Q. In hydrogen atom, the electron is moving round the nucleus with velocity $2.18 \times 10^{6}m/s$ in an orbit of radius $0.528\mathring{\mathrm{A}}$. The acceleration of the electron is
A) $9 \times 10^{18}m/s^{2}$
B) $9 \times 10^{22}m/s^{2}$
C) $9 \times 10^{- 22}m/s^{2}$
D) $9 \times 10^{12}m/s^{2}$
Explanation
(b) The electron in a hydrogen atom, moves with constant acceleration, called centripetal acceleration, round the nucleus. Acceleration of electron $a = \frac{v^{2}}{r}$
Given, $v = 2.18 \times 10^{6}m/s$
$r = 0.528\ \mathring{\mathrm{A}} = 0.528 \times 10^{- 10}m$
$\therefore a = \frac{\left( 2.18 \times 10^{6} \right)^{2}}{0.528 \times 10^{- 10}} = 9 \times 10^{22}\ m/s^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.