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 Main Online Paper (Held On 09 April 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 :

                As, \[i=\frac{e}{T}\]                                 and magnetic moment                                  \[M=iA\]                              \[(\because \,A=\pi {{r}^{2}})\] \[\therefore \] \[M=\frac{e}{T}\cdot \,\pi {{r}^{2}}\]                                      ?(i)                 Now,     \[T=\frac{2\pi r}{v}\]                 It becomes, \[M=\frac{\frac{e}{2\pi r}\cdot \,\pi {{r}^{2}}}{v}=\frac{evr}{2}\]      ?(ii)                 Also,      \[mvr=\frac{nh}{2\pi }\]                 \[vr=\frac{nh}{2\pi m}\]                 Putting this value in Eq. (ii), we get                 \[M=\frac{e\cdot nh}{2.2\pi m}=\left( \frac{e}{2m} \right)\frac{nh}{2\pi }\]