Question

lii the Bohr model an election of mass $m$ 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 $n^{\text{th}}$ excited state, is:
(Assume$h=$ Planck's constant)

• 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)

If $R$ be the radius of circular path, then magnetic moment,
$M=i\times A=(e\times f)\times \left(\pi R^2\right)$
$\left(\because i=\frac{e}{T}=ef\right)$
$=e\times \left(\frac{v}{2\pi R}\right)\times \left(\pi R^2\right)$
$\left[\because \omega =2\pi f\Rightarrow \frac{V}{R}=2\pi f\Rightarrow f=\frac{V}{2\pi R}\right]$
$\Rightarrow M=\frac{evR}{2}\dots .$ (i)
Now, using Bohr's principle,
Angular momentum, $L=mvR=\frac{nh}{2\pi }$
$\Rightarrow vR=\frac{nh}{2\pi m}\dots (ii)$
From Eqs. (i) and (ii), we get
$M=\frac{e}{2}\times \left(\frac{nh}{2\pi m}\right)$
$=\left(\frac{e}{2m}\right)\times \left(\frac{nh}{2\pi }\right)$