Thomson’s Experiment to Determine Specific Charge (e/m) of Electrons

Thomson’s Experiment to Determine Specific Charge (e/m) of Electrons

The ratio of charge to mass is called the specific charge.

Principle: If a beam of electrons is subjected to electric and magnetic fields it experiences forces. By adjusting the magnitude and direction of the two fields, the net force on the electron is made zero.

Construction or Experimental Set-up

The apparatus consists of a cathode C and anode A which are enclosed in an evacuated discharge tube. When a high potential difference is applied between cathode and anode, a narrow beam of cathode rays emerges from a small hole in the anode and passes between two parallel plates P₁ and P₂. A uniform electric field is applied to the plane of paper with the help of horizontal plates P₁ and P₂ and the electrons will deflect upwards (i.e. towards point S₂) in the plane of the paper. On the other hand, a uniform magnetic field is also applied in this region perpendicular to the plane of the paper and inward direction. The magnetic field will deflect the electrons downwards (towards point S₁) in the plane of the paper.

Theory

If the uniform magnetic field is applied, then force on electron on electron due to the magnetic field, F_m = Bev, where e is the changes on an electron, B is the magnetic field strength and v is the velocity of the electron.

The force Be v is always normal to the path of electron beam and therefore moves the electron along circular path. Hence, the force on electron due to magnetic field provides necessary centripetal force and therefore,

\begin{align*} Bev &= \frac {mv^2}{r} \\ \text {or,} \: \frac em &= \frac {v}{rB} \\ \end{align*}

Therefore, to find \( \frac em \) , we have to determine B, v and r.

1. Measurement of v
   To measure the velocity of the electron, when both the fields are applied at the same time, the intensities of the electric and magnetic field can be adjusted so that the electron beam remains undeflected i.e. remains at point S. If E is the electric field strength, F_e = eE.
   As the electron spot remain at S, then the force on electron due to electric field is equal and opposite then the force on electron due to magnetic field.
   \begin{align*} \text {i.e.} \: F_m &= F_e \\ \text {or,} \: Bev &= eE \\ \text {or,} v &= \frac EB \\ \end{align*}
   Since, E = V/d and B is measured already, then the value of v can be found.
2. To Measure the B
   The strength of the applied magnetic field is measured with the help of a search coil or flux meter.
3. To Measure 'r'
   

   

   
   The electron beam follows a circular path in the region of magnetic field. As the electrons out of the magnetic field, they move along the tangent AS₁ to the circular path OA at point A as shown in a figure. Draws normals to the circular path OA at points O and A. In the diagram the point C, where the two normals meet is the center of the circular path. Let ÐOCA = Ï´. Produce tangents AS₁ backward to meet OS in point Q. Then SQS₁ is equal to Ï´.
   Let y be the displacement of the electron spot on the screen due to a magnetic field, L be the distance of the screen from the center Q of the magnetic field and r be the radius of the circular path. Then from figure SS₁ = y, QS = L, and CO = CA = r.

Now, from the right angled DQSS₁

\begin{align*} \tan \theta &= \frac {SS_1}{QS} = \frac yL \\ \text {Since} \: \theta \text {is small} \\ \therefore \tan \theta \approx \theta \\ \therefore \theta &= \frac yL \dots (i) \\ \text {OA is an arc whose length is nearly equal to l,} \\ \therefore \theta = \frac {\widehat {OA}}{OC} = \frac {\text {arc length}}{\text {radius}} \\ \text {or,} \: \theta &= \frac lr \dots (ii) \\ \text {Therefore, from equation} \: (i) \text {and} \: (ii), \text {we have} \\ \frac lr &= \frac yL \\ \text {or,} \: r &= \frac {l \times L}{y} \dots (iii) \\ \end{align*}

From the experimental arrangement the value of l, L and y can be measured. Thus, from the relation value of r can be calculated.

Knowing the values of B, v and r, the value of e/m of an electron can be calculated which equal to 1.77 × 10¹¹ C kg^-1.

Reference

Manu Kumar Khatry, Manoj Kumar Thapa, et al. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.

S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.