JEE Main 2023 · 24 January · evening · Single correct · +4 / −1

A photon is emitted in transition from n = 4 to n = 1 level in hydrogen atom. The corresponding wavelength for this transition is (given, h = 4 × 10⁻¹⁵ eVs) :

  1. (a) 99.3 nm
  2. (b) 94.1 nm
  3. (c) 974 nm
  4. (d) 941 nm
Answer and solution

Answer (b) 94.1 nm

Solution

Physics sir · 3 steps

Find the energy gap in eV, then use hc = 1200 eV nm with the given h to get the wavelength.

  1. Energy of photon. In hydrogen, Eₙ = -13.6/n² eV. The photon carries away the gap between n = 4 and n = 1: Δ E = 13.6(1 - 1/16) = 13.6 × 15/16 = 12.75 eV.
  2. Value of hc. With the given h = 4 × 10⁻¹⁵ eV s and c = 3 × 10⁸ m s⁻¹, hc = 1.2 × 10⁻⁶ eV m = 1200 eV nm.

    λ = hc/(Δ E) = (1200 eV nm)/(12.75 eV) ≈ 94.1 nm

  3. So. The wavelength is 94.1 nm, in the ultraviolet Lyman series. The answer is (b).

The idea to keep. For photon problems, work in eV and nm: λ (nm) = hc/E with hc ≈ 1240 eV nm, or whatever value the given h produces. Use the h the question gives you.

Answer from the NTA final answer key. Solution by PrepWiser.

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Asked about 110 times in JEE Main since 2020. This question is from 2023.