Question

A hydrogen-like atom (atomic number $Z$) is in a higher excited state of quantum number ${}^{\prime }n{}^{\prime }$. This excited atom can make a transition to the first excited state by successively emitting two photons of energies $10.2eV$ and $17.0eV$, respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies $4.25eV$ and $5.95eV$, respectively. The values of $n$ and $Z$ are, respectively

• A. $\text{6 and 6}$
• B. $\text{3 and 3}$
• C. $\text{6 and 3}$
• D. $\text{3 and 6}$

Answer 1

The correct option is C $\text{6 and 3}$

We know that, $\Delta E=13.6Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]eV$

In first case, $(10.2+17.0)=13.6Z^2[\frac{1}{2^2}-\frac{1}{n^2}]...\left(1\right)$

In second case, $(4.25+5.95)=13.6Z^2[\frac{1}{3^2}-\frac{1}{n^2}]...\left(2\right)$

$\left(1\right)/\left(2\right)\Rightarrow \frac{27.2}{10.2}=\frac{9(n^2-4)}{4(n^2-9)}$

$\Rightarrow 1.18=\frac{(n^2-4)}{(n^2-9)}$

$\Rightarrow n^2-4=1.18n^2-10.66$

$\Rightarrow 0.18n^2=6.66$ or $n^2\sim 36$

$\Rightarrow n=6$

Put the value of $n$ in (1), we get $Z=3$