Q. In hydrogen atom, the electron is moving round the nucleus with velocity $2.18\ \times \ 10^{6}\ ms^{- 1}$ in an orbit of radius $0.528\ \mathring{\mathrm{A}}.\ $The acceleration of the electron is
A) $9\ \times 10^{18}\ ms^{- 2}$
B) $9\ \times \ 10^{22}\ ms^{- 2}$
C) $9\ \times \ 10^{- 22}\ ms^{- 2}$
D) $9\ \times \ 10^{12}\ ms^{- 2}$
Explanation
(b) Given, $v = 2.18\ \times 10^{6}\ {ms}^{- 1},\ r = 0.528\ \times \ 10^{- 10}m$
Acceleration of electron moving round the nucleus
$\ \ \ \ \ \ \ \ \ \ \ \ \ a\ =$ $\frac{\left( 2.18\ \times 10^{6} \right)^{2}}{0.528\ \times 10^{- 10}} \approx$ $9\ \times 10^{22}\ ms^{- 2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.