# Find Threshold Frequency and Incident Frequency - Physics

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 ]

#### 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
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2015-2016 (March)

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