Question

If the binding energy of $2^{nd}$ excited state of a hydrogen-like sample is 24 eV approximately, then the atomic number Z of the H-like species is:

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is B $4$

Given, the binding energy of $2^{nd}$ excited state $=24$ eV.
For hydrogen like species, the expression of energy is written as:

$E_n=E_1\times \frac{Z^2}{n^2}$

Where $E_1=13.6eV$

For $2^{nd}$ excited state, $n=3$.

$24=13.6\times \frac{Z^2}{3^2}$

$Z^2=\frac{9\times 24}{13.6}=15.8824$

$\Rightarrow Z=3.9853=4$.