Photoelectric Effect: Laws, Examples & Applications

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Arpita Srivastava

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The photoelectric effect refers to the phenomenon wherein electrically charged particles are released within a material when it absorbs electromagnetic radiation. The effect involves ejection of electrons from a metal plate when light falls on it. 

  • The photoelectric effect has been found useful in electronic devices that are specialized for light detection. 
  • The ejected electrons in the effect are called photoelectrons.
  • It was discovered in 1887 by the German physicist Heinrich Rudolf Hertz.
  • In this process, the energy of photons is greater than the work potential.
  • The electrons are ejected by a process called thermionic emission.
  • The photoelectric effect in chemistry is used to draw inferences about the properties of atoms.
  • It is used in the fields of material science and astrophysics.

Key Terms: Photoelectric Effect, Electrons, Metals, Electromagnetic Radiation, Planck’s Constant, Frequency, Photon, Thermionic Emission


What is Photoelectric Effect?

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Photoelectric effect is the phenomenon through which metals release electrons when they are exposed to the light of the appropriate frequency. The electrons emitted during the process are known as photoelectrons. It can be expressed as:

hν = W + E

where,

Photoelectric Effect

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Laws of Photoelectric Effect

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The laws of photoelectric effect in chemistry are as followed:

  • The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
  • A certain minimum (energy) frequency exists for a given material, called threshold frequency.
  • Below the level of threshold frequency, the discharge of photoelectrons stops completely.
  • The maximum kinetic energy of the photoelectrons increases with the increase in the frequency.
  • It means that frequency γ > γ Th exceeds the threshold limit of the incident light. 
  • The maximum kinetic energy is free from the intensity of light. 
  • The process of photo-emission is an instantaneous process.

History of Photoelectric Effect

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In the year 1887, Wilhelm Ludwig Franz Hallwachs introduced the photoelectric effect in chemistry, and Heinrich Rudolf Hertz undertook the experimental verification process. When a material absorbs electromagnetic radiation, it releases electrically charged particles. 

  • The effect is defined as the emission of electrons in a metal plate when exposed to direct light. 
  • This light or radiant energy can be gamma rays, UV light, normal light, x-rays, or infrared. 
  • The material can either be liquid, solid, or gas. 
  • The particles released can be electrons or ions (i.e. atoms or molecules which are electrically charged). 
  • These emitted electrons are defined as photoelectrons.
  • The energy from the light is absorbed by the electrons which are present in the metal.
  • It uses energy to bypass the forces that create constraints.
  • When compared with other electrons, photoelectrons are no different.

Photoelectric Effect


Photoelectric Effect: Concept of Photons

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Instead of considering light as a wave, we have to visualize light as a stream of particles (i.e., electromagnetic energy). So, these light ‘Particles’ are considered Photons. Max Planck defines in his equation that the frequency of light is related to the energy held by photons.

  • Different frequencies of light carry photons of varying energies. 
  • Red light has less frequency than blue light because the wavelength of red light is larger when compared with the wavelength of blue light. 
  • So, the photon’s energy in blue light is greater than the photon’s energy in red light.

Max Planck’s equation

E = hv = hc / λ

  • where, E = Photon’s energy
  • h = Constant
  • v = Light’s frequency
  • c = Light’s speed (in vacuum)
  • λ = Light’s Wavelength

Max Planck’s equation

Properties of Photon

The various properties of photons are as follows:

  • The quantum number is zero for a photon
  • Photon has no mass and no charge.
  • It is not reflected in an electric field and a magnetic field 
  • Speed of the light in which the photon moves in empty space.
  • Photon energy is directly proportional to photon frequency.
  • It is also inversely proportional to proton wavelength.

The momentum of Photon and Energy of Photon

E = p.c

  • where, p = Momentum magnitude
  • c = Light’s speed 

Threshold Energy for Photoelectric Effect

Threshold energy is the minimum energy that all particles in an atom possess to make the collision between molecules effective. Photons which are present on a metal surface must incline sufficient energy to bypass the forces that coerce the electrons to the metal’s core. 

  • We need to know the minimum amount of energy required to remove an electron from a metal. 
  • This minimum amount of energy is known as Threshold energy.
  • For the photoelectric effect to occur, we must know the minimum frequency of light required. 
  • This is known as a threshold frequency
  • There is a wavelength associated with the frequency of light. 
  • This is considered as the threshold wavelength.

Photoelectric Effect: Threshold energy

Φ = hvth = hc / λth

  • where, Φ = Threshold energy
  • Vth= threshold frequency
  • λth= threshold wavelength

Threshold Energy

Kinetic Energy Emitted Photoelectron and Frequency of Incident Photon

The frequency and the kinetic energy can be derived as,

Ephoton = Φ + Eelectron

hv = hvth + ½ mev2

  • Ephoton = Incident Photon’s energy (equal to hv)
  • Φ = Threshold energy (equal to hvth)
  • Eelectron= Photoelectron’s kinetic energy equal to ½ mev2 ;
  • me=9.1x10-31kg (mass of electron) 

Photoelectron’s emission will not take place if the threshold energy is higher than the photon’s energy. It will happen when the threshold frequency and the photon’s frequency are equal. At this point, the kinetic energy will be zero

Kinectic Energy of Photon


Einstien’s Contribution to Photoelectric Effect

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The emission of electrons in a metal plate's surface, when exposed to direct light, is said to be the process of the photoelectric effect. This effect helps us identify and understand light's electrons and quantum nature. 

  • After conducting in-depth research, Albert Einstein states, "In separate quantized packets light energy is being carried thus as a result Photoelectric Effect occurs". 
  • In the year 1921, he was awarded the Nobel Prize (Physics) in this regard.

Einstien’s equation

E = hv

  • where, E = Photon’s energy (Joule)
  • h = Planks Constant ( 6.62 x 10-34J.s)
  • v = Photon’s frequency (Hz)

Einstien’s equation


Principles of Photoelectric Effect

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The principle of photoelectric effect are as follows:

  • In a specific electronic configuration, the electrons bound to atoms occur.
  • The valence band is the highest energy band, occupied by the material's electrons. 
  • The electrons are half-filled in a valence band if it is a typical metal conductor. 
  • It will move immediately from atom to atom in the case of the conductor.
  • The electrons are full in a valence band if it is a good insulator like rubber. 
  • It will have very little movement in the case of an insulator.
  • The positive charge flow and the light's electrons release cause photoconductivity.
  • The radiation of higher frequencies rays also causes photoelectric effects.
  • When a photon of X-ray collides with an electron, the Compton effect takes place.

Applications of Photoelectric Effect

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The various applications of photoelectric effect are as follows:

Solar Panels

The photoelectric effect is being used in solar panels to generate electricity. 

Position and Motion Sensors

In front of IR LED a photoelectric material is placed. This will result in a light cut-off and a change will be registered by the electronic circuit. 

Smartphone

Smartphone screen brightness adjusts automatically, due to the intensity of light (a current generated through photoelectric effect) touching the sensor on the mobile.

XPS (X-ray Photoelectron Spectroscopy)

XPS (X-ray Photoelectron Spectroscopy) includes irradiating the surface of the device, the kinetic energy in electrons (emitted) is measured. 


Things to Remember

  • Photoelectric effect is the phenomenon through which metals release electrons.
  • The effects take place when metals are exposed to the light of the appropriate frequency.
  • Photoelectric effect can be expressed as: hν = W + E
  • Max Planck’s equation can be written as E = hv = hc / λ
  • Equation for threshold energy for photoelectric effect is written as Φ = hvth = hc / λth
  • Einstein’s equation of photoelectric effect is written as E = hv
  • The frequency and the kinetic energy can be derived as, Ephoton = Φ + Eelectron and hv = hvth + ½ mev2

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Sample Questions

Ques. Based on the photoelectric effect, an incident light has a threshold wavelength value of 260nm and energy (eV) is 1237 / λ (nm). Identify the Kinetic energy of electrons emitted? (2 marks)

Ans. Kmax = hc/λ - hc/λ0

= hc x [(λ0 - λ) / λλ0]

Kmax = (1237) x [(380 - 260) / 380 x 260]

Kmax = 1.5 eV

Hence, Based on the photoelectric effect, the maximum kinetic energy of electrons emitted is 1.5 eV.

Ques. A photon’s frequency is 5x1014, Calculate the energy of one mole of Photon? (2 marks)

Ans. E = hV

V = 5x1014

h = 6.626 x 10-34J.s

E = (6.626 x 10-34J.s) x (5x1014)

E = 3.313 x 10-19J

One mole of photon’s energy

= (3.313 x 10-19J) x (6.022 x 1023 mol -1)

= 199.51 kJ mol-1

Ques. 400 nm of a wavelength of monochromatic light is emitted from a 100-watt bulb. Find how many protons are emitted every second by the 100-watt bulb.? (2 marks)

Ans. Bulbs power = 100 Js-1

One Photon Energy, E = hV = hc / λ 

= 6.626 x 10-34J.s x 3 x 108 ms / 400 x 10-9 m

= 4.969 x 10-19 J

= 100 J s-1 / 4.969 x 10-19 J

= 2.012x1020s-1

Ques. Metal has a threshold frequency v0 of 7.0x1014 s-1. Find the electron’s kinetic energy when frequency radiation is v = 1.0x1015 s-1? (2 marks)

Ans. Kinetic energy = ½ mev2=h(v-v0)

= (6.626 x 10-34J.s) (1.0 x 1015 s-1 - 7.0x1014 s-1)

= (6.626 x 10-34J.s) (10.0 x 1014 s-1 - 7.0x1014 s-1)

= (6.626 x 10-34J.s) x (3.0 x 1014 s-1)

= 1.988 x 10-19J

Ques. Justify the possibility of the photoelectric effect. The threshold energy of copper metal is Φ = 7.53 x 10-19J and the frequency of light is 3.0 x 1016 Hz? (2 marks)

Ans. We have to calculate the energy of Photon using Planck’s equation,

Ephoton = hv

Ephoton = (6.626 x 10-34J.s) (3.0 x 1016 Hz)

Ephoton = 2.0 x 10-17J

 Φ = 7.53 x 10-19J < Ephoton = 2.0 x 10-17J

Now if we compare the energy of the photons with the threshold of copper metal, the photon energy is greater. Hence we have witnessed the possibility of photoelectrons released from the copper. 

Ques. State the laws of the photoelectric effect. (3 marks)

Ans. The laws of the photoelectric effect are:

  1. The photoelectric current is in direct proportion to the intensity of light, for a light of any given frequency; (γ > γ Th).
  2. There exists a certain minimum (energy) frequency for a given material, called threshold frequency, below which the discharge of photoelectrons stops completely, irrespective of how high the intensity of incident light is.
  3. The maximum kinetic energy of the photoelectrons increases with the increase in the frequency (provided frequency γ > γ Th exceeds the threshold limit) of the incident light. The maximum kinetic energy is free from the intensity of light. 
  4. The process of photo-emission is an instantaneous process.

Ques. What are the applications of the photoelectric effect? (3 marks)

Ans. The applications of the photoelectric effect are:

  • The photoelectric effect is being used in Solar panels to generate electricity. 
  • Position and Motion Sensors: In front of IR LED a photoelectric material is placed. This will result in a light cut-off and a change will be registered by the electronic circuit. 
  • Based on the available lighting, a smartphone screen brightness adjusts automatically, this is due to the intensity of light (a current generated through photoelectric effect) touching the sensor on the mobile.
  • XPS (X-ray Photoelectron Spectroscopy): By irradiating the surface, the kinetic energy in electrons (emitted) is measured. 
  • Burglar alarms use Photoelectric cells.

Ques. What are the principles of the Photoelectric effect? (4 marks)

Ans. The principles of the Photoelectric effect are as follows:

  • In a specific electronic configuration, the electrons bound to atoms occur.
  • The valence band is said to be the highest energy band which is occupied by the electrons of a prescribed material. 
  • The electrons are half-filled in a Valence band if it is a typical metal conductor. It will move immediately from atom to atom.
  • The electrons are full in a Valence band if it is a good insulator like rubber. It will have very little movement.
  • Based on the flow of positive charges and the electrons released from the light causes photoconductivity.
  • The radiation of higher frequencies like rays also causes photoelectric effects.
  • When a photon of x-ray collides with an electron arises the Compton effect.

Ques. Write the principles of the photon. (4 marks)

Ans. The principles of photons are given below:

  • The quantum number is zero for a photon
  • Photon has no mass, no charge.
  • Photons are not reflected in an electric field and a magnetic field 
  • Speed of the light in which the photon moves in empty space.
  • Photons energy = directly proportional to photon frequency
  • Photon energy = inversely proportional to proton wavelength

Ques. What are the characteristic features of the photoelectric effect? (3 marks)

Ans. The characteristic features of the photoelectric effect are:

  • The threshold frequency changes with material which means it is different for different materials.
  • The photoelectric current is in direct proportion with the light intensity.
  • The kinetic energy of the photoelectrons is in direct proportion with the light frequency.
  • The stopping potential is in direct proportion with the frequency and the process is prompt.

Ques. In a photoelectric effect experiment, the threshold wavelength of incident light is 200 nm and E (in eV) = 1237/λ (nm). Find the maximum kinetic energy of emitted electrons? (3 marks)

Ans. Kmax = hc/λ – hc/λ= hc × [(λ0 – λ)/λλ0]

⇒ Kmax = (1237) × [(300 – 200)/300×200] = 2.06 eV

Therefore, the maximum kinetic energy of emitted electrons in the photoelectric effect is 2.06 eV.

Ques. In a photoelectric experiment, the wavelength of the light incident on metal is changed from 100 nm to 200 nm and (hc/e = 1240 nm-V). Find the decrease in the stopping potential? (3 marks)

Ans. The wavelength of the light incident on metal is changed from 100 nm to 200 nm

hc/λ= ϕ + eV1 . . . . (i)

hc/λ= ϕ + eV2 . . . . (ii)

Equation (i) – (ii)

hc(1/λ1 – 1/λ2) = e × (V1 – V2)

⇒V– V= (hc/e) × [(λ2 – λ1)/(λ1 λ2)]

= (1240 nm V) × 100nm/(100nm × 200nm)

=12.4/ 2 ≈ 6.2V

Therefore, the decrease in the stopping potential during the photoelectric experiment is 6.2V.

Ques. In a photoelectric experiment, the wavelength of the light incident on metal is changed from 400 nm to 500 nm and (hc/e = 1240 nm-V). Find the decrease in the stopping potential? (3 marks)

Ans. The wavelength of the light incident on metal is changed from 100 nm to 200 nm

hc/λ= ϕ + eV1 . . . . (i)

hc/λ= ϕ + eV2 . . . . (ii)

Equation (i) – (ii)

hc(1/λ1 – 1/λ2) = e × (V1 – V2)

⇒V– V= (hc/e) × [(λ2 – λ1)/(λ1 λ2)]

= (1240 nm V) × 100nm/(400nm × 500nm)

≈ 0.62 V

Therefore, the decrease in the stopping potential during the photoelectric experiment is 0.62 V

Ques. 500 nm of a wavelength of monochromatic light is emitted from a 100-watt bulb. Find how many protons are emitted every second by the 100-watt bulb.? (2 marks)

Ans. Bulbs power = 100 Js-1

One Photon Energy, E = hV = hc / λ 

= 6.626 x 10-34J.s x 3 x 108 ms / 500 x 10-9 m

= 3.975 x 10-19 J

= 100 J s-1 / 3.975 x 10-19 J

= 2.515x1020s-1

Ques. In a photoelectric experiment, the wavelength of the light incident on metal is changed from 700 nm to 800 nm and (hc/e = 1240 nm-V). Find the decrease in the stopping potential? (3 marks)

Ans. The wavelength of the light incident on metal is changed from 100 nm to 200 nm

hc/λ= ϕ + eV1 . . . . (i)

hc/λ= ϕ + eV2 . . . . (ii)

Equation (i) – (ii)

hc(1/λ1 – 1/λ2) = e × (V1 – V2)

⇒V– V= (hc/e) × [(λ2 – λ1)/(λ1 λ2)]

= (1240 nm V) × 100nm/700nm × 800nm)

≈ 0.22 V

Therefore, the decrease in the stopping potential during the photoelectric experiment is 0.22 V

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