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Bohr model is applied to a particle of mass ' $\mathrm{m}$ ' and charge ' $\mathrm{q}$ ' moving in a plane under the influence of a transverse magnetic field ' $\mathrm{B}$ '. The energy of the charged particle in the $\mathrm{n}^{\text {th }}$ level will be $[\mathrm{h}=$ Planck's constant $]$
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Verified Answer
The correct answer is:
$\frac{\mathrm{nhqB}}{4 \pi \mathrm{m}}$
We know,
$\begin{aligned}
\mathrm{mvr} & =\frac{\mathrm{nh}}{2 \pi} \\
\therefore \quad \mathrm{vr} & =\frac{\mathrm{nh}}{2 \pi \mathrm{m}}... (i)
\end{aligned}$
Also,
$\begin{aligned}
\mathrm{qvB} & =\frac{\mathrm{mv}^2}{\mathrm{r}} \\
\therefore \quad \mathrm{mv} & =\mathrm{qBr} ... (ii)\\
& \mathrm{mv}^2 \mathrm{r}=\mathrm{qBr} \times \frac{\mathrm{nh}}{2 \pi \mathrm{m}} \quad \ldots . . .(\text { Multiplying (i) with (ii)) } \\
\mathrm{E} & =\frac{1}{2} \mathrm{mv}^2=\mathrm{n}\left[\frac{\mathrm{qBh}}{4 \pi \mathrm{m}}\right]
\end{aligned}$
$\begin{aligned}
\mathrm{mvr} & =\frac{\mathrm{nh}}{2 \pi} \\
\therefore \quad \mathrm{vr} & =\frac{\mathrm{nh}}{2 \pi \mathrm{m}}... (i)
\end{aligned}$
Also,
$\begin{aligned}
\mathrm{qvB} & =\frac{\mathrm{mv}^2}{\mathrm{r}} \\
\therefore \quad \mathrm{mv} & =\mathrm{qBr} ... (ii)\\
& \mathrm{mv}^2 \mathrm{r}=\mathrm{qBr} \times \frac{\mathrm{nh}}{2 \pi \mathrm{m}} \quad \ldots . . .(\text { Multiplying (i) with (ii)) } \\
\mathrm{E} & =\frac{1}{2} \mathrm{mv}^2=\mathrm{n}\left[\frac{\mathrm{qBh}}{4 \pi \mathrm{m}}\right]
\end{aligned}$
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