Magnetics 4 Question 17
18. An electron in the ground state of hydrogen atom is revolving in anti-clockwise direction in a circular orbit of radius $R$.
(1996, 5M)
(a) Obtain an expression for the orbital magnetic moment of the electron.
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Answer:
Correct Answer: 18. (a) $M=\frac{e h}{4 \pi m}$ (b) $\tau=\frac{e h B}{8 \pi m}$, perpendicular to both $\mathbf{M}$ and $\mathbf{B}$.
Solution:
- In ground state $(n=1)$ according to Bohr’s theory
$$ m v R=\frac{h}{2 \pi} \text { or } v=\frac{h}{2 \pi m R} $$
Now, time period,
$$ T=\frac{2 \pi R}{v}=\frac{2 \pi R}{h / 2 \pi m R}=\frac{4 \pi^{2} m R^{2}}{h} $$
Magnetic moment, $M=i A$
$$ \begin{aligned} \text { where, } i & =\frac{\text { charge }}{\text { time period }}=\frac{e}{\frac{4 \pi^{2} m R^{2}}{h}} \\ & =\frac{e h}{4 \pi^{2} m R^{2}} \text { and } A=\pi R^{2} \\ \therefore \quad M & =\left(\pi R^{2}\right) \frac{e h}{4 \pi^{2} m R^{2}} \quad \text { or } \quad M=\frac{e h}{4 \pi m} \end{aligned} $$
Direction of magnetic moment $\mathbf{M}$ is perpendicular to the plane of orbit.
(b)
$$ \tau=\mathbf{M} \times \mathbf{B} \Rightarrow|\tau|=M B \sin \theta $$
where, $\theta$ is the angle between $\mathbf{M}$ and $\mathbf{B}$
$$ \theta=30^{\circ} $$
$$ \begin{aligned} & \therefore \quad \tau=\frac{e h}{4 \pi m}(B) \sin 30^{\circ} \\ & \therefore \quad \tau=\frac{e h B}{8 \pi m} \end{aligned} $$
The direction of $\tau$ is perpendicular to both $\mathbf{M}$ and $\mathbf{B}$.
