The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz. If light of frequency 8.2 × 1014 Hz is incident on this metal, the cut- off voltage for the photoelectron emission is nearly (a) 1 V (b) 2 V (c) 3 V (d) 5 V

Answer: (b). eV₀ = h(ν - ν₀) V₀ = 6.63 × 10⁻³⁴ × (8.2 - 3.3) × 10¹⁴1.6 × 10⁻¹⁹ = 3.25 × 10⁻¹⁹1.6 × 10⁻¹⁹ V₀ ≈ 2.0 V, so the answer is (b) 2 V.

The threshold frequency for a photosensitive metal is 3.3 × 1014 Hz… | NEET 2011 (AIPMT, Mains)

Q154NEET 2011 · AIPMT · MAINSEasy

Photoelectric Effect: Einstein's Equation

The threshold frequency for a photosensitive metal is 3.3 $\times$ 1014 Hz. If light of frequency 8.2 $\times$ 1014 Hz is incident on this metal, the cut- off voltage for the photoelectron emission is nearly

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Answer: (b)

$eV_0 = h(\nu - \nu_0)$
$V_0 = \frac{6.63 \times 10^{-34} \times (8.2 - 3.3) \times 10^{14}}{1.6 \times 10^{-19}} = \frac{3.25 \times 10^{-19}}{1.6 \times 10^{-19}}$
$V_0 \approx 2.0\text{ V}$ , so the answer is (b) 2 V.

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