question_answer

In the Bohr model an electron moves in a circular orbit around the proton. Considering the orbiting electron to be a circular current loop, the magnetic moment of the hydrogen atom, when the electron is in \[{{\operatorname{n}}^{\operatorname{th}}}\] exited state, is:              [JEE ONLINE 09-04-2013]

A) \[\left( \frac{e}{2m} \right)\frac{{{\operatorname{n}}^{2}}h}{2\pi }\]

B) \[\left( \frac{e}{m} \right)\frac{\operatorname{n}h}{2\pi }\]

C)             \[\left( \frac{e}{2m} \right)\frac{\operatorname{n}h}{2\pi }\]

D) \[\left( \frac{e}{m} \right)\frac{{{\operatorname{n}}^{2}}h}{2\pi }\]  

Correct Answer: C

Solution :

[c] As,

and magnetic moment

(i)

Now,

It becomes,          (ii)

Also,

Putting this value in Eq. (ii), we get