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. The ionization energy of the hydrogen atom is $13.6$ eV. The value of Z is

Answer 1

Energy of $n_{th}$ shell is
$E_n=13.6\frac{Z^2}{n^2}$
$△E_{2-1}=13.6\times z^2[1-\frac{1}{4}]=13.6\times z^2\left[\frac{3}{4}\right]$

$△E_{3-2}=13.6\times z^2[\frac{1}{4}-\frac{1}{9}]=13.6\times z^2\left[\frac{5}{36}\right]$

Given $△E_{2-1}=△E_{3-2}+74.8$

$⟹13.6\times z^2\left[\frac{3}{4}\right]=13.6\times z^2\left[\frac{5}{36}\right]+74.8$
$=13.6\times z^2[\frac{3}{4}-\frac{5}{36}]=74.8$
$z^2=9$

$z=+3$ (z can't be - ve)