Question

In the Bohr model, an electron moves in circular orbit around the proton. Considering an orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in n\({}^{th}\) state, is

• A. \(\left(\frac{e}{m}\right)\frac{n^{2}h}{2\pi }\)
• B. \(\left(\frac{e}{2m}\right)\frac{nh}{2\pi }\)
• C. \(\left(\frac{e}{2m}\right)\frac{n^{2}h}{\pi }\)
• D. \(\left(\frac{e}{m}\right)\frac{nh}{2\pi }\)

Hint:

Treat the orbiting electron as a current loop, so the magnetic moment is current times area, then apply Bohr's angular momentum condition.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
A circulating electron acts as a current loop of current \(I=\dfrac{e}{T}\) and area \(A=\pi r^{2}\).

Step 2: Magnetic Moment
\(\mu=IA=\dfrac{e}{T}\pi r^{2}=\dfrac{e v r}{2}\), since \(T=\dfrac{2\pi r}{v}\).
Writing \(vr=\dfrac{L}{m}\) gives \(\mu=\dfrac{e}{2m}L\).

Step 3: Bohr's Condition
In the nth orbit the angular momentum is \(L=\dfrac{nh}{2\pi}\).

Step 4: Result
\(\mu=\left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}\).

Final Answer:
The magnetic moment is \(\dfrac{e}{2m}\) times \(\dfrac{nh}{2\pi}\). \[ \boxed{\text{(B) }\left(\dfrac{e}{2m}\right)\dfrac{nh}{2\pi}} \]