Social Sciences, asked by kushal9581, 1 year ago

Explain the effect of intensity and frequency on the photoelectric effect

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Answered by sushmita7
2
Effect Of Intensity On The Photoelectric Effect

It increases or decreases the number of electrons ejected from the metal surface. In other words intensity of light (number of photons) incident on the metal surface has no effect on the energy of the ejected electrons. This can be understood by the photoelectric equation :

11sdshν=ϕ+KEmax
or,KEmax=hν−ϕ
where,
KEmax→ maximum Kinetic Energy of ejected electron
h11111→ Planck's Constant
ν11111→frequency of radiation and
ϕ11111→work function of the metal

Effect Of Frequency On The Photoelectric Effect

The stopping potential Vc is found to be changing linearly with frequency of incident light being more negative for high frequency. An increase in frequency of the incident light increases the kinetic energy of the emitted electrons, so greater retarding potential is required to stop them completely. For a given frequency v, Vc measures the maximum kinetic energy Emax of photoelectrons that can reach plate C.

 eVc = 1/2 mV2max 

Where m is the mass of electron, e is charge of electron and Vmax is maximum velocity of electron. This means that work done by stopping potential must just be equal to maximum kinetic energy of an electron.

The effect of changing incident frequency v can also be studied from the plot of photoelectric current Vs potential applied across CD, keeping the intensity of incident radiation same.

From graph, we see that imax is same in all cases (for same intensity).

From graph, as v increases, Vc becomes more negative.





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