Question

Under the action of a given coulombic force the acceleration of an electron is $2.5 \times 10^{22}\, m \,s^{-2}$. Then the magnitude of the acceleration of a proton under the action of same force is nearly

• A. $1.6 \times 10^{19}\, m \,s^{-2}$
• B. $9.1 \times 10^{31}\, m \,s^{-2}$
• C. $1.5 \times 10^{19}\, m \,s^{-2}$
• D. $1.6 \times 10^{27}\, m \,s^{-2}$

Answer 1

Answer: (C)

The acceleration due to given coulombic force $F$ is
$a=\frac{F}{m}$ or $a\propto \frac{1}{m}\,...\left(i\right)$
$\therefore \frac{a_{p}}{a_{e}}=\frac{m_{e}}{m_{p}} $, where $m_{e}$ and $m_{p}$ are masses of electron and proton respectively
$a_{p}=\frac{a_{e}m_{e}}{m_{p}}=\frac{\left(2.5\times10^{22}\,m\,s^{-2}\right)\left(9.1\times10^{-31}\,kg\right)}{\left(1.67\times10^{-27}\,kg\right)}$
$=13.6\times10^{18}\,m\,s^{-2}\approx1.5\times10^{19}\,m\,s^{-2}$