The minimum energy of emitted photon is — Atomic Structure Chemistry Question

💡 Solution & Explanation

The minimum energy emitted from transitions between $n=4,3,2,1$ is the transition $4 \to 3$. $\Delta E = I.E. \times (\frac{1}{9} - \frac{1}{16}) = 14.4 \times (\frac{7}{144}) = 0.1 \times 7 = 0.7$ eV. Therefore, correct answer is D.