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f the frequency of the incident radiation on the cathode of a photocell is doubled, how will the following change? (a) Kinetic energy of electrons (b) Photoelectric current © Stopping potential .

f the frequency of the incident radiation on the cathode of a photocell is doubled, how will the following change? (a) Kinetic energy of electrons (b)
Photoelectric current © Stopping potential .

 

Grade:12

1 Answers

Arun
25750 Points
5 years ago

I will give here a ' master ' equation for photoelectric effect from which you can get almost all informations about photoelectric effect.

eVo=(1/2)mv( max)^2=h(f-fo)=hf-W=hc[1/(Lambda)-1/(Lambda,o)]. Here,

e is electronic charge.

m is mass of electron.

v(max) is maximum speed of photo electrons.

h is Planck's constant.

f is frequency of the incident radiation.

fo is threshold frequency( minimum frequency at which photoelectric effect starts.)

W(=hfo) is work function of photo sensitive surface.( The minimum energy to be spent by electron before emission.)

c is velocity of light.

Lambda is wave length of incident radiation.

Lambda,o is wave length corresponding to threshold frequency.

From the above equation, it can be seen immediately that when frequency of incident radiation is f, the maximum kinetic energy is

(1/2)mv(max)^2=K1=hf-W………..(1).

When frequency is doubled,

K2=2hf-W……………….(2).

Change in kinetic energy,

K2-K1=hf……………….(3)

The number of photo electrons determines photo electric current.

The number of photo electrons depends on the intensity of incident radiation and NOT on the frequency of incident radiation.

Hence, change in frequency will not change the photoelectric current.

We know that incident photons with energy hf are absorbed by the electrons of the photosensitive surface. So, electrons get this much energy in the form of kinetic energy,but they are not emitted with the same kinetic energy. This is because electrons are bound in the surface by a kind of potential well. So, before electrons are emitted they have to spend energy to come out from potential well. The minimum amount of energy that electrons have to spend is called work function (W) of the metal. Then, account of energies is

hf(gain by electron)-W( spent by electron)=(1/2)mv( max)^2 (maximum kinetic energy of emitted electron ).

Note that in the general formula nowhere number of electrons appear.

According to photon aspect of light, more the number of photons, more the intensity of light. If the frequency of incident is less than the threshold frequency then with any number of photons( with any intensity) the photo electrons will not be emitted. Also, intensity of light has nothing to do with the frequency of light. The photoelectric effect can be obtained with low enough intensity of light, but, having frequency sufficiently large.

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