Q. 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 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) $1.6 \times 10^{27}m/s^{2}$
Explanation
(c) Coulombic force on electron, $F_{e} = a_{e}m_{e}\ $
Similarly, for proton $F_{p} = a_{p}m_{p}$
Here $F_{e} = F_{p}$
$\therefore\ \ \ \ \ \ \ a_{e}m_{e} = a_{P}m_{p}$
$\Longrightarrow \ \ \ \ \ \ \ \ \ \ a_{p} = \frac{a_{e}m_{e}}{m_{p}}$
$\ \ \ \ \ \ \ \ \ = \frac{\left( 2.5 \times 10^{22} \right)(9.1 \times 10^{- 31})}{1.6 \times 10^{- 27}}$
$\ \ \ \ \ \ \ \ \ = 1.42 \times 10^{19}$
$= 1.5 \times 10^{19}$ m/$s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.