JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Atomic Structure

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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 \times 1 0^{15} s^{- 1}

and

h = 6.6 \times 1 0^{- 34} J s

(A) $3.21 \times 1 0^{- 14} J$
(B) $6.24 \times 1 0^{- 16} J$
(C) $8.58 \times 1 0^{- 19} J$ ✓ Correct
(D) $9.76 \times 1 0^{- 20} J$

Answer: (C)

Solution

The minimum energy required for the photoelectric effect is the work function, given by

W = h ν_{0}

, where

ν_{0}

is the threshold frequency. Substituting the given values:

W = 6.6 \times 1 0^{- 34} \times 1.3 \times 1 0^{15} = 8.58 \times 1 0^{- 19} J

JEE Main · 2022 · 25 Jun 2022 · Shift 2 · Single correct option

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