JEE MAIN - Physics (2025 - 22nd January Evening Shift - No. 6)
An electron projected perpendicular to a uniform magnetic field B moves in a circle. If Bohr's quantization is applicable, then the radius of the electronic orbit in the first excited state is :
$\sqrt{\frac{h}{\pi e B}}$
$\sqrt{\frac{4 h}{\pi e B}}$
$\sqrt{\frac{\mathrm{h}}{2 \pi e \mathrm{eB}}}$
$\sqrt{\frac{2 h}{\pi e B}}$
Explanation
$$\begin{aligned}
&\begin{aligned}
& \mathrm{r}=\frac{\mathrm{mv}}{\mathrm{eB}} \& \mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi} \Rightarrow(\mathrm{eBr}) \mathrm{r}=\frac{\mathrm{nh}}{2 \pi} \\
& \Rightarrow \mathrm{r}=\sqrt{\frac{\mathrm{nh}}{2 \pi \mathrm{eB}}}
\end{aligned}\\
&\text { first excited state } \mathrm{n}=2 \therefore \mathrm{r}=\sqrt{\frac{\mathrm{h}}{\pi \mathrm{eB}}}
\end{aligned}$$
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