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. $\left(\frac{e}{2 m} \frac{n^{2} h}{2 \pi}\right)$
• B. $\left(\frac{e}{m}\right) \frac{n h}{2 \pi}$
• C. $\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$
• 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}=e f\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{e v R}{2} \ldots . $ (i)
Now, using Bohr's principle,
Angular momentum, $L=m v R=\frac{n h}{2 \pi}$
$\Rightarrow v R=\frac{n h}{2 \pi m} \ldots (ii) $
From Eqs. (i) and (ii), we get
$M=\frac{e}{2} \times\left(\frac{n h}{2 \pi m}\right)$
$=\left(\frac{e}{2 m}\right) \times\left(\frac{n h}{2 \pi}\right)$