Q 12-12-034JEE MainJEE Main 2025 (22 Jan, Shift 2)Medium
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
Answer: (A) $\sqrt{\dfrac{h}{\pi eB}}$
In the magnetic field, $r = \dfrac{mv}{eB}$, so $mv = eBr$.
Bohr's quantization: $mvr = \dfrac{nh}{2\pi}$
$$eBr^2 = \frac{nh}{2\pi} \Rightarrow r = \sqrt{\frac{nh}{2\pi eB}}$$
First excited state, $n = 2$:
$$r = \sqrt{\frac{2h}{2\pi eB}} = \sqrt{\frac{h}{\pi eB}}$$
Solution by Sreeraj P, M.Sc Physics