Question

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 $n^{th}$ excited state, is :

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

Answer 1

Answer: (C)

Magnetic moment of the hydrogen atom, when the electron is in $n^{th}$ excited state, i.e.,
$n^{\prime }=\left(n+1\right)$
As magnetic moment $M_n=I_{n}A=i_n\left(\pi r_n^2\right)$
$i_n=e{V}_n=\frac{m{z}^2{e}^5}{4{\epsilon }_0^2{n}^3{h}^3}$
$r_n=\frac{n^2{h}^2}{4{\pi }^{2}kzm{e}^2}\left(k=\frac{1}{4\pi {\in }_0}\right)$
Solving we get magnetic moment of the hydrogen atom for $n^{th}$ excited state
$M_{n^{\prime }}=\left(\frac{e}{2m}\right)\frac{nh}{2\pi }$