Q. In a hydrogen atom, the electron is making $6.6\ \times \ 10^{15}revs^{- 1}$ around the nucleus in an orbit of radius $0.528\ \mathring{\mathrm{A}}.$The magnetic moment ($Am^{2}$) will be
A) $1\ \times \ 10^{- 15}$
B) $1\ \times \ 10^{- 10}$
C) $1\ \times \ 10^{- 23}$
D) $1\ \times \ 10^{- 27}$
Explanation
(c) Current , $I = 6.6\ \times \ 10^{15}\ \ \times \ 1.6\ \times \ 10^{- 19}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10.5\ \times \ 10^{- 4}\ A$
Area $\ A = \pi$R$^{2}$ = $3.142\ \times \ (\ 0.528)^{2} \times \ 10^{- 20}\ m^{2}$
So, magnetic moment $M = IA = 10.5 \times 10^{- 4} \times 3.142$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \times (0.528)^{2} \times 10^{- 20}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 10 \times 10^{- 24} = 10^{- 23}\ units$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.