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