Q. 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) $\sqrt{\frac h{\pi eB}}$
B) $\sqrt{\frac{2h}{\pi eB}}$
C) $\sqrt{\frac h{2\pi eB}}$
D) $\sqrt{\frac{4h}{\pi eB}}$
Explanation
[Solution by Dr. Abubacker] $r=\frac{mv}{eB}$, so $mvr=eBr^2$. Bohr's rule: $mvr=\frac{nh}{2\pi}$, so $r^2=\frac{nh}{2\pi eB}$. For the first excited state, $n=2$: $r=\sqrt{\frac h{\pi eB}}$. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.