JEE Mains · Physics · STD 12 - 12. atoms
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 :
- A \(\sqrt{\frac{\mathrm{h}}{\pi \mathrm{eB}}}\)
- B \(\sqrt{\frac{2 h}{\pi e B}}\)
- C \(\sqrt{\frac{h}{2 \pi e B}}\)
- D \(\sqrt{\frac{4 \mathrm{~h}}{\pi e B}}\)
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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