Q 12-13-132JEE MainJEE Main 2018 (16 Apr, Shift 1)Medium
Both the nucleus and the atom of some element are in their respective first excited states. They get de-excited by emitting photons of wavelengths $\lambda_N$, $\lambda_A$ respectively. The ratio $\dfrac{\lambda_N}{\lambda_A}$ is closest to:
Answer: (B) $10^{-6}$
Photon energy $E = \dfrac{hc}{\lambda}$, so $\dfrac{\lambda_N}{\lambda_A} = \dfrac{E_A}{E_N}$.
Atomic excitation energies are of the order of eV, while nuclear excitation energies are of the order of MeV:
$$\frac{\lambda_N}{\lambda_A} \sim \frac{1\ \text{eV}}{1\ \text{MeV}} = 10^{-6}$$
Solution by Sreeraj P, M.Sc Physics