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 \ \ \ \ .

Physics · Atomic Physics

JEE Advanced 2018 — Paper 2 — Question 13

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 $\mathrm{n}=2$ to $\mathrm{n}=1$ transition has energy 74.8 eV higher than the photon emitted in the $\mathrm{n}=3$ to $\mathrm{n}=2$ transition. The ionization energy of the hydrogen atom is 13.6 eV . The value of Z is $\_\_\_\_$ .

Answer: 3

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{E}_{\mathrm{n}}=13.6 \frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}$

$\mathrm{E}_{2}-\mathrm{E}_{1}=13.6 \mathrm{Z}^{2}\left(1-\frac{1}{4}\right)=13.6 \mathrm{Z}^{2} \times \frac{3}{4}$ .

$\mathrm{E}_{3}-\mathrm{E}_{2}=13.6 \mathrm{Z}^{2}\left(\frac{1}{4}-\frac{1}{9}\right)=13.6 \mathrm{Z}^{2} \times \frac{5}{36}$

Now, $13.6 Z^{2}\left(\frac{3}{4}-\frac{5}{36}\right)=74.8$

$\Rightarrow \quad \mathrm{Z}=3$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2018
Paper: Paper 2
Subject: Physics
Chapter: Atomic Physics
Topic: Hydrogen Spectrum

In a photoelectric experiment a parallel beam of monochromatic light with power of 200 W is incident on a perfectly absorbing cathode of…

Answer: 3

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