Question

A hydrogen-like atom $( Z )$ is in a higher excited state of quantum number $n$. The excited atom can make a transition to the first excited state by successively emitting two photons of energies $10.20 eV$ and $17.00 eV$ 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 $4.25 eV$ and $5.95 eV$, respectively. Determine the values of $n$ and $Z$. (Ionization energy of hydrogen atom $=13 \cdot 6 eV$).

Answer 1

SOLUTION — First excited state, $n=2$ and second excited state, $n=3$.

Total energy emitted from given $n$ to $n=2$,

$\Delta E_{1}=(10 \cdot 20+17 \cdot 00) cV =27 \cdot 20 cV , \quad \because \quad E_{n}=-13 \cdot 6 Z^{2} / n^{2}$

Hence $\Delta E_{1}=13 \cdot 6 Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{n^{2}}\right)$

Again, total energy emitted from given $n$ to $n=3$,

$\begin{array}{l}\Delta E_{2}=(4 \cdot 25+5 \cdot 95) cV \\\Delta E_{2}=13 \cdot 6 Z^{2}\left(\frac{1}{9}-\frac{1}{n^{2}}\right)\end{array}$

Pulting the values of $\Delta E_{1}$ and $E_{2}$ in (I) and (II), we have

$27 \cdot 2=13 \cdot 6 Z^{2}\left(\frac{1}{4}-\frac{1}{n^{2}}\right) \text { and } 10 \cdot 2=13 \cdot 6 Z^{2}\left(\frac{1}{9}-\frac{1}{n^{2}}\right)$

From these two relations $\frac{27 \cdot 2}{10 \cdot 2}=\frac{n^{2}-4}{4} \times \frac{9}{n^{2}-9}=\frac{8}{3}$ 

or, $\frac{32}{27}=\frac{n^{2}-4}{n^{2}-9}$

or, $\frac{5}{27}=\frac{5}{n^{2}-9}$

or, $n^{2}-9=27$,

or $n^{2}=36 \therefore n=6$

Again, $\quad 27 \cdot 2=13 \cdot 6 Z^{2}\left(\frac{1}{4}-\frac{1}{n^{2}}\right)^{n^{2}-9}$

or, $\quad 2=Z^{2}\left(\frac{1}{4}-\frac{1}{36}\right) \quad=Z^{2} \times \frac{8}{36}$

$\therefore \quad Z^{2}=\frac{36 \times 2}{8}=9 \quad \therefore \quad Z=3 .$