Chapter 20, Problem 58P

To determine

To show: Magnetic dipole moment of an electron is $M=\frac{e}{2m}L$

Explanation of Solution

Given:

Orbital angular momentum is L .

Atom is hydrogen.

Formula used:

In a circular orbit, angular momentum L is given by

  $L=mvr$

Here, m is mass of electron, v is speed and r is radius of electron.

Magnetic dipole moment is given by

  $M=iA=\frac{q}{t}\cdot A$

Calculation:

Time of travel is

  $\begin{array}{c}t=\frac{d}{v}\\ =\frac{2\pi r}{v}\end{array}$

Here,

v is the velocity

d is the circumference and r is the radius of an orbit.

For a proton nucleus of hydrogen atom,

  $\begin{array}{c}q=ne\\ =1\times e\\ =e\end{array}$

Substituting the values,

  $\begin{array}{c}M=\frac{e}{\left(\frac{2\pi r}{v}\right)}\times \pi r^2\\ =\frac{evr}{2}\end{array}$

To deduce the above relation in terms of orbital angular momentum, multiply and divide by m ,

  $M=\frac{emvr}{2m}=\frac{e}{2m}L$