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 $10.20\text{eV}$ and $17.0\text{eV}$ 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 $4.25\text{eV}$ and $5.95\text{eV}$ respectively. Determine the value of $n$.

• A. $3$
• B. $2$
• C. $6$
• D. $4$

Answer 1

The correct option is C $6$

Total energy emitted by photo-electron during transition from $n=n$ to $n=2$,
$=(10.2+17)\text{eV}=27.20\text{eV}$
$\therefore E_1=\frac{hc}{{\lambda }_1}=27.20eV$
We have,
$\frac{1}{{\lambda }_1}=R_h{Z}^2(\frac{1}{2^2}-\frac{1}{n^2})$
$\Rightarrow E_1=\frac{hc}{{\lambda }_1}=\left(hc\right)R_h{Z}^2(\frac{1}{2^2}-\frac{1}{n^2})$
$\therefore 27.20=\left(\text{hc}\right)R_h{Z}^2(\frac{1}{4}-\frac{1}{n^2}).....\left(1\right)$
Similarly, total energy liberated during transition of electron from $n=n$ to $n=3$ is,
$E_2=\frac{hc}{{\lambda }_2}=(4.25+5.95)\text{eV}=10.20\text{eV}$
$\Rightarrow 10.20=\left(hc\right)R_h{Z}^2(\frac{1}{9}-\frac{1}{n^2}).....\left(2\right)$
Dividing Equation $\left(1\right)$ by $\left(2\right)$,
$\frac{27.20}{10.20}=\frac{4-n^2}{4n^2}\times \frac{9n^2}{9-n^2}$
$\Rightarrow 36\times 27.20-36\times 10.20=4\times 27.20\times n^2-9\times 10.20\times n^2$
$\Rightarrow 36(27.20-10.20)=17n^2$
$\Rightarrow n^2=36$
$\Rightarrow n=6$