The excitation energy of an electron from second orbit to third orbit of a hydrogen-like atom or ion — Atomic Structure Chemistry Question

💡 Solution & Explanation

$\Delta E = 13.6 Z^2 (\frac{1}{2^2} - \frac{1}{3^2}) = 13.6 Z^2 (\frac{5}{36})$. Setting this to 47.2 gives $13.6 Z^2 (\frac{5}{36}) = 47.2 \implies Z^2 = \frac{47.2 \times 36}{13.6 \times 5} = \frac{1699.2}{68} = 24.98 \approx 25$. Thus $Z=5$. Therefore, correct answer is B.