Modern Physics 1 Question 42
39. Consider a hydrogen-like ionised 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 ionisation energy of the hydrogen atom is 13.6 $eV$. The value of $Z$ is
(2018 Adv.)
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Solution:
- $\Delta E _{2-1}=13.6 \times Z^{2} 1-\frac{1}{4}=13.6 \times Z^{2} \frac{3}{4}$
$$ \begin{aligned} & \Delta E _{3-2}=13.6 \times Z^{2} \frac{1}{4}-\frac{1}{9}=13.6 \times Z^{2} \frac{5}{36} \\ & \therefore \quad \Delta E _2=\Delta E _{3-2}+74.8 \\ & 13.6 \times Z^{2} \frac{3}{4}=13.6 \times Z^{2} \frac{5}{36}+74.8 \\ & 13.6 \times Z^{2} \frac{3}{4}-\frac{5}{36}=74.8 \\ & Z^{2}=9 \\ & \therefore \quad Z=3 \end{aligned} $$
