question_answer

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

A)  \[10\]           

B)  \[5\]

C)  \[50\]

D)  \[20\]

Correct Answer: B

Solution :

In Millikan's oil drop method, when gravitational force balances the forces 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   \[~q=ne\] \[\therefore \] \[ne\times \frac{V}{d}=mg\] \[\Rightarrow \] \[n=\frac{mgd}{eV}\] Given, \[m=1.8\times {{10}^{-14}}kg,\,\,\,g=10m/{{s}^{2}},\] \[d=0.9\,cm=0.9\times {{10}^{-2}}\,m,\] \[e=1.6\times {{10}^{-19}}C,\] \[V=2000volt\] \[\therefore \] \[n=\frac{1.8\times {{10}^{-14}}\times 10\times 0.9\times {{10}^{-2}}}{1.6\times {{10}^{-19}}\times 2000}\] \[n=\frac{81}{16}=5\] electrons