When an electron jumps from the second orbit to fourth orbit, its distance from nucleus increases by — Atomic Structure Chemistry Question

💡 Solution & Explanation

Change in radius $\Delta r = r_4 - r_2 = 0.529 \frac{4^2 - 2^2}{Z} = 0.529 \frac{12}{Z}$. Given $\Delta r = 2.116$ Å. So $0.529 \times 12 / Z = 2.116 \implies 6.348 / Z = 2.116 \implies Z = 3$. This corresponds to Li$^{2+}$. Therefore, correct answer is C.