Question

In a Millikan's oil drop experiment the charge on an oil drop is calculated to be $6.35 ? 10^{-19}\, C$. The number of excess electrons on the drop is

• A. $32$
• B. $4$
• C. $42$
• D. $6$

Answer 1

Answer: (B)

$4$

Answer 2

As we know, $Q = ne$
Number of electron $n=\frac{Q}{e}$
$\frac{6.35\times10^{-19}}{1.6\times10^{-19}}$
$=3.9=4$