In electrolysis, we assume that the carriers of current through an acid or salt solution are ions, which may be positively and negatively charged. From Faraday’s law of electrolysis, the charge carried by each ion is proportional to its valency.
When the element is monoatomic, the number of ions of one kind carrying the charge is equal to the number of molecules. Thus, the charge on each ion is given by 1.6 ×10^{-19} C. If 1.6 × 10^{-19} C is denoted by e and e is the basic unit of charge. All charges, whether produced in electrostatics current electricity or any other method, are multiples of the basic unit 'e'.
J.J. Thomson discovered electron while studying the conduction of electricity of electricity through gasses at low pressure. R.A Millikan measured the charge of an electron using a simple method known as Millikan’s oil drop experiment.
Millikan determines the value of charge on an electron using oil drop method discussed below. This method was based upon the measurement of
The experiment arrangement used by Millikan to determine the charge on an electron is shown in the figure. It consists of two metal circular plates A and B about 20 cm in diameter and 1.5 cm apart with a small hole H in the center of the upper plate A.
The upper plate A is connected to a high tension battery while the lower plate B is earthed. The plates are arranged inside a double walled chamber.
Through the hole in the upper plate H, clock oil is sprayed by means of an atomizer. These drops get charged due to friction and carry a few electronic charges. The window W_{1} is used to illuminate the oil drops by providing enough light. The window W_{2} is used to let X-rays pass into the space between the plates in order to ionize the oil drops in case the oil drop are not ionized by friction. The microscope is provided with a crosswire and a micrometer so that motion of the oil can be observed and measured.
Theory 1
Suppose the electric field is not applied. As the oil drop falls under gravity its velocity goes on increasing. A stage comes when the viscous force on the oil drop becomes equal to its resultant weight. The oil drop now moves with a constant velocity v_{1} called terminal velocity.
\begin{align*} \text {Let,} \: r &= \text {radius of oil drop} \\ m &= \text {mass of oil drop} \\ \rho &= \text {density of the oil} \\ \sigma &= \text {density of air} \\ \text {Then,} \\ \text {Volume of the oil drop} &= \frac 43 \pi r^3 \\ \text {Weight of the oil drop} (W) &= \frac 43 \pi r^3 \rho g \\ \text {Upthrust due the air, U} &= \text {weight of the air displaced by a drop} \\ &= \frac 43 \pi r^3\sigma g \\ \end{align*} The viscous force on the oil drop is upward direction \(F = 6\pi \eta rv_1 \) Stoke's Law. When the oil drop is moving with terminal velocity \( V_1,\) then \begin{align*}F + U &= W \\ 6\pi \eta rv_1 + \frac 43 \pi r^3 \sigma g = \frac 43 \pi r^3 \rho g \\ \text {or,} \: \frac 43 \pi r^3 (\rho - \sigma) g = 6\pi \eta r v_1 \dots (i) \\ \therefore r &= \sqrt {\frac {9\eta v_1}{2(\rho -\sigma) g}} \\ \end{align*}
Knowing the values of \(\rho, \sigma , \eta , v_1 \text {and}\: g\)the radius of the drop can be calculated.
Theory 2:
A strong electric field is applied between the plates in such a direction that force on the negatively charged oil drop starts moving upward and soon attains a terminal velocity v_{2} in an upward direction.
Let, E be the strength of the electric field. As the drop carriers a charge Q , then electrostatic force on oil drop in
\begin{align*} \text {Upward direction} (F_e) &= QE \\ \text {Viscous force in downward}(F) &= 6\pi \eta r v_2 \\ W – U &= \frac 43 \pi r^3 (\rho - \sigma ) g \\ \end{align*} When the oil drop attains terminal velocity \begin{align*} \: v2, \text {then} \\ F_e + U &= F + W \\ \text {or,} \: F_e = (W – U) + F \\ \text {or,} \: QE &= \frac 43 \pi r^3 (\rho - \sigma ) g + 6\pi \eta r v_2 \\ \therefore QE &= 6\pi \eta r v_1 + 6\pi \eta r v_2 \\ \text {or,} \: Q &= \frac {6\pi \eta r (v_1 +v_2)}{E} \\ &= \frac {6\pi \eta (v_1 +v_2)}{E} \times \sqrt {\frac {9\eta v_1}{2(\rho -\sigma) g}} \dots (ii) \\ \end{align*}
Knowing all the quantities on the right hand side, the value of charge Q on the oil drop can be determined.
Estimate of Mass of an Electron
If we assume that the electron which are the constituents of cathode rays carry the fundamental charge as calculated by Millikan’s oil drop experiment, then from the Thomson’s measured value of e/m, we can estimate the mass of the electron.
\begin{align*} \therefore m &= \frac {e}{e/m} = \frac {1.6 \times 10^{-19} \: c}{1.75 \times 10^{11} c/kg} = 9.11 \times 10^{-31} \: kg \end{align*}
Importance of Millikan’s Experiment
Reference
Manu Kumar Khatry, Manoj Kumar Thapa, Bhesha Raj Adhikari, Arjun Kumar Gautam, Parashu Ram Poudel. Principle of Physics. Kathmandu: Ayam publication PVT LTD, 2010.
S.K. Gautam, J.M. Pradhan. A text Book of Physics. Kathmandu: Surya Publication, 2003.
From Faraday’s law of electrolysis, the charge carried by each ion is proportional to its valency.
R.A Millikan measured the charge of an electron using a simple method known as Millikan’s oil drop experiment.
Milikan's oil drop method was based upon the measurement of the terminal velocity of an oil drop under the influence of gravity alone and the terminal velocity under the joint action of gravity and an electric field opposing the gravity.
.
ASK ANY QUESTION ON Millikan’s Oil Drop Experiment
You must login to reply
ananta dhakal
What is the use of x-rays in milikan's oil drop experiment ?
Dec 29, 2016
4 Replies
Successfully Posted ...
Please Wait...