JEE Main 2022 · 25 June · evening · Single correct · +4 / −1

The minimum energy that must be possessed by photons in order to produce the photoelectric effect with platinum metal is :
[Given : The threshold frequency of platinum is 1.3 × 10¹⁵ s⁻¹ and h = 6.6 × 10⁻³⁴ J s.]

  1. (a) 3.21 × 10⁻¹⁴ J
  2. (b) 6.24 × 10⁻¹⁶ J
  3. (c) 8.58 × 10⁻¹⁹ J
  4. (d) 9.76 × 10⁻²⁰ J
Answer and solution

Answer (c) 8.58 × 10⁻¹⁹ J

Solution

Chemistry ma’am · 3 steps

The threshold frequency is the bare minimum, so the minimum photon energy is just hν₀.

  1. What minimum means. Electrons leave the metal only if each photon carries at least the work function, W₀ = hν₀. A photon at exactly the threshold frequency ν₀ has this minimum energy.

    Eₘᵢₙ = hν₀ = (6.6 × 10⁻³⁴ J s) × (1.3 × 10¹⁵ s⁻¹)

  2. Multiply. Number part: 6.6 × 1.3 = 8.58. Powers of ten: 10⁻³⁴ × 10¹⁵ = 10⁻¹⁹. The seconds cancel, which leaves joules.
  3. So. Eₘᵢₙ = 8.58 × 10⁻¹⁹ J, which is option (c).

The idea to keep. Threshold frequency always means minimum energy: W₀ = hν₀. Multiply the number parts and add the powers of ten separately, then check that the units come out in joules.

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.