Question

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

The mass of an electron is

• A. 9.1094 $\times$ ${10}^{-31}$
• B. 9.1094 $\times$ ${10}^{-21}$
• C. 6.1094 $\times$ ${10}^{-31}$
• D. 6.1094 $\times$ ${10}^{-21}$

Answer 1

Answer: (A)

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

Hence, the mass of electron is $9.1094\times {10}^{-28}g$ .