Question

H-atom is exposed to electromagnetic radiation of 1026 Å and gives out induced radiations (radiations emitted when e returns to ground state). Calculate $\lambda$ (in Å) of induced radiations.

• A. 1026
• B. 1216
• C. 656
• D. 100

Answer 1

Answer: (B)

1216

Answer 2

The incident photon energy is $E_{incident} = \frac{hc}{\lambda_{incident}} \approx \frac{12400 \text{ eV-Å}}{1026 \text{ Å}} \approx 12.086 \text{ eV}$. This excites the electron from $n=1$ to $n=3$ ($E_3 \approx -1.511$ eV). Induced radiations ending at the ground state ($n=1$) are from transitions $3 \to 1$ ($\lambda \approx 1026$ Å) and $2 \to 1$ ($\lambda \approx 1216$ Å). The question asks for the induced radiation, implying a new emission, hence $\lambda \approx 1216$ Å (Lyman-alpha).