Photoelectric Effect and light frequency

[wikidrox]

What happens when you increase the frequency of the light incident on a metal surface?

Does it increase the threshold frequency for the emission of photoelectrons? Does it increase the Kinetic energy of some energetic photoelectrons? I can't figure this out.

 

[jdavel]

wikidrox,

Starting at very low frequency (low energy) nothing happens. Then as the frequency is increased, the energy quanta of the photons reaches a high enough value for an electron which absorbs it to accelerate and escape from the surface of the metal. The threshold energy (called the work function) of the surface doesn't change, but now a few electrons have enough energy to get over it. Now as the frequency of the incident light increases still further, electrons that absorb these quanta have enough energy to get over the barrier with a little kinetic energy left over. As the frequncy continues to increase, the kinetic energy of the emitted electrons increases proportionally according to E = hf.

 

[lakshmi]

the threshold frequency is the minimum frequency of the radiation for which there will be emission of electrons so for a given metal it is constant then acccording to Einstein's equation E=WORK FUNCTION+KINETIC ENERGY OF PHOTONS.hv*=w+kinetic energy.w is work function which is the mnimum energy for electrons to be liberated.so it is constant as h represents Planck's constant kinetic energy mainly depends on frequency(v)

 

[maverick280857]

Remember this,

$$h\nu = h\nu_{0} + eV_{0}$$

$$\nu$$ is the frequency of incident light, $$\nu_{0}$$ is the threshold frequency below which photoelectrons will not be emitted and $$V_{0}$$ is the stopping potential. Note also that $$eV_{0}$$ is equal to the kinetic energy of the photoelectron.

So if you increase $$\nu$$, that is, if you increase the frequency of incident light then you are increasing the kinetic energy of the emitted photoelectrons. The threshold frequency is a material characteristic and is a constant so all the increase on the left hand side of the equation goes into increasing this kinetic energy. You can think that by increasing the incident frequency, you are making the electrons more energetic since you are giving more energy than they need to break free from the surface of the metal.

Hope that helps...

Cheers
Vivek