Question

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

• A. $2$
• B. $4$
• C. $5$
• D. None of these

Answer 1

The correct option is A $2$

Maximum energy is liberated for transition $E_n\to 1$ and minimum energy for $E_n\to E_{n-1}$
Hence, $\frac{E_1}{n^2}-E_1=52.224eV...\left(i\right)$
and $\frac{E_1}{n^2}-\frac{E_1}{{(n-1)}^2}=1.224eV...\left(ii\right)$
Solving Eqs. (i) and (ii) we get
$E_1=-54.4eV;$
and $n=5$
But $E_1=-\frac{13.6Z^2}{1^2}$
$\Rightarrow Z=2$