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

The Correct Option is D

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. 

Answer 2

Electron Orbit and Magnetic Moment
An electron in a circular orbit around a nucleus creates a magnetic moment (M) due to its orbital motion.
Magnetic Moment (M):

• M is given by M = e * v * R / (2m), where:
• e = electron charge,
• v = electron velocity,
• R = radius of the orbit,
• m = electron mass.

Relation to Angular Momentum (L):

• The angular momentum L = n * h / (2π), where n is the principal quantum number and h is Planck's constant.
• The velocity v of the electron is v = n * h / (2π * m * R).

Derivation of Magnetic Moment (M):

• Substituting v into the expression for M gives M = e * n * h / (2m).
• This shows M is proportional to the principal quantum number n.

Constant of Proportionality:

• The constant e / (2m) * h / (2π) relates M to fundamental constants of the electron.