Question

A hydrogen like atom of atomic number $Z$ is in an excited state of quantum number $2n$. It can emit a photon of maximum energy 204 $eV$. If it makes a transition to quantum state $n$, a photon of energy 40.8 $eV$ is emitted. The value of $n$ will be.

Answer 1

Let ground state energy (in $eV$) be $E_1$
Then from the given condition
$E_{2n}-E_1=204eV$
As $E_1=-13.6Z^2$
Therefore, $\frac{E_1}{4n^2}-E_1=204eV$
$\Rightarrow E_1(\frac{1}{4n^2}-1)=204eV$ ......(i)
and $E_{2n}-E_n=40.8eV$
$\Rightarrow \frac{E_1}{4n^2}-\frac{E_1}{n^2}=E_1(-\frac{3}{4n^2})=40.8eV$ ....(ii)
Divide equation (i) by (ii), $\frac{\frac{1}{4n^2}-1}{-\frac{3}{4n^2}}=5\Rightarrow n=2$