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 $Z$.

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (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, $\frac{1}{{\lambda }_1}=R_H.Z^2\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$
or $\frac{hc}{{\lambda }_1}=\left(hc\right)R_H.Z^2\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$
$\therefore 27.20=\left(hc\right)R_H.Z^2\left(\frac{1}{2^2}-\frac{1}{4}\right)\cdots \left(1\right)$
Similarly, total energy liberated during transition of electron from $n=n$ to $n=3$ is
$E_2=\frac{hc}{{\lambda }_2}=(4.25+5.95)=10.20eV$
$\therefore 10.20=\left(hc\right)R_H{Z}^2\left(\frac{1}{9}-\frac{1}{n^2}\right)\cdots \left(2\right)$
Dividing Eq.$\left(1\right)$ by $\left(2\right)$, we get $n=6$ and putting $n=6$ in Eq.$\left(1\right)$ or $\left(2\right)$, we get. $Z=3$