Question

Under the action of a given coulombic force the acceleration of an electron is $\times {10}^{22}m{s}^{-2}$. 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. $0.6\times {10}^{27}m{s}^{-2}$

Answer 1

The correct option is C $1.5\times {10}^{19}m{s}^{-2}$

The acceleration due to given coulombic force $F$ is

$a=\frac{F}{m}ora\propto \frac{1}{m}......\left(i\right)$

$\therefore \frac{a_p}{a_e}=\frac{m_e}{m_p}$, where $m_e$ and $m_p$ are masses of electron and proton respectively

$a_p=\frac{a_e{m}_e}{m_p}=\frac{(2.5\times {10}^{22}m{s}^{-2})(9.1\times {10}^{-31}kg)}{1.67\times {10}^{-27}kg}=13.6\times {10}^{18}m{s}^{-2}\simeq 1.5\times {10}^{19}m{s}^{-2}$