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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answer is 3.

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Detailed Solution

As described in question, we have
$Δ E_{2 - 1} = 74.8 + Δ E_{3 - 2}$
Transition energy can be calculated as
$Δ E = 13.6 Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} e V)$
$\Rightarrow 13.6 Z^{2} [1 - \frac{1}{4}] = 74.8 + 13.6 Z^{2} [\frac{1}{4} - \frac{1}{9}]$
$\Rightarrow Z = 3$