Question

A photon of energy $12.75eV$ is completely absorbed by a hydrogen atom initially in the ground state. The principle quantum number of the excited state is:

• A. $1$
• B. $3$
• C. $4$
• D. $\infty$

Answer 1

Answer: (C)

$4$

The energy of an electron in nth orbit of H atom is $E=\frac{-13.6}{n^2}$ eV/atom.

For the transition from the ground state to the nth state, the energy difference is

$\Delta E=-13.6(\frac{1}{1^2}-\frac{1}{n^2})$

But this corresponds to 12.75 eV.

Hence, $-12.75=-13.6(\frac{1}{1^2}-\frac{1}{n^2})$

$0.9375=(\frac{1}{1^2}-\frac{1}{n^2})=1-\frac{1}{n^2}$

Thus, $n=4$

Hence, the correct option is $C$