JEE Main 2022 · 24 June · evening · Single correct · +4 / −1

The energy of one mole of photons of radiation of wavelength 300 nm is
(Given : h = 6.63 × 10⁻³⁴ J s, N A = 6.02 × 10²³ mol⁻¹, c = 3 × 10⁸ m s⁻¹)

  1. (a) 235 kJ mol⁻¹
  2. (b) 325 kJ mol⁻¹
  3. (c) 399 kJ mol⁻¹
  4. (d) 435 kJ mol⁻¹
Answer and solution

Answer (c) 399 kJ mol⁻¹

Solution

Chemistry ma’am · 3 steps

Find the energy of one photon with E = hc/λ, then multiply by Avogadro's number for a mole.

  1. Convert wavelength. Put λ in metres first: 300 nm = 300 × 10⁻⁹ m = 3 × 10⁻⁷ m.
  2. One photon. E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(3 × 10⁻⁷) = 6.63 × 10⁻¹⁹ J. The 3 on top cancels the 3 below, which keeps the arithmetic short.

    Eₘₒₗₑ = N_A hc/λ = 6.02 × 10²³ × 6.63 × 10⁻¹⁹ = 3.99 × 10⁵ J mol⁻¹

  3. So. 3.99 × 10⁵ J mol⁻¹ = 399 kJ mol⁻¹, which is option (c).

The idea to keep. Photon energy questions are always hc/λ for one photon. For a mole, multiply by N_A. Convert nm to m before you start, and J to kJ at the end.

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

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Asked about 170 times in JEE Main since 2020. This question is from 2022.