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Question
The work function for a metal surface is 2.2eV. If the light of wavelength 5000Å is incident on the surface of the metal, find the threshold frequency and incident frequency. Will there be an emission of photoelectrons or not? (c = 3 x 108 m/ s, 1eV = 1.6x10-19 J , h = 6.63 x 10-34 J.s.)
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Solution
Given work function = `Φ_0` = 2.2 eV
= `2.2 xx 1.6 xx 10^(-19) J`
= `3.52 xx 10^(-19)`
λ = 500 Å = 5000 x 10-10 m, c = `3xx10^8` m/s, h = `6.63 xx 10^(-34) Js`
Incident frequency v = `c/λ = (3xx10^8)/(5000 xx 10^(-10)`
= `(3xx10^8)/(5xx10^(-7) )= 3/5 xx 10^(-15) Hz`
=`30/5 xx 10^(14) Hz = 6 xx 10^(14) Hz`
Threshold frequency v0 = `Φ_0/h = (3.52 xx 10^(-19) J)/(6.63 xx 10^(-34) Js`
= `0.53 xx 10^15 Hz = 5.3 xx 10^14 Hz`
Emission of photoelectron takes place only if incident frequency is greater than the threshold frequency. But, in this case n>n0 therfore, photoelectric emission takes place.
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