. [HSEB 2061]</li> <li>What do you mean by matter waves? [HSEB 2062]</li> <li>What are the differences between matter wave and electromagnetic wave? [HSEB 2067]</li> <li>Define population inversion and optical pumping. [HSEB 2068]</li> <li>A proton and an electron have the same kinetic energy. Which has longer de-Broglie wavelength?&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; [HSEB 2068 Old]</li> <li>When x-rays are produced only about 10% of the initial input energy appears as x -ray energy. Explain what has happened to the other 90% of the energy. [HSEB 2068]</li> <li>A stone is dropped from the top of a building. How does its de Broglie wavelength change? [HSEB 2069 Set A]</li> <li>What do you mean by ionization energy and ionization potential? [HSEB 2069 Set A Old]</li> <li>What is laser? On what principle it works? [HSEB 2069 Set A Old]</li> </ol> <p>&nbsp;</p> <p><strong>Long Answer Questions</strong></p> <ol> <li>State and explain uncertainty principle. [HSEB 2052]</li> <li>What are Bohr's postulates of hydrogen atom? Derive an expression for the radius of Bohr's orbit. [HSEB 2053]</li> <li>What are the X-rays? Confirm with experiment the wave nature of X-rays. [HSEB 2055]</li> <li>Derive an expression for the energy of an electron in a hydrogen atom. [HSEB 2057]</li> <li>Derive Bragg's law and explain how is this law used to determine the crystal plane spacing. [HSEB 2058]</li> <li>X-ray diffraction has been very useful in determining the structure of a crystallize substance. Use this concept to determine the distance between two planes. [HSEB 2062]</li> <li>Explain how Bohr modified the Rutherford model of an atom to explain the emission of radiation from atoms: (only quantitative discussion is required). [HSEB 2062]</li> <li>Obtain the expression of energy of electron in n<sup>th</sup> orbit of hydrogen atom. [HSEB 2066]</li> <li>What do you mean by laser? Describe the He-Ne laser. [HSEB 2067]</li> <li>Explain Bragg's law of diffraction of x-rays. [HSEB 2067]</li> <li>What is laser? Describe the construction and working principle of He-Ne laser. [HSEB 2067]</li> <li>Starting from Bohr's postulates, obtain an expression for the energy of the electron in nth orbit of the hydrogen atom. [HSEB 2068 Old]</li> <li>Describe the modern method of productions of x-rays. Discuss crystal diffraction. [2068]</li> <li>On the basis of Bohr's theory of hydrogen atom, derive an expression for the energy of an electron in the nth orbit of a hydrogen atom. [HSEB 2068 Old]</li> <li>Describe the construction and working principle of He-Ne laser. Write some important uses of laser. [HSEB 2069 Set A]</li> <li>Describe the construction and working principle of He-Ne laser. Also write its important uses. [HSEB 2069 Set B]</li> </ol> <p><strong>&nbsp;</strong></p> <p><strong>Numerical Problems</strong></p> <p>&nbsp;</p> <ol> <li>X-ray beam of wavelength 2.9 is diffracted from the plane of cubic crystal. The first order diffraction is obtained at angle 35&deg;. Calculate the spacing between the planes. [HSEB 2052]</li> <li>Calculate the wavelength of the first line of the Balmer series, if the wavelength of the second line of this series is 4.86 &times; 10-7m. [HSEB 2054]</li> <li>An electron of energy 20eV comes into collision with a hydrogen atom in its ground state. The atom is excited into a higher state and the electron is scattered with reduced velocity. The atom subsequently returns to its ground state with the emission of photon of wavelength 1.216 &times; 10-7m. Determine the velocity of the scattered electron. [HSEB 2055]<br /> (mass of electron = 9.1 &times; 10-31 kg).</li> <li>If an electron position can be measured to an accuracy of 10&ndash;9 m. How accurately can its velocity be measured? (me = 9.1 &times; 10&ndash;31 kg.) [HSEB 2062]</li> <li>X-rays of wavelength 0.36 A0 are diffracted by a Bragg's crystal spectograph at a glancing angle of (4.8)&deg;. Find the spacing of the atomic planes in the crystal. [HSEB 2065]</li> <li>Obtain the de Broglie wavelength of a neutron of Kinetic energy 150 ev. (Given mass of neutron = 1.675 &times; 10-27 kg) [HSEB 2066 Supp]</li> <li>An x-ray tube operated at a d.c. potential difference of 40 KV, produces heat at the rate of 720 w assuming 0.5% of the energy of the incident electrons converted into x-radiation. Calculate (i) number of electrons per second striking the target. (ii) the velocity of the incident electrons. [Given e/m = 1.8 &acute; 1011C/kg]. [HSEB 2067]</li> <li>Find the wavelength of the radiation emitted from a hydrogen atom when an electron jumps from third orbit to second orbit. [HSEB 2067 Sup]</li> </ol> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; (Given &ndash; E0 = 8.854&times;10-12C2 N-1 m-2, h=6.62&times;10-34 Js, me = 9.1&times;10-31 kg.)</p> <ol start="9"> <li>X-ray beam of wavelength 2.9AU (&Aring;) is diffracted from the plane of cubic crystal. The first order diffraction is obtained at an angle of 35&deg;. Calculate the spacing between the planes. [HSEB 2069 (Set A) Old]</li> <li>An x-ray tube, operated at a dc potential difference of 10 KV, produces heat at the target at the rate of 720w. Assuming 0.5% of the energy of incident electrons is converted into x-radiation, calculate the tube current and velocity of the incident electrons. (e/m = 1.8 &times; 1011 Ckg-1) [HSEB 2069 (Set A)]</li> <li>Calculate the Q-value of the reaction and mention the type of reaction (endothermic or exothermic)</li> </ol> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 2HC<sup>4 </sup>= 4.00377 amu &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 80<sup>17</sup> = 17.00450 amu</p> <p>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 7N<sup>14</sup> = 14.00783 amu&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; 1H<sup>1</sup> = 1.00814 amu&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; <strong>Ans: 0.96824 Mev</strong></p> <ol start="12"> <li>Calculate the speed of particle if the mass of it is equal to 5 times its rest mass.</li> </ol> <p>Ans: 2.68 &times; 10<sup>8</sup> ms<sup>&ndash;1</sup></p> <p>&nbsp;</p> <p><strong>Objective Questions</strong></p> <p>&nbsp;</p> <ol> <li>The ratio of the energy of the orbital electron in first orbit of Bohr and that of second orbit of Bohr is</li> <li>2 b. &nbsp;&nbsp;&nbsp; 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; 9&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 6</li> <li>The ratio of the radius of the Bohr&rsquo;s first orbit to that of second orbit is</li> <li>2 b. &nbsp;&nbsp;&nbsp; 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c. &nbsp;&nbsp;&nbsp; 1/4 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 1/8</li> <li>Electrons in the atom are held in the atom because of</li> <li>Nuclear force b.&nbsp;&nbsp;&nbsp;&nbsp; gravitational force</li> <li>electrostatic force &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; weak force</li> <li>The energy of lowest level of hydrogen atom is -13.6 eV. The energy of photon emitted in transition from n=4 to n=2 is</li> <li>1.2 eV b.&nbsp;&nbsp;&nbsp;&nbsp; 2.55 eV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; 1.85 ev&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 2.8 eV</li> <li>The electron is in third excited state of hydrogen atom. How many photon transitions are possible</li> <li>3 b.&nbsp;&nbsp;&nbsp;&nbsp; 4&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c. &nbsp;&nbsp;&nbsp; 5&nbsp; &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d. &nbsp;&nbsp;&nbsp; 6</li> <li>The angular momentum of electron in n<sup>th</sup> orbit is given as</li> <li>nh/2p b. &nbsp;&nbsp;&nbsp; 2pr&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c. &nbsp;&nbsp;&nbsp; h/2p&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; p/2h</li> <li>The ratio of the velocity of the electron in the first orbit and the second orbit is</li> <li>4 b.&nbsp;&nbsp;&nbsp;&nbsp; 8&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; 3 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 2</li> <li>The ratio of frequencies o longest wavelength limit of Lyman and Balmer series in hydrogen atom is</li> <li>5:27 b.&nbsp; &nbsp;&nbsp; 27:5&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c. &nbsp;&nbsp;&nbsp; 4:1&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d: &nbsp;&nbsp;&nbsp; 1:9</li> <li>Matter waves are similar in nature to</li> <li>X-rays b. &nbsp;&nbsp;&nbsp; g-rays&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c. &nbsp;&nbsp;&nbsp; cathode rays &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d. &nbsp;&nbsp;&nbsp; none</li> <li>The energy of electron in hydrogen atom in its ground state is -13.6eV. The energy of electron corresponding to level n = 5 is</li> <li>-5.40 eV b.&nbsp;&nbsp;&nbsp;&nbsp; -0.85 eV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; -0.54 eV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 5.4 eV</li> <li>To produce hard X-rays we should increase</li> <li>Tension applied in filament b.&nbsp;&nbsp;&nbsp;&nbsp; Current in filament</li> <li>Potential across cathode and target d.&nbsp;&nbsp;&nbsp;&nbsp; None</li> <li>Penetrating power of X-rays actually depends on</li> <li>Velocity b.&nbsp;&nbsp;&nbsp;&nbsp; intensity&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; frequency&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; none</li> <li>The maximum energy of X-ray is 2 KeV. What is the accelerating potential applied across the filament and target</li> <li>1 KV b.&nbsp;&nbsp;&nbsp;&nbsp; 4 KV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; 3 KV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 2 KV</li> <li>An X-ray tube is operated by a million volt. The wavelength of shortest wave produced will be</li> <li>0.01&acute;10<sup>-10</sup> m b.&nbsp;&nbsp;&nbsp;&nbsp; 0.06&acute;10<sup>-10</sup> m&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; 3&acute;10<sup>-10</sup> m&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 1 &acute; 10<sup>-10</sup> m</li> <li>The difference between hard and soft X-rays is</li> <li>Velocity &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; b.&nbsp;&nbsp;&nbsp;&nbsp; frequency&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; intensity&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; polarization</li> <li>X-rays can&rsquo;t penetrate bone because bones have</li> <li>a high mol. Wt &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; b.&nbsp;&nbsp;&nbsp;&nbsp; low mol. Wt&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;</li> <li>hard crystalline structure &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d. &nbsp;&nbsp;&nbsp; amorphous nature</li> <li>Which of the following principle is involved in the generation of X-rays?</li> <li>Conversion of electrical energy into radiant energy</li> <li>Conversion of kinetic energy into potential energy</li> <li>Conversion of electrical energy into electromagnetic wave</li> <li>Conversion of mass into energy</li> <li>The minimum wavelength of X-rays produced by electrons accelerated by a potential difference of V volts is</li> <li>eV/hc b.&nbsp;&nbsp;&nbsp;&nbsp; hc/eV&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; h/c&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; none</li> <li>An X-ray tube is operating at 60 KV. The wavelength of X-rays produced is</li> <li>0.621&acute;10<sup>-10</sup> m b.&nbsp;&nbsp;&nbsp;&nbsp; 23.3&acute;10<sup>-10</sup> m&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; 20.6&acute;10<sup>-10</sup> m&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 0.206&acute;10<sup>-10</sup> m</li> <li>The inter planner distance of a crystal is 0.4 nm. X-rays are incident on the crystal and first order reflection is obtained at the glancing angle of 30&deg;. What is the wavelength of incident X-rays?</li> <li>0.1 nm b.&nbsp;&nbsp;&nbsp;&nbsp; 0.3 nm&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; c.&nbsp;&nbsp;&nbsp;&nbsp; 0.04 nm&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; d.&nbsp;&nbsp;&nbsp;&nbsp; 0.4 nm</li> </ol> <p>&nbsp;</p> <p><strong>Answers:</strong></p> <p>&nbsp;</p> <table> <tbody> <tr> <td width="27"> <p><strong>1.</strong></p> </td> <td width="30"> <p>b</p> </td> <td width="35"> <p><strong>2.</strong></p> </td> <td width="37"> <p>c</p> </td> <td width="42"> <p><strong>3.</strong></p> </td> <td width="30"> <p>c</p> </td> <td width="38"> <p><strong>4.</strong></p> </td> <td width="35"> <p>d</p> </td> <td width="39"> <p><strong>5.</strong></p> </td> <td width="39"> <p>b</p> </td> <td width="40"> <p><strong>6.</strong></p> </td> <td width="39"> <p>a</p> </td> <td width="38"> <p><strong>7.</strong></p> </td> <td width="29"> <p>d</p> </td> <td width="38"> <p><strong>8.</strong></p> </td> <td width="29"> <p>a</p> </td> </tr> <tr> <td width="27"> <p><strong>9.</strong></p> </td> <td width="30"> <p>c</p> </td> <td width="35"> <p><strong>10.</strong></p> </td> <td width="37"> <p>c</p> </td> <td width="42"> <p><strong>11.</strong></p> </td> <td width="30"> <p>c</p> </td> <td width="38"> <p><strong>12.</strong></p> </td> <td width="35"> <p>c</p> </td> <td width="39"> <p><strong>13.</strong></p> </td> <td width="39"> <p>d</p> </td> <td width="40"> <p><strong>14.</strong></p> </td> <td width="39"> <p>a</p> </td> <td width="38"> <p><strong>15.</strong></p> </td> <td width="29"> <p>b</p> </td> <td width="38"> <p><strong>16.</strong></p> </td> <td width="29"> <p>c</p> </td> </tr> <tr> <td width="27"> <p><strong>17.</strong></p> </td> <td width="30"> <p>a</p> </td> <td width="35"> <p><strong>18.</strong></p> </td> <td width="37"> <p>b</p> </td> <td width="42"> <p><strong>19.</strong></p> </td> <td width="30"> <p>d</p> </td> <td width="38"> <p><strong>20.</strong></p> </td> <td width="35"> <p>d</p> </td> <td width="39"> <p><strong>&nbsp;</strong></p> </td> <td width="39"> <p>&nbsp;</p> </td> <td width="40"> <p><strong>&nbsp;</strong></p> </td> <td width="39"> <p>&nbsp;</p> </td> <td width="38"> <p><strong>&nbsp;</strong></p> </td> <td width="29"> <p>&nbsp;</p> </td> <td width="38"> <p><strong>&nbsp;</strong></p> </td> <td width="29"> <p>&nbsp;</p> </td> </tr> </tbody> </table> <p>q&nbsp; q&nbsp; q</p>



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