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Find Threshold Frequency and Incident Frequency - Physics

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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.

 

Concept: Photoelectric Effect - Hertz’S Observations
  Is there an error in this question or solution?

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