If one were to apply the Bohr… - JEE Main 2015 Physics PYQ

JEE Main 2015 — Physics Modern Physics

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If one were to apply the Bohr model to a particle of mass $'m'$ and charge $'q'$ moving in a plane under the influence of a magnetic field 'B', the energy of the charged particle in the ${n}^{th}$ level will be:

Official previous-year question

Held on 10 Apr 2015 · Verified 6 Jul 2026.

Options

A
$n(\frac{hqB}{\pi m})$
B
$n(\frac{hqB}{4\pi m})$
C
$n(\frac{hqB}{2\pi m})$
D
$n(\frac{hqB}{8\pi m})$

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Solution

For a charge $q$ moving in a uniform magnetic field $B$

JEE Main 2015 Physics, Atomic Physics — solution figure

${f}_{m}=\frac{m{v}^{2}}{R}$

$qVB= \frac{m{v}^{2}}{R}$

$\Rightarrow m{v}^{2}=qVBR$

$\Rightarrow \frac{1}{2} m{v}^{2}=\frac{qVBR}{2}$

$\Rightarrow Energy=\frac{qVBR}{2} (1)$

By Bohr's quantization condition,

Angular momentum $L=n\frac{h}{2\pi }$

$\Rightarrow mvR=\frac{nh}{2\pi }$

$\Rightarrow vR=\frac{nh}{2\pi m} (2)$

Put $(2)$ in $(1)$

$\Rightarrow Energy=\frac{qB}{2} (\frac{nh}{2 \pi m})$

$= \frac{qB nh}{4 \pi m}$

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Solution

For a charge $q$ moving in a uniform magnetic field $B$

JEE Main 2015 Physics, Atomic Physics — solution figure

${f}_{m}=\frac{m{v}^{2}}{R}$

$qVB= \frac{m{v}^{2}}{R}$

$\Rightarrow m{v}^{2}=qVBR$

$\Rightarrow \frac{1}{2} m{v}^{2}=\frac{qVBR}{2}$

$\Rightarrow Energy=\frac{qVBR}{2} (1)$

By Bohr's quantization condition,

Angular momentum $L=n\frac{h}{2\pi }$

$\Rightarrow mvR=\frac{nh}{2\pi }$

$\Rightarrow vR=\frac{nh}{2\pi m} (2)$

Put $(2)$ in $(1)$

$\Rightarrow Energy=\frac{qB}{2} (\frac{nh}{2 \pi m})$

$= \frac{qB nh}{4 \pi m}$

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics