Question

The photon radiated from hydrogen corresponding to $2^{\text {nd }}$ line of Lyman series is absorbed by a hydrogen like atom ' $X$ ' in $2^{\text {nd }}$ excited state. As a result the hydrogen like atom ' $X$ makes a transition to $n^{\text {th }}$ orbit. Then,

• A. $X= He ^{+}, n=4$
• B. $X= Li ^{++}, n=6$
• C. $X= He ^{+}, n=6$
• D. $X=Li ^{+4}, n=9$

Answer 1

Answer: (D)

Energy of $n^{\text {th }}$ sate in Hydrogen is same as energy of $3 n^{\text {th }}$ state in $Li ^{++}$.
$\Rightarrow$ For $3 \rightarrow 1$ transition in $H$ would it give same energy as the $3 \times 3 \rightarrow 1 \times 3$ transition in $Li ^{++}$.