Question:

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 photoelectric emission is nearly:

From photoelectric equation, K.E. = hv − hvth = eV0 (V0 = cutoff voltage)
V0 = h/e(8.2×10¹⁴ − 3.3×10¹⁴) = 6.6×10⁻³⁴ × 4.9×10¹⁴ / 1.6×10⁻¹⁹ ≈ 2V.