Question

A hydrogen-like atom (atomic number $Z$) is in a higher excited state of quantum number $n$. 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

Answer: (C)

$\text{6 and 3}$

Given, $E_{n⟶2}=E_n-E_2=10.2+17=27.2eV$
and $E_{n⟶3}=E_n-E_3=4.25+5.95=10.2eV$
Now, from above two equations:
$E_{3\to 2}=E_3-E_2=27.2-10.2=17.0eV$
$E_{3\to 2}=E_3-E_2=27.2-10.2=17.0eV\Rightarrow \frac{13.6Z^2}{4}-\frac{13.6Z^2}{9}=\frac{5}{36}13.6Z^2\therefore Z^2=\frac{17\times 36}{5\times 13.6}=9\Rightarrow Z=3$

Now, considering $E_{n\to 2}=E_n-E_2=10.2+17=27.2eV$
$\Rightarrow \frac{13.6\times 9}{4}-\frac{13.6\times 9}{n^2}=27.2\Rightarrow n=6$