Class 12 Physics
CBSE and UP Board
Dual Nature of Radiation and Matter
Chapter-11 Exercise 11.8
NCERT Solutions for Class 12 Physics Chapter 11 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 cutoff voltage for the photoelectric emission
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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 x 10-19 C
Planck’s constant, h = 6.626 x 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.