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sudarsan
sudarsan
10:15
. Give reason for maximum kinetic energy of the
photoelectrons is independent of the intensity of
incident radiation.
For a given frequency of the incident radiation, the
stopping potential is independent of its intensity, i.e.,
the maximum kinetic energy of photoelectrons depends
on the light source and emitter plate material but is
independent of intensity of radiation. As Kmax= eV0 where
V0 = Stopping potential
this the answr for that question i dont know why they are saying the same thing said teh question and not telling the reason
@bml
answer this question please
Bml
Bml
@sudarsan Are you referring to me? I have only studied little Quantum Mechanics, so I am not sure I can provide the answer you want, but if you want I can try....
sudarsan
sudarsan
@bml
yes i want it :-)
Bml
Bml
10:33
@sudarsan Ok, I will try. This question explains the failures of classical wave theory of light to explain photoelectric effect. Referring to the classical wave theory, if intensity of the wave is increased, the oscillating electric field vector E of the light increases in amplitude (I ∝ A^2). Yes?
OK, you've disappeared, but I will go ahead. The force applied to the electron in the metal due to the falling radiation will be eE and this force will increase on increasing the intensity of light. So the kinetic energy of the emitted photoelectrons should also increase due to the larger force applied in emitting them out as the light beam is more intense.
Now, photons do not give their energy in parts, they either will give all the energy or none at all. The energy given to the electrons should be more than or equal to the work function of the metal, if the electron is to be ejected. The excess energy will become the electron's kinetic energy.
So, observations show that the maximum kinetic energy is independent of light's intensity (at constant frequency). If h\nu is the incident energy and h\nu_{\circ} Is the work function, we Will have KE_{max}=h\nu - h\nu_{\circ}. As we can see, maximum kinetic energy depends only on the frequency of the incident radiation, not in its intensity.
According to the wave theory, the photoelectric effect should take place at any frequency of light, provided that the light is intense enough to supply the energy needed to eject the photoelectrons.
But, however, observation show that the photoelectric effect does not occur if the frequency of the falling radiation is less than the threshold frequency, even if the light is highly intense.
Let N be the number of photons falling on metal plate per unit area per unit time; then,
N=\frac{IA}{h\nu}.
Now, if we double the intensity, we double the number of photons which consequently doubles the photocurrent. This is done without changing the energy of the individual photon.
Hence, kinetic energy does not increase when you double the intensity of light (keeping energy of individual photon constant).
If we consider the case KE_{max}=0, we have,
Workfunction (ϕ)=h\nu_{\circ}.
This tells us that the photon has just enough to eject the photoelectrons and no extra energy appear as kinetic energy.
If frequency is reduced below threshold frequency (\nu_{\circ}), no photoelectrons will be ejected no matter how intense the radiation may be.
So, the fact that "maximum kinetic energy of the photoelectrons is independent of the intensity of
incident radiation" is quite important, because it is one of the selections which show that in photoelectric effect light exhibits its particle-like nature.
 
4 hours later…
John Rennie
John Rennie
15:23
@sudarsan The reason is that one electron interacts with one photon, so the energy the electron can get is the photon energy h𝑣.
It doesn't matter how intense you make the light, one photon still has energy h𝑣 and that's the maximum energy the photoelectron can get from the light.

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