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The work function for a metal surface is 2.2eV. If the light of wavelength 5000Å is incident on - Physics

Numerical

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 v= `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>ntherfore, photoelectric emission takes place.

Concept: Einstein’s Equation - Particle Nature of Light
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