Question

An electron in an atom revolves around the nucleus in an orbit of radius $0.53\stackrel{\circ }{A}$. Calculate the equivalent magnetic moment, if the frequency of revolution of electron is $6.8\times {10}^9$ MHz(Given : $e=1.6\times {10}^{-19}$)

Answer 1

Given, frequency ie; angular velocity v= $6.8\times {10}^{9}MHz$

Current due to revolving electron I=$\frac{ev}{2\pi r}$------- (1),

where r= orbital radius.

Area of the orbit, A=$\pi r^2$

A=$3.14\times {(0.53\times {10}^{-10})}^2{m}^2$

=$8.8247\times {10}^{-21}{m}^2$

magnetic moment

$M=IA$-------- (2)

From (1), I=$\frac{ev}{2\pi r}=\frac{1.6\times {10}^{-19}\times 6.8\times {10}^9}{2\times \pi \times 0.53\times {10}^{-10}}$

= 3.267 A

(2) $=>M=IA$= $3.267\times 8.824\times {10}^{-21}$

M= $2.882\times {10}^{-20}$ $Amp.m^2$