Question

An electron in an atom revolves around the nucleus in an orbit of radius $0.7\mathring{A}.$ Calculate the equivalent magnetic moment,if the frequency of revolution of electron is $2\times {10}^{10}$ MHz.

• A. $4.92\times {10}^{-23}A{m}^2$
• B. $4.92\times {10}^{-24}A{m}^2$
• C. $5.21\times {10}^{-23}A{m}^2$
• D. $5.21\times {10}^{-24}A{m}^2$

Answer 1

The correct option is A $4.92\times {10}^{-23}A{m}^2$

Given: Radius of an orbit= $0.7A^{\circ }$

Frequency of revolution= $2\times {10}^{10}MHz$

Solution: The revolving electron can be considered as a current loop. So, its magnetic moment can be given by,

$M=NIA$ (1)

where, N is the number of turns,

I is the current in the loop and,

A is the area of the loop.

Also, $I=ev$

where v is the frequency and e is the electron charge.

Substituting the value of current in equation (1) we get,

$M=1\times \left(ev\right)\times \pi r^2$

$M=1\times (1.6\times {10}^{-19}\times {10}^{10})\times (3.14\times (7\times {10}^{-10}{)}^2)$

On solving we get,

$M=4.92\times {10}^{-23}A{m}^2$

Hence, the correct option is (A).