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Question
The photoelectric work function for a metal surface is 2.3 eV. If the light of wavelength 6800A is incident on the surface of metal, find threshold frequency and incident frequency. Will there be an emission of photoelectrons or not?
[Velocity of light c = 3 x 108 m/s,
Planck’s constant, h = 6.63 * 10-34 Js ]
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Solution
Given:
`phi_0=2.3 eV=2.3*1.6*10^-19 J=3.68*10^-19J`
λ=6800 Å=6800*10-10m, c=3*108 m/s, h=6.63*10-34 Js
We know that the incident frequency is given as
`v=c/lambda`
∴`v=(3*10^8)/(6800*10^-10)=4.41*10^14Hz`
Now, if the incident frequency is greater than the threshold frequency, then photoelectrons will be emitted from the metal surface. The threshold frequency is given from work function as
`v_0=phi_0/h`
`v_0=(3.68*10^-19)/(6.63*10^-34)=5.55*10^14Hz`
Since, photoelectrons `v<v_0` will not be emitted.
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