Question

A hydrogen-like atom of atomic number $Z$ is in an excited state of quantum number $2n$ . It can emit a maximum energy photon of $204 \, eV$ . If it makes a transition to the quantum state $n$ , a photon of energy $40.8 \, eV \, $ is emitted. Calculate the atomic number $Z$ . Ground state energy of hydrogen atom is $-13.6 \, eV$

Answer 1

Let ground state energy (in eV) be E₁
Then, from the given condition
$E_{2 n}- \, E_{1}= \, 204 \, eV$
Or $\frac{\text{E}_{1}}{4 \text{n}^{2}} - \text{E}_{1} = \text{204 eV}$
Or $E _{1}\left(\frac{1}{4 n ^{2}}-1\right)=204 eV \ldots$ (i)
And $E_{2 n}-E_{n}=40.8 eV$
Or $\frac{E_{1}}{4 n^{2}}-\frac{E_{1}}{n^{2}}=40.8 e V$
$orE _{1}=\left(\frac{-3}{4 n ^{2}}\right)=40.8 eV \ldots .$ (ii)
From Eqs. (i) and (ii),
$\frac{1-\frac{1}{4 n}}{\frac{3}{4 n^{2}}}=5$
Or $1=\frac{1}{4 n^{2}}+\frac{15}{4 n^{2}}$
Or $\frac{4}{n^{2}}=1$
or $n=2$
From Eq. (ii),
$E _{1}=-\frac{4}{3} n ^{2}(40.8) eV$
$=-\frac{4}{3}(2)^{2}(40.8) eV$
or $E_{1}=-217.6 eV$
$ E_1=-\left(\begin{array}{ll} 13.6 \end{array}\right) Z_2 $
Therefore $Z^2=\frac{E_1}{-13 \cdot 6}=\frac{-217-6}{-13 \cdot 6}=16$
Therefore $Z=4$