Question

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

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

Hint:

Orbital magnetic moment equals e L / 2m, and L is n h over 2 pi.

Answer 1

The Correct Option is A

Step 1: Understand the concept
An electron orbiting in a circle is a current loop. Its magnetic moment is \(\mu = \dfrac{e}{2m}L\), where \(L\) is the orbital angular momentum.

Step 2: Use Bohr's quantisation
In the \(n\)th Bohr orbit, \(L = \dfrac{nh}{2\pi}\).

Step 3: Substitute
\[ \mu = \frac{e}{2m}\cdot\frac{nh}{2\pi} = n\,\frac{eh}{4\pi m} \]

Step 4: Result
The magnetic moment is proportional to \(n\), option (A). The factor \(\dfrac{eh}{4\pi m}\) is the Bohr magneton, which is the same for every orbit.

Final Answer:
The magnetic moment is proportional to n. This is option (A). \[ \boxed{\text{(A) }n} \]