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The threshold frequency for a photosensitive metal is 3.3 × 10^14 Hz. If light of frequency 8.2 × 10^14 Hz is incident on this metal, the cut-off voltage for the photoelectric emission is nearly

1. 2V
2. 3V
3. 5V
4. 1V

Correct answer: 2V

Solution

The energy of the incident photon is given by E = h*f, and the work function is W = h*f₀. The kinetic energy of the emitted photoelectron is KE = E - W = h*(f - f₀). The cut-off voltage is related to the maximum kinetic energy by eV₀ = KE. Substituting values: h = 6.63 × 10^-34 J·s, f = 8.2 × 10^14 Hz, f₀ = 3.3 × 10^14 Hz, and e = 1.6 × 10^-19 C, we get V₀ ≈ 2V.

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