Question

A hydrogen sample is prepared in a particular excited state $A$. Photons of energy $2.55eV$ gets absorbed in the sample to take some electrons to a further excited state $B$. What are the quantum numbers of states $A$ and $B$?

• A. 1 and 2
• B. 2 and 3
• C. 2 and 4
• D. 3 and 4

Answer 1

The correct option is C 2 and 4

$E=2.55eV$
$2.55=13.6\times 1^2[\frac{1}{n_A^2}-\frac{1}{n_B^2}]$

$\frac{n_A^2{n}_B^2}{n_A^2-n_B^2}=5.33$

Checking values from options given in above question,
we get, $n_A=2$ and $n_B=4$