Question

Consider a hydrogen-like ionized atom with atomic number $Z$ with a single electron. In the emission spectrum of this atom, the photon emitted in the $n=2 \, to \, n= \, 1$ transition has energy $74.8 \, eV$ higher than the photon emitted in the $n=3 \, to \, n=2$ transition. Given that the ionization energy of the hydrogen atom is $13.6 \, eV$ , what is the value of $Z$ ?

Answer 1

$\Delta E_{2 \rightarrow 1}=13.6\times Z^{2}\left[1 - \frac{1}{4}\right]=13.6\times Z^{2}\left[\frac{3}{4}\right]$
$\Delta E_{3 \rightarrow 2}=13.6\times Z^{2}\left[\frac{1}{4} - \frac{1}{9}\right]=13.6\times Z^{2}\left[\frac{5}{36}\right]$
$\Delta E_{2 \rightarrow 1}=\Delta E_{3 - 2}+74.8$
$13.6\times Z^{2}\left[\frac{3}{4}\right]=13.6\times Z^{2}\left[\frac{5}{36}\right]+74.8$
$13.6\times Z^{2}\left[\frac{3}{4} - \frac{5}{36}\right]=74.8$
$Z^{2}=9$
$Z=+3$