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, \[m=1.8\times {{10}^{-14}}kg,\,\,\,g=10m/{{s}^{2}},\] \[d=0.9\,cm=0.9\times {{10}^{-2}}\,m,\] \[e=1.6\times {{10}^{-19}}C,\] \[V=2000volt\] \[\therefore \] \[n=\frac{1.8\times {{10}^{-14}}\times 10\times 0.9\times {{10}^{-2}}}{1.6\times {{10}^{-19}}\times 2000}\] \[n=\frac{81}{16}=5\] electronsYou need to login to perform this action.
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