question_answer

In Millikans oil drop experiment, a charged drop of mass \[1.8\times {{10}^{-14}}kg\] is stationary between the plates. The distance between the plates 0.9 cm and potential difference between the plates is 2000 V. The number of electons on the oil drop is:

A)  10                                         

B)  5

C)  50                                         

D)  20

Correct Answer: B

Solution :

                In Millikans oil drop method, when gravitational force balances the force due to electric field, then drop becomes stationary. Therefore, \[qE=mg\] where E is electric field intensity, q is charge on drop, m the mass, g the acceleration due to gravity. Also,    \[E=\frac{V}{d}\]and \[q=ne\]                 \[\therefore \]  \[ne\times \frac{V}{d}=mg\] Given, \[m=1.8\times {{10}^{-14}}kg,\text{ }g=10\text{ }m/{{s}^{2}},\] \[d=0.9\text{ }cm=0.9\times {{10}^{-2}}m,\] \[e=1.6\times {{10}^{-19}}C,\] \[V=2000\text{ }Volt\] \[\therefore \]  \[n=\frac{1.8\times {{10}^{-14}}\times 10\times 0.9\times {{10}^{-2}}}{1.6\times {{10}^{-19}}\times 2000}\] \[n=\frac{81}{16}\approx 5\text{ }electrons\]