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

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Solution

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

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

Given $△ E_{2 - 1} = △ E_{3 - 2} + 74.8$

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

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

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Energy of nth shell is En=13.6Z2n2 △E2−1=13.6×z2[1−14] =13.6×z2[34] △E3−2=13.6×z2[14−19] =13.6×z2[536] Given △E2−1=△E3−2+74.8 ⟹13.6×z2[34]=13.6×z2[536]+74.8 = 13.6×z2[34−536]=74.8 z2=9 z=+3 (z can't be - ve)