Question

Consider a hydrogen like atom whose energy in $n^{th}$ excited state is given by $E_n=-\frac{13.6}{n^2}{Z}^2$. When this excited atom makes a transition from an excited state to ground state. The most energetic photons have energy $E_{max}=52.224eV$ and the least energetic photons have energy $E_{min}=1.224eV$. Find the atomic number of atom.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

The correct option is B 2

The most energetic photons have energy $E_{max}=52.224eV$.

This corresponds to transition from infinity to ground state.

$E_{max}=52.224eV=[\frac{13.6}{1^2}{Z}^2-\frac{13.6}{{\infty }^2}{Z}^2]$

$52.224eV=13.6Z^{2}eV$

$Z^2=\frac{52.224}{13.6}=3.84\simeq 4$

$Z=2$