Question

The magnetic moment of electron due to orbital motion is proportional to (n = principal quantum number)

• A. $\frac{1}{n^2}$
• B. $\frac{1}{n}$
• C. $n^2$
• D. $n$

Answer 1

Answer: (D)

$n$

Answer 2

From the image below, we can see that an electron (e) is moving in an orbit with radius (Rn).
Assuming, Rn = R

The magnetic moment (M) of an electron due to its orbital motion arises from the current loop formed by the electron moving around the nucleus.

The magnetic moment of a current loop is given by:
$M = I\times A$,
Where, I is current flowing and A is area of circle

Now, for an electron orbiting a nucleus in a circular orbit, the current (I) can be related to the electron's charge (e) and its velocity (v):
$I = \frac{e}{T}$
where, e = rate of flow of charge and T = time taken by the charge to travel one round around a circle.
$A =\pi R^2$
The time period (T) of the orbit can be related to the circumference (2πr) of the orbit and the velocity (v) of the electron:
$T=\frac{2\pi R}{V}$
therefore, on putting the value of T in I we get,
$I=\frac{ev}{2\pi R}$

So magnetic moment (M):
$M=\frac{ev}{2\pi R}\times \pi R^2$

$M=\frac{evR}{2}$
Now, let's relate the velocity (v) of the electron to its angular momentum (L). For a circular orbit, the angular momentum is given by:
L = mvR, where m = mass of the electron, R = radius of the orbit
Since L is the quantised unit of h , we have
$L=n\times h$, where n is the principal quantum number.
Now we can express v in terms of n and r:
$v=\frac{L}{mR}=\frac{nh}{mR}$
Substituting this expression for v into the equation for M, we get:
$M=\frac{e(\frac{nh}{mR})R}{2}=\frac{enh}{2m}$
The constant of proportionality is ($\frac{e}{2m}$), where e is the charge of the electron, is the reduced Planck constant and m is the mass of the electron.
So, the magnetic moment (M) of an electron due to its orbital motion is proportional to n, the principal quantum number.