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Einstein's photoelectric equation: energy quantum of radiation,threshold frequency, threshold wavelength, practice problems, FAQs

Einstein's photoelectric equation: energy quantum of radiation,threshold frequency, threshold wavelength, practice problems, FAQs

Christian Huygen proposed propagation of light in the form of waves. But it was unable to describe the motion of light in vacuum which was later explained by wave theory proposed by Maxwell. Wave theory further was unable to describe the photoelectric effect discovered by Hertz. Max Plank described the propagation of light in the form of packets of energy called quanta. Einstein further used this theory to propose the photoelectric equation which we will discuss here.

Table of contents

  • Einstein's photoelectric equation
  • Threshold wavelength
  • Threshold frequency
  • Practice problems
  • FAQs

Einstein's photoelectric equation

In 1905, Einstein (1879-1955) proposed a completely different picture of electromagnetic waves. He picturises the EM waves as energy discrete units called photons and photoelectric emission does not take place by continuous absorption of energy from radiation. These discrete units are also called quanta of energy of radiation. Each quantum of energy is h, where is the frequency of light and h is Planck's constant.

In photoelectric effect, an electron absorbs a photon (hv) of radiation. If this absorbed energy exceeds the threshold (work function image of the metal), the most loosely bound electron will emerge with maximum kinetic energy, and the most tightly bound electron will emerge with almost zero kinetic energy.

Einstein's photoelectric equation,


The observations of a photoelectric effect can be explained easily using the above equation,

  • Kmax is independent of intensity of radiation but depends linearly on which is as per our observation because the photoelectric effect arises from the absorption of a single quantum of radiation by a single electron. It is evident that the number of photons incident per second is enough to determine the intensity of light. 
  • Photoelectric emission is possible only if hv>image, because Kmax must be non-negative hv >image


where vo is threshold frequency. No photoelectric emission is possible below , even if the incident radiation of high intensity and long duration falls on the surface.

  •  intensity of radiation is directly proportional to the number of energy quanta per unit area per unit time. Higher the intensity means, more number of photons hits per unit area per unit time, hence more no of electron will absorb photons and no of photoelectrons generated is higher that increase the rate of flow of charge (for v >vo). This is the reason, for v>vphotoelectric current is directly proportional to intensity.
  • In photoelectric effect the elementary process is involved in absorption of a light photon by an electron. This process takes place instantly and does not depend on intensity, if the frequency of radiation is greater than the threshold frequency photoelectric effect takes place instantly .The time order of the emission of the photoelectron from the surface as it is hit by radiation is in order of nanoseconds.
  • There is zero chance of two photons hitting a single electron at a time. Means at an instant only one photon can hit an electron. So, the electron can only take energy equal to ℎ𝜈 and not multiples of ℎ𝜈. Conclusion:- either an electron does not receive any energy or if it is receiving energy it must be ℎ𝜈.

Threshold wavelength

The maximum wavelength of the incident radiation above which photoelectric emission is not possible. If the photoelectric effect has to happen then the energy of the photon must be greater than the work function of metal.


Maximum K.E of the electrons emitted can be calculated as,


Threshold frequency

The minimum frequency of the incident radiation below which photoelectric emission is not possible. If the photoelectric effect has to happen then the energy of the photon must be greater than the work function of metal.


Maximum KE of the emitted electrons will be,


Practice problems

Q1. In a photoelectric experiment, the collector plate is at 2.0 𝑉 with respect to the emitter plate made of copper (𝜙=4.5 𝑒𝑉). The emitter is illuminated by a source of monochromatic light of wavelength 200 𝑛𝑚. What will be the minimum and maximum K.E of the photoelectrons that are reaching the collector.


Answer: The maximum K.E. with which the photoelectrons can leave the emitter plate is


Therefore, the K.E. of photoelectrons leaving the emitter plate lies between 

0<K<1.7 eV

The direction of the electric field is from A to C, this means that the force acting on the electrons is in the opposite direction to the direction of the electric field i.e. towards A due to which the electrons get accelerated in the direction of force. The electron is going to gain kinetic energy and lose potential energy.

The change in potential energy when electron goes from C to A, imageU=-2eV-0

The increase in K.E is equal to this decrease in P.E., imageK=2eV

As, the K.E. of photoelectrons leaving the emitter plate lies between 0<K<1.7 eV

Therefore, the K.E. of photoelectrons reaching the collector plate lies between 

2 eV <Kcollector<3.7 eV

KEmin=2eV KEmax=3.7 eV

Q2. The threshold frequency of a metal for photoelectric effect to take place is 𝑓o. The maximum velocity of electrons emitted is 𝑣1, if the light incident on the metal plate is having frequency 2𝑓o. When the frequency of the incident radiation is increased to 5𝑓o, the maximum velocity of electrons emitted is 𝑣2. Find the ratio of 𝑣1 to 𝑣2.

Answer: The maximum K.E. of emitted electrons is


Q3. The photoelectric threshold wavelength of silver is 3250×10−10 𝑚. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength 2536×10−10 𝑚 is 

[Given ℎ=4.14×10−15 𝑒𝑉 𝑠 & 𝑐=3×108 𝑚 /s]

A. 0.6×105 𝑚/s
B. 61×103 𝑚/s
C. 0.3×106 𝑚/s
D. 6×105 𝑚/s

Answer: We know that the maximum K.E. of a photoelectron is given by


Q4. The graph is showing variation between stopping potential (𝑉o)with frequency (𝜈) for two photosensitive metals 𝑃 and 𝑄. Which metal will be having a smaller threshold wavelength and why?



The x-intercept on Vo vs 𝜈 graph gives the threshold frequency of the metal.


Hence metal Q has smaller threshold wavelength.


Question 1. On what factor does the work function of a metal depend upon?
Answer: Work function depends on the nature of the metal surface.

Question 2. If a light of frequency less than threshold frequency is incident on a metal surface, what will be the kinetic energy of the electrons emitted?
Answer: Zero. Because no electron will get emitted for frequency less than threshold frequency.

Question 3. How does the kinetic energy of the electrons emitted during a photoelectric effect vary with the frequency of light?
Answer: Kinetic energy varies linearly with the frequency of the incident light.

Question 4. What is the dimension of work function?
Answer: It is same as that of energy i.e. [M1L2T-2].

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