If $\frac{e}{m_e}$ = 1.758820 $\times$ ${10}^{11}$ C ${kg}^{-1}$

The mass of an electron is

The correct option is A

9.1094 × 10^-31

What do we know from Millikan oil drop experiment?

Charge of electron(e) = 1. 6 $\times$ ${10}^{11}$ C/kg

Therefore,

Mass of electron($m_e$) = $\frac{e}{\frac{e}{m_e}}$

$m_e$ = $\frac{1.6022\times {10}^{-19}C/electron}{1.75822\times {10}^{11}C/kg}$

= 9.1094 $\times$ ${10}^{-31}$kg

= 9.1094 $\times$ ${10}^{-28}$g