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.,
$n' = \left(n + 1\right)$
As magnetic moment $M_{n} = I_{n}A = i_{n}\left(\pi r^{2}_{n}\right)$
$i_{n} = eV_{n} = \frac{mz^{2}e^{5}}{4\varepsilon^{2}_{0}n^{3}h^{3}}$
$r_{n} = \frac{n^{2}h^{2}}{4\pi^{2}kzme^{2}}\left(k = \frac{1}{4\pi\in_{0}}\right)$
Solving we get magnetic moment of the hydrogen atom for $n^{th}$ excited state
$M_{n'} = \left(\frac{e}{2m}\right) \frac{nh}{2\pi}$