Question 8

The threshold frequency for a certain metal is 3.3 × 1014 Hz. If light of frequency 8.2×1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Answer

Threshold frequency of the metal, V0 = 3.3 x 1014 Hz

Frequency of light incident on the metal, V = 8.2 x 1014 Hz

Charge on an electron, e = 1.6 × 10−19 C

Planck’s constant, h = 6.626 × 10−34 Js

Cut-off voltage for the photoelectric emission from the metal = V0

The equation for the cut-off energy is given as: eV0 = h (v - v0)

∴ v0 = h(v - v0) / e = 6.626 x 10-34 x (8.2 x 1014 - 3.3 x 1014) / 1.6 x 10-19 = 2.0292 V

Therefore, the cut-off voltage for the photoelectric emission is 2.0292 V

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