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