d
Millikan's oil drop expeniment is used to measure chrorge of an electron.

We have, \(q=n e \quad \cdots\) where \(n=1,2,3 \cdots\). it was conclucle that, as charge is quantized it should be pritegral multiple of \(1.6 \times 10^{-19}\).

Fon given options,

Excess \(e^{-}=\frac{\text { charge on millikan's oil drop }}{\text { charge on } 1 e^{-}}\)

\(\therefore\) for option \(A\)

Excess \(e^{-}=\frac{4.0 \times 10^{-19}}{1.6 \times 10^{-19}}=2.5\).

for option \(B\)

Excesse \(e^{-}=\frac{6.0 \times 10^{-19}}{1.6 \times 10^{-9}}=3.75\)

for option \(C\)

Excess \(e^{-}=\frac{10.0 \times 10^{-19}}{1.6 \times 10^{-19}}=6.25\)

Here, as all values comes in decimal values and as charge is quantized, we need integral multiple of \(1.6 \times 10^{-19}\).