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The photon radiated from hydrogen corresponding to $2^{n d}$ line of Lyman series is absorbed by a hydrogen like atom ' $X$ ' in $2^{n d}$ excited state. As a result, the hydrogen like atom ' $X$ ' excites to $n^{t h}$ orbit. Then which of the following can be correct?

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$X = H e^{+}, n = 4$

$X = L i^{+ +}, n = 6$

$X = L i^{+ +}, n = 9$

$X = H e^{+}, n = 6$

answer is D.

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Detailed Solution

$13.6 [\frac{1}{1^{2}} - \frac{1}{3^{2}}] = 13.6 \times z^{2} [\frac{1}{3^{2}} - \frac{1}{n^{2}}]$

from the options $X = L i^{+ +}, n = 9$ satisfies the relation.

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