Physics · Atomic Physics

NEET (UG) 2025 — Question 17

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)

Option A:
$\frac{h e}{\pi m}$
Option B:
$\frac{h e}{2 \pi m}$
Correct
Option C:
$\frac{h e B}{\pi m}$
Option D:
$\frac{h e B}{2 \pi m}$

Answer: B

Step-by-step solution

Magnetic moment

$\mathrm{M}=\mathrm{IA}=\mathrm{I}\left(\pi \mathrm{r}^{2}\right)$

M=\left(\frac{e u}{2 \pi r}\right)\left(\pi r^{2}\right) …(1)

Given $B\left(\pi r^{2}\right)=n\left(\frac{h}{e}\right)$

$\Rightarrow \mathrm{r}^{2}=\frac{\mathrm{h}}{\mathrm{B} \pi e} \ldots(2)$

$(\because \mathrm{n}=1)$

And when charge moving in external magnetic field then $r=\frac{m u}{q B}$

$\frac{\mathrm{v}}{\mathrm{r}}=\frac{\mathrm{eB}}{\mathrm{m}} \ldots(3)$

$(\because \mathrm{q}=\mathrm{e})$

Put value from equation (2) and (3) in equation (1)

$M=\left(\frac{e v}{2 \pi r}\right)\left(\pi r^{2}\right)$

$\mathrm{M}=\frac{e}{2 \pi}\left(\frac{e \mathrm{~B}}{\mathrm{~m}}\right) \pi\left(\frac{\mathrm{h}}{\mathrm{B} \pi e}\right)$

$M=\frac{e h}{2 \pi m}$

Answer key and solution verified before publishing.

Practise Atomic Physics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: NEET (UG) 2025
Paper: 2025 paper
Subject: Physics
Chapter: Atomic Physics
Topic: Dual Nature of Matter

Answer: B

Step-by-step solution

Practise Atomic Physics

More Atomic Physics questions from this paper