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 $$\frac{\lambda_N}{\lambda_A}$$ is closest to:

We recall the photon-energy relation $$E = \frac{hc}{\lambda}$$, where $$h$$ is Planck’s constant and $$c$$ is the speed of light. Thus, for two photons, the ratio of their wavelengths equals the inverse ratio of the corresponding energy gaps:

$$\frac{\lambda_N}{\lambda_A} \;=\; \frac{E_A}{E_N}.$$

Now we estimate the typical magnitudes of the first‐excited-state energy gaps for an atom and for a nucleus.

For an atom in its first excited state the gap is normally of the order of a few electron-volts; we may write

$$E_A \approx 1\;{\rm eV}\;{\rm to}\;10\;{\rm eV}.$$

For a nucleus, the first excited level lies in the mega-electron-volt range; we write

$$E_N \approx 1\;{\rm MeV}\;=\;10^{6}\;{\rm eV}\;{\rm to}\;{\rm a\;few}\times10^{6}\;{\rm eV}.$$

Taking representative values, let us set $$E_A \sim 1\;{\rm eV}$$ and $$E_N \sim 10^{6}\;{\rm eV}$$. Substituting these into the wavelength ratio gives

$$\frac{\lambda_N}{\lambda_A} \;=\; \frac{E_A}{E_N} \;\approx\; \frac{1\;{\rm eV}}{10^{6}\;{\rm eV}} \;=\; 10^{-6}.$$

Hence, the ratio is nearest to $$10^{-6}$$.

Hence, the correct answer is Option B.