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In the Bohr model an electron 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 $n$th excited state. (Assume, $h=$ Planck's constant)
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The correct answer is:
$\left(\frac{e}{2 m}\right) \frac{n h}{2 \pi}$
If $R$ be the radius of circular path, then magnetic moment,
$M=i \times A=(e \times f) \times\left(\pi R^2\right) \quad\left(\because i=\frac{e}{T}=e f\right)$
$=e \times\left(\frac{v}{2 \pi R}\right) \times\left(\pi R^2\right)$
$\left[\because \omega=2 \pi f \Rightarrow \frac{V}{R}=2 \pi f \Rightarrow f=\frac{V}{2 \pi R}\right]$
$\Rightarrow \quad M=\frac{e v R}{2}$ $\ldots$ (i)
Now, using Bohr's principle,
Angular momentum, $L=m v R=\frac{n h}{2 \pi}$
$\Rightarrow \quad v R=\frac{n h}{2 \pi m}$ $\ldots$ (ii)
From Eqs. (i) and (ii), we get
$M=\frac{e}{2} \times\left(\frac{n h}{2 \pi m}\right)=\left(\frac{e}{2 m}\right) \times\left(\frac{n h}{2 \pi}\right)$
$M=i \times A=(e \times f) \times\left(\pi R^2\right) \quad\left(\because i=\frac{e}{T}=e f\right)$
$=e \times\left(\frac{v}{2 \pi R}\right) \times\left(\pi R^2\right)$
$\left[\because \omega=2 \pi f \Rightarrow \frac{V}{R}=2 \pi f \Rightarrow f=\frac{V}{2 \pi R}\right]$
$\Rightarrow \quad M=\frac{e v R}{2}$ $\ldots$ (i)
Now, using Bohr's principle,
Angular momentum, $L=m v R=\frac{n h}{2 \pi}$
$\Rightarrow \quad v R=\frac{n h}{2 \pi m}$ $\ldots$ (ii)
From Eqs. (i) and (ii), we get
$M=\frac{e}{2} \times\left(\frac{n h}{2 \pi m}\right)=\left(\frac{e}{2 m}\right) \times\left(\frac{n h}{2 \pi}\right)$
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