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 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) $1.6 \times 10^{27}m/s^{2}$
Explanation
(c) The acceleration of the electron due to given coulombic
force $F$ is $a_{e} = \frac{F}{m_{e}}$ $\ldots(i)$
where $m_{e}$ is the mass of the electron
The acceleration of the proton due to same force $F$ is
$a_{p} = \frac{F}{m_{p}}$ ….(ii)
Where $m_{p}$ is the mass of the proton
Divide (ii) by (i), we get $\frac{a_{p}}{a_{e}} = \frac{m_{e}}{m_{p}}$
$a_{p} = \frac{a_{e}m_{e}}{m_{p}} = \frac{(2.5 \times 10^{22}ms^{- 2})(9.1 \times 10^{- 31}kg)}{(1.67 \times 10^{- 27}kg)}$
$= 13.6 \times 10^{18}ms^{- 2} \approx 1.5 \times 10^{19}ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.