Chap.16 Photoelectric effect

Summary

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💢1. What is the photoelectric effect? 

It’s the emission of electrons from the surface of a metal when illuminated by a suitable radiations.

💢2. Corpuscular theory of light ; what mean this?

That means: a radiation is constituted of quantum of energy called photon. 

💢 3.Energy of Photon

 A photon is characterized by either a wavelength, denoted by λ or equivalently an energy, denoted by E

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       Question 3-1 :

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💢 4. The work function W₀

is the minimum energy required to extract an electron from the surface of a material.

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💢 5.  Threshold wavelength length λ0

is the maximum wavelength of incident radiations capable of ejecting photoelectrons from a photosensitive surface.

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 If the wavelength of incident photon is more than threshold wavelength, then photoelectron emission will not take place.

The relationship between the threshold wavelength (λ₀) and the work function W₀  can be expressed using the equation :

Question 5-1 :

💢 6. Threshold frequency of a material ν0 

or cut-off frequency

is the minimum frequency of incident radiation needed to cause 

photoelectric emission from the surface of that material.

It is different for different metals.

💢 7. Conditions for the Photoelectric Effect

W ≥ W₀

If the energy of the photon is less than the threshold energy,(W < W₀ ) 

there will be no emission of photoelectrons.

💢 8.Kinetic energy KE(Max) of a free electron  

Einstein represented the equation with the formula of

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💢 9.Effective photons

In reality, indefinite number of photon didn’t interact with the metal. For this reason; the number of liberated electrons is always much smaller than the number of real photons received during the same duration.

We say :     

💢 10. Source of radiation and power

The notion of power of a source is related to the energy transmitted by a number N of photons emitted each second.

Application :

       Question 10-1 :  How can we calculate the power of a source of radiation  ? 

Given «frequency of radiations 6.5×10¹⁴ Hz » and the «  number of emitted photons

 each 2 min. is equal to 2×10¹¹ ».

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      Question 10-2 : Knowing the average current i = 3.33×10^-9 A, of the electric current 

               due to the emitted electrons from the metal of a photocell.

The wavelength of the used radiation is 500 nm; deduce the average power of the source 

of radiation.

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💢 11. Efficiency of a cell

The quantum efficiency is given by :

  ( during the same interval of time).

      Question 11-1: We use a radiation of frequency ν= 2.2×10¹⁴ Hz

Calculate the quantum efficiency of a photocell, knowing that the number 

of liberated electrons from the metal during 5 minutes is 3×10¹²  ;  

and the power of the source of radiation   is P=3×10^-5 W.  

Answer ⇨

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💢 12.  How determine grafically the cut-off frequency from the graph (KECMax ; ν ).

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💢 13.Effect of frequency of photons on the value of the max. kinetic energy KECMax of ejected electrons.

        Effect of intensity (or power) of the light source on KECMax .

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      Question 13_1 :

Answer ⇨

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    Question 13_2 :

Answer ⇨

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    Question 13_3 :What is the effect of the modification of the power 

                         of the source on the speed of an emitted electron?

  Answer ⇨ 

    Question 13_4 :What should we do to increase the KE of emitted electrons?

 Answer ⇨

Exercise 1  :

Exercise  2 :

Exercise  3 :

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