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 $n=2ton=1$ transition has energy $74.8eV$ higher than the photon emitted in the $n=3ton=2$ transition. Given that the ionization energy of the hydrogen atom is $13.6eV$ , what is the value of $Z$ ?

Answer 1

Answer: 3

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