Question 22 of 239

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:

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

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Answer: (B) $n\left(\dfrac{hqB}{4\pi m}\right)$

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



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

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

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

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

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

By Bohr's quantization condition,

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

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

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

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

$\Rightarrow Energy=\dfrac{qB}{2}\left(\dfrac{nh}{2\pi m}\right)$

$=\dfrac{qBnh}{4\pi m}$