Chemistry · Structure of Atom

JEE Advanced 2023 — Paper 2 — Question 33

For $\mathrm{He}^{+}$ , a transition takes place from the orbit of radius 105.8 pm to the orbit of radius 26.45 pm . The wavelength (in nm ) of the emitted photon during the transition is $\_\_\_\_$ _. [Use: Bohr radius, $\mathrm{a}=52.9 \mathrm{pm}$ Rydberg constant, $R_{H}=2.2 \times 10^{-18} \mathrm{~J}$ Planck's constant, h $=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ Speed of light, $\mathrm{c}=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$ ]

Answer: 30

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{r}_{\mathrm{n}}=\frac{52.9 \times \mathrm{n}^{2}}{\mathrm{Z}} \mathrm{pm}$ $105.8=\frac{52.9 \times \mathrm{n}_{1}^{2}}{2} \therefore \mathrm{n}_{1}^{2}=4, \mathrm{n}_{1}=2$ $26.45=\frac{52.9 \times \mathrm{n}_{2}^{2}}{2} \therefore \mathrm{n}_{2}=1$ $\frac{1}{\lambda}=109677 \times 4 \times \frac{3}{4}$ $\lambda=\frac{4}{109677 \times 4 \times 3} \mathrm{~cm}$ $=\frac{10^{7}}{109677 \times 3}=\frac{10^{7}}{329031} \mathrm{~nm}$ $\lambda=30.3 \mathrm{~nm} \approx 30 \mathrm{~nm}$

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Exam: JEE Advanced 2023
Paper: Paper 2
Subject: Chemistry
Chapter: Structure of Atom
Topic: Bohr's Model of Atom