Other / Mixed: Atoms & Bohr Links question from JEE Main 2019 (12 Jan, Shift 2)

Answer: (b)

1. Energy given to a mercury atom $= 5.6 - 0.7 = 4.9\text{ eV}$.
2. The atom emits this as a photon: $\lambda = \frac{1240}{4.9}\text{ nm}$
3. $\lambda \approx 253\text{ nm} \approx 250\text{ nm}$
4. So the answer is (b) 250 nm.