In a Millikans oil drop experiment the charge on an oil drop is calculated to be\[6.35\times {{10}^{-19}}C\]. The number of excess electrons on the drop is
A) 3.2
B) 4
C) 4.2
D) 6
Correct Answer:
B
Solution :
Number of electron \[n=\frac{Q}{e}\] \[=\frac{6.35\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}\] \[=3.9=4\]