Question

A hydrogen atom in a state having a binding energy of $0.85eV$ makes transition to a state with excitation energy $10.2eV$
Identify the quantum numbers $n$ of the upper and the lower energy states involved in the transition.

Answer 1

Initially, the electron is at $n_1=4$ shell

As, $\frac{13.6}{n^2}=B.E=0.85,n_1=4$

After deexcitation, it is at $n_2=2$ shell.

As,

Excitation Energy = $13.6-\frac{13.6}{n_2^2}=10.2eV$

So, the wavelength of emitted photon is,

$\frac{he}{\lambda }=[\frac{13.6}{n_2^2}-\frac{13.6}{n_1^2}]1.6\times {10}^{-19}$

$\lambda =\frac{6.6\times {10}^{-34}\times 3\times {10}^{-8}\times 16}{1.6\times {10}^{-19}\times 13.6\times 3}$

$\lambda =4852\mathring{A}$