Question

The photon radiated from hydrogen corresponding to the second line of Lyman series is absorbed by a hydrogen-like atom $X$ in the second excited state. Then, the hydrogen-like atom $X$ makes a transition of $n^{th}$ orbit

• A. $X=H{e}^{+},n=4$
• B. $X=L{i}^{++},n=6$
• C. $X=H{e}^{+},n=6$
• D. $X=L{i}^{++},n=9$

Answer 1

The correct option is D $X=L{i}^{++},n=9$

Second line of Lyman series is obtained by transition from $n=3$ to $n=1$.
Energy obtained is $\frac{8\times 13.6}{9}$
Checking all the options, for D:
${Li}^{++}$ from $n=3$ to $n=9$
Energy required is same $=\frac{8\times 13.6}{9}$