Q103 / 180 · NEET 2025 · Physics
A model for quantized motion of an electron in a uniform magnetic field $B$ states that the flux passing through the orbit of the electron is $n(h / e)$ where $n$ is an integer, $h$ is Planck's constant and $e$ is the magnitude of electron's charge. According to the model, the magnetic moment of an electron in its lowest energy state will be ( $m$ is the mass of the electron)
- A$\frac{h e}{\pi m}$
- Correct answerB$\frac{h e}{2 \pi m}$
- C$\frac{h e B}{\pi m}$
- D$\frac{h e B}{2 \pi m}$
Solution
Magnetic moment
$\begin{aligned}
& M=I A=I\left(\pi r^2\right) \\
& M=\left(\frac{e \mathrm{u}}{2 \pi r}\right)\left(\pi r^2\right)
\end{aligned}$
Given $B\left(\pi r^2\right)=n\left(\frac{h}{e}\right)$
$\Rightarrow \mathrm{r}^2=\frac{\mathrm{h}}{\mathrm{~B} \pi \mathrm{e}} \quad \ldots(2) \quad(\because \mathrm{n}=1)$
And when charge moving in external magnetic field then $r=\frac{m u}{q B}$
$\frac{\mathrm{v}}{\mathrm{r}}=\frac{e \mathrm{~B}}{\mathrm{~m}} \ldots(3) \quad(\because \mathrm{q}=e)$
Put value from equation (2) and (3) in equation (1)
$\begin{aligned}
& M=\left(\frac{e v}{2 \pi r}\right)\left(\pi r^2\right) \\
& M=\frac{e}{2 \pi}\left(\frac{e B}{m}\right) \pi\left(\frac{h}{B \pi e}\right) \\
& M=\frac{e h}{2 \pi m}
\end{aligned}$
Hence, the correct answer is option: 'B'.