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 :

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Answer & Solution

Correct Answer

(A) \(\sqrt{\frac{\mathrm{h}}{\pi \mathrm{eB}}}\)

Step-by-step Solution

Detailed explanation

\(m v r=\frac{n h}{2 \pi}\) ...(i)
\(r=\frac{v m}{B q}\) ....(ii)
\(\begin{aligned} & n=2 \\ & m r\left(\frac{r B q}{m}\right)=\frac{2 h}{2 \pi} \\ & r=\sqrt{\frac{h}{\pi B q}} \\ & q=e \\ & r=\sqrt{\frac{h}{\pi B e}}\end{aligned}\)

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