Albert Einstein was awarded the Nobel Prize in Physics (1921) for his explanation of photoelectric effect and not for his much more famous Theory of Relativity. You will come across questions based on Einstein’s photoelectric equation in almost all Medical and Engineering Entrance Tests.

You can remember Einstein’s photoelectric equation in different forms:

Let us consider the following question which appeared in Kerala Medical Entrance 2000 test paper:

You can remember Einstein’s photoelectric equation in different forms:

**(1) hν = KE**where ‘h’ is Planck’s constant, ‘ν’ is the frequency of the incident photon, KE_{max}+ Φ_{max}is the maximum kinetic energy of the electron emitted from the surface on which the photon is incident and ‘Φ’ is the work function of the surface.**(2) KE**where ν_{max}= hν–hν_{0}= h(ν–ν_{0})_{0}is the threshold frequency which is the minimum frequency (of the photon) required for photoelectric emission. Note that we have written the work function in terms of the threshold frequency as Φ = hν_{0}which can also be written as Φ = hc/λ_{0}where λ_{0}is the threshold wave length and ‘c’ is the speed of light in free space. Therefore, KE_{max}= hc[1/λ – 1/λ_{0}].**(3) ½ m(V****where V**_{max})^{2}= h(ν–ν_{0})_{max}is the maximum velocity of the photoelectron.**(4) eV**where ‘e’ is the electronic charge and Vs is the stopping potential which is the negative voltage to be applied on the anode to prevent photo electric emission._{s}= h(ν–ν_{0})Let us consider the following question which appeared in Kerala Medical Entrance 2000 test paper:

**If the photo electric work function is Φ eV, the threshold wave length is**

(a) eΦ/h (b) hc/eΦ (c) h/Φ (d) hc/Φ (e) eΦ/hc

Since the work function is given in electron volts you have to write its value in joule and equate to hc/λ(a) eΦ/h (b) hc/eΦ (c) h/Φ (d) hc/Φ (e) eΦ/hc

_{0}. You know that Φ electron volt is equal to Φe joule where ‘e’ is the electronic charge. So, we have Φe = hc/λ_{0}from which λ_{0}= hc/eΦ.The following MCQ appeared in the

In the second case, we have eV

Subtracting the first equation from the second, e(V

[If you find this more convenient, you may do it this way: The

Now consider the following question:

The correct option is (e) since the energy of the incident photon is 12400/6200 = 2 eV, which is insufficient for photoemission in the present case. [Note that the product of the energy in electron volts and the wave length in Angstrom Units in the case of photons is equal to 12400. Remember this to facilitate easy calculation of energy or wave length when one quantity is given].

Now consider the following M.C.Q. which is aimed at gauging your understanding of certain fundamental principles of photoelectric effect:

The kinetic energy of the emitted electron is independent of the intensity of light as is evident from Einstein’s photo electric equation. (It depends only on the frequency of the incident light and the nature of the emitting surface). The number of electrons emitted is directly proportional to the intensity of incident light (number of incident photons). Since the source of light is a point source, the intensity is governed by the inverse square law according to which the intensity is inversely proportional to the square of the distance. When the distance is quadrupled, the intensity becomes one sixteenths and hence the number of electrons emitted also becomes one sixteenths.[Option (d)]. Let us now discuss a question involving stopping potential:

The solution becomes quite simple if you express the energy of the photon in electron volt. Since the product of the wave length in Angstrom and the energy in electron volt is 12400 for any photon, the energy of the photon in the problem is 12400/2000 = 6.2 eV. In accordance with Einstein’s equation, 1.2 eV [which is the work function] is spent for removing the photoelectron from the surface and the remaining 5 eV is transferred to the photo electron (as its kinetic energy). To ‘kill’ this energy of the photo electron you have to apply a retarding potential of 5 volts. So, the correct option is (b).

Here is a ‘different’ question which appeared in the AIEEE 2003 question paper:

To conclude this discussion for the time being, let me ask you a simple question:

You might have heard of the thermionic work function. Is it greater than or equal to or less than the photo electric work function?

You should note that these two work functions are the minimum energy required to remove the electron from the surface of the emitter. In thermionic emission, the electron is removed by supplying heat energy. In photo electric effect, the required energy is supplied by the incident photon. In both cases the energy required for removing the electron from a given surface is the same. Therefore,

**question paper:***Kerala Engineering Entrance 2006***A metallic surface is irradiated by monochromatic light of frequency ν**

(a) V

In the first case, we have eV_{1}and stopping potential is found to be V_{1}. If light of frequency ν_{2}irradiates the surface, the stopping potential will be(a) V

_{1}+ (h/e) (ν_{1}+ ν_{2}) (b) V_{1}+ (h/e) (ν_{2}– ν_{1}) (c) V_{1}+ (e/h) (ν_{2}– ν_{1}) (d) V_{1}- (h/e) (ν_{1}+ ν_{2}) (e) V_{1}- (e/h) (ν_{2}– ν_{1})_{1}= h(ν_{1}–ν_{0}) where ν_{0}is the threshold frequency.In the second case, we have eV

_{2}= h(ν_{2}–ν_{0}).Subtracting the first equation from the second, e(V

_{2}- V_{1}) = h(ν_{2}–ν_{1}), from which V_{2}= V_{1}+ h/e(ν_{2}–ν_{1}). So, the correct option is (b).[If you find this more convenient, you may do it this way: The

*extra*energy available with the photon of frequency ν_{2}= hν_{2}- hν_{1}. So the*extra*stopping potential (ΔV) required is given by e(ΔV) = hν_{2}- hν_{1}, from which ΔV = h/e(ν_{2}–ν_{1}). Therefore V_{2}= V_{1}+ h/e(ν_{2}–ν_{1})].Now consider the following question:

**The photoelectric work function of a surface is 2.2 eV. The maximum kinetic energy of photo electrons emitted when light of wave length 6200 A.U. is incident on the surface is**

(a) 1.6 eV (b) 1.4 eV (c) 1.2 eV (d) 0.4 eV (e) photo electrons are not emitted(a) 1.6 eV (b) 1.4 eV (c) 1.2 eV (d) 0.4 eV (e) photo electrons are not emitted

The correct option is (e) since the energy of the incident photon is 12400/6200 = 2 eV, which is insufficient for photoemission in the present case. [Note that the product of the energy in electron volts and the wave length in Angstrom Units in the case of photons is equal to 12400. Remember this to facilitate easy calculation of energy or wave length when one quantity is given].

Now consider the following M.C.Q. which is aimed at gauging your understanding of certain fundamental principles of photoelectric effect:

**A photo cell is illuminated by a point source of light 50cm away. When the source is shifted to 2m, then**

(a) each emitted electron carries a quarter of the initial kinetic energy (b) The number of electrons emitted is a quarter of the initial number (c) each emitted electron carries one sixteenths the initial kinetic energy (d) the number of electrons emitted is one sixteenths the initial number (e) the number of electrons emitted and the energy of each electron are one sixteenths the initial values.(a) each emitted electron carries a quarter of the initial kinetic energy (b) The number of electrons emitted is a quarter of the initial number (c) each emitted electron carries one sixteenths the initial kinetic energy (d) the number of electrons emitted is one sixteenths the initial number (e) the number of electrons emitted and the energy of each electron are one sixteenths the initial values.

The kinetic energy of the emitted electron is independent of the intensity of light as is evident from Einstein’s photo electric equation. (It depends only on the frequency of the incident light and the nature of the emitting surface). The number of electrons emitted is directly proportional to the intensity of incident light (number of incident photons). Since the source of light is a point source, the intensity is governed by the inverse square law according to which the intensity is inversely proportional to the square of the distance. When the distance is quadrupled, the intensity becomes one sixteenths and hence the number of electrons emitted also becomes one sixteenths.[Option (d)]. Let us now discuss a question involving stopping potential:

**The retarding potential required to stop the emission of photoelectrons when a photosensitive material of work function 1.2 eV is irradiated with ultraviolet rays of wave length 2000 A.U. is**

(a) 4V (b) 5V (c) 6V (d) 8V (e)10V(a) 4V (b) 5V (c) 6V (d) 8V (e)10V

The solution becomes quite simple if you express the energy of the photon in electron volt. Since the product of the wave length in Angstrom and the energy in electron volt is 12400 for any photon, the energy of the photon in the problem is 12400/2000 = 6.2 eV. In accordance with Einstein’s equation, 1.2 eV [which is the work function] is spent for removing the photoelectron from the surface and the remaining 5 eV is transferred to the photo electron (as its kinetic energy). To ‘kill’ this energy of the photo electron you have to apply a retarding potential of 5 volts. So, the correct option is (b).

Here is a ‘different’ question which appeared in the AIEEE 2003 question paper:

**Two identical photo cathodes receive light of frequencies f**

(a) v

The energies of the photons are hf_{1}and f_{2}. If the velocities of the photoelectrons (of mass m) coming out are respectively v_{1}and v_{2}, then(a) v

_{1}- v_{2}= [2h(f_{1}- f_{2})/m]^{½}(b) v_{1}^{2}- v_{2}^{2}= 2h(f_{1}- f_{2})/m (c) v_{1}+ v_{2}= [2h(f_{1}+ f_{2})/m]^{½}(d) v_{1}^{2}+ v_{2}^{2}= 2h(f_{1}+ f_{2})/m_{1}and hf_{2}and therefore the kinetic energies of the photo electrons are hf_{1}-Φ and hf_{2}-Φ respectively, where Φ is the work function. Therefore, ½ mv_{1}^{2}- ½ mv_{2}^{2}= hf_{1}- hf_{2}, from which v_{1}^{2}- v_{2}^{2}= 2h(f_{1}- f_{2})/m. So, the correct option is (b).To conclude this discussion for the time being, let me ask you a simple question:

You might have heard of the thermionic work function. Is it greater than or equal to or less than the photo electric work function?

You should note that these two work functions are the minimum energy required to remove the electron from the surface of the emitter. In thermionic emission, the electron is removed by supplying heat energy. In photo electric effect, the required energy is supplied by the incident photon. In both cases the energy required for removing the electron from a given surface is the same. Therefore,

*thermionic work function of a given surface is exactly equal to its photo electric work function*.
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