Q. 83.6( 16 Votes )

# The threshold frequency for a certain metal is 3.3 × 10^{14} Hz. If the light of frequency 8.2 × 10^{14} Hz is incident on the metal, predict the cutoff voltage for the photoelectric emission.

Answer :

__Photoelectric equation__

eV_{0} = hv - ϕ_{0 ………………} equation(1)

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

V_{0} = Stopping Potential

h = plank’s constant

v = frequency of photon

ϕ_{0} = 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 v _{0.}__

__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.__

ϕ_{0} = hv_{0} ....................equ.02

V_{0} = Threshold frequency

Putting value of ϕ_{0} from equ.02 to equ.01

eV_{0} = hv – hv_{0}

∴ V_{0} = h (v-v_{0})/ e = 6.63 × 10^{-34} × (8.2-3.3) × 10^{14}/1.6 × 10^{-19}

= 2.03 V

__For photo emission cut of potential is equal to 2.03 V__

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