Bohr Model: Orbits, Speed & Angular Momentum question from MHT CET 2026 (14 May, Shift 1)

Answer: (c)

1. Magnetic force gives circular motion: $qvB = \frac{mv^2}{r} \Rightarrow r = \frac{mv}{qB}$.
2. Bohr condition: $mvr = \frac{m^2 v^2}{qB} = \frac{nh}{2\pi} \Rightarrow v^2 = \frac{nhqB}{2\pi m^2}$.
3. Magnetic field does no work, so energy is only kinetic: $E = \frac{1}{2}mv^2 = \frac{nhqB}{4\pi m}$
4. So the answer is (c) $\frac{nhqB}{4\pi m}$.