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JEE Advanced Chemistry Structure of Atom 2023 JEE Advanced 2023 (Paper 2)

JEE Advanced Chemistry Question (2023) — Solution

Question

For 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 Bohr radius, a = 52 . 9   pm Rydberg constant, R H = 2 . 2 × 10 - 18   J Planck's constant, h = 6 . 6 × 10 - 34 Js Speed of light, c = 3 × 10 8   ms - 1 ]

Options

  1. A. A
  2. B. B
  3. C. C
  4. D. D

Answer

A. A

Step-by-step solution

The radius of the n th  orbit can be represented as, r = 52 . 9 × n 2 z pm ∴    105 . 8 = 52 . 9 × n 2 2    ∴ n 2 = 2 and  26 . 45 = 52 . 9 × n 2 2    ∴ n 1 = 1 ∵ ΔE = R H hC × z 2 1 n 1 2 - 1 n 2 2 hc λ = R H hC × z 2 1 n 1 2 - 1 n 2 2 6 . 6 × 10 - 34 × 3 × 10 8 λ = 2 . 2 × 10 - 18 × 4 × 1 1 - 1 4 6 . 6 × 10 - 34 × 3 × 10 8 λ = 2 . 2 × 10 - 18 × 4 × 3 4 ∴ λ = 300   A ∴ λ = 30   nm

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