Q. 83.6( 16 Votes )

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

Answer :

Photoelectric equation


eV0 = hv - ϕ0 ……………… equation(1)


Where e = Charge on an electron = 1.6 × 10-19 C


V0 = Stopping Potential


h = plank’s constant


v = frequency of photon


ϕ0 = Work function of the metal


Threshold frequency is the minimum frequency at which photoemission starts taking place from a metal surface. It is represented by v0.


At threshold frequency, stopping potential is zero and energy of the emitted electron is also zero. Energy provided by the incident photon is just sufficient to eject out electrons from the metal surface.


ϕ0 = hv0 ....................equ.02


V0 = Threshold frequency


Putting value of ϕ0 from equ.02 to equ.01


eV0 = hv – hv0


V0 = h (v-v0)/ e = 6.63 × 10-34 × (8.2-3.3) × 1014/1.6 × 10-19


= 2.03 V


For photo emission cut of potential is equal to 2.03 V


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