SATHEE: Modern Physics 1 Question 42

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 $74.8 e V$ higher than the photon emitted in the $n = 3$ to $n = 2$ transition. The ionisation energy of the hydrogen atom is 13.6 $e V$ . The value of $Z$ is

(2018 Adv.)

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Solution:

$Δ E_{2 - 1} = 13.6 \times Z^{2} 1 - \frac{1}{4} = 13.6 \times Z^{2} \frac{3}{4}$

$\begin{aligned} Δ E_{3 - 2} = 13.6 \times Z^{2} \frac{1}{4} - \frac{1}{9} = 13.6 \times Z^{2} \frac{5}{36} \\ ∴ Δ E_{2} = Δ E_{3 - 2} + 74.8 \\ 13.6 \times Z^{2} \frac{3}{4} = 13.6 \times Z^{2} \frac{5}{36} + 74.8 \\ 13.6 \times Z^{2} \frac{3}{4} - \frac{5}{36} = 74.8 \\ Z^{2} = 9 \\ ∴ Z = 3 \end{aligned}$

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Modern Physics 1 Question 42

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