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JEE MAIN Physics QUESTION #989
Question 1
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:
  • \(\sqrt{\dfrac{h}{\pi eB}}\)✔️
  • \(\sqrt{\dfrac{2h}{\pi eB}}\)
  • \(\sqrt{\dfrac{h}{2\pi eB}}\)
  • \(\sqrt{\dfrac{4h}{\pi eB}}\)
Correct Answer Explanation
By Bohr quantization: \(mvr = \frac{nh}{2\pi}\). For circular motion in \(B\): \(mv = eBr\). So \(eBr^2 = \frac{nh}{2\pi} \Rightarrow r = \sqrt{\frac{nh}{2\pi eB}}\). First excited state: \(n=2\) gives \(r=\sqrt{\frac{2h}{2\pi eB}}=\sqrt{\frac{h}{\pi eB}}\).