Question

A hydrogen like species (atomic number $Z$) is present in a higher excited state of quantum number $n$. This excited atom can make a transition to the first excited state by successive emission of two photons of energies $10.20eV$ and $17.0eV$ respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successive emission of two photons of energy $4.25eV$ and $5.95eV$ respectively. Determine the value of $Z$.
$(Given:\frac{27.2}{10.2}=2.66,\frac{60}{1.66}\approx 36,hc{R}_H=13.6eV)$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

Total energy emitted by photo-electron
$=10.2+17=27.20eV$
Since , $E_1$ = Photon of energy emitted through the transition
$n=n$ to $n=2\Rightarrow \frac{hc}{{\lambda }_1}=27.20eV$
We have $\frac{1}{{\lambda }_1}=R_H.Z^2(\frac{1}{2^2}-\frac{1}{n^2})$
or $\frac{hc}{{\lambda }_1}=\left(hc\right)R_H.Z^2(\frac{1}{2^2}-\frac{1}{n^2})$
$\therefore 27.20eV=\left(hc\right)R_H{Z}^2(\frac{1}{4}-\frac{1}{n^2})....\left(1\right)$
Similarly, total energy liberated during transition of electron from $n=nton=3$ is
$E_2=\frac{hc}{{\lambda }_2}=(4.25+5.95)=10.20eV$
$\therefore 10.20eV=\left(hc\right)R_H{Z}^2(\frac{1}{9}-\frac{1}{n^2})...\left(2\right)$
Dividing Eq. $\left(1\right)$ by $\left(2\right)$, we get
$\frac{27.2}{10.2}=\frac{n^2-4}{4n^2}\times \frac{9n^2}{n^2-9}\Rightarrow 2.66=\frac{n^2-4}{4}\times \frac{9}{n^2-9}\Rightarrow 10.66n^2-96=9n^2-36\Rightarrow 1.66n^2=60\Rightarrow n^2\approx 36\Rightarrow n=6$
$\therefore n=6$
Putting $n=6$ in Eq . $\left(1\right)$ or $\left(2\right)$, we get, $Z=3$.