Particle Nature of Light: Photon Properties

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The particle nature of light came into the picture when Albert Einstein looked up to the experiment conducted by Max Planck and observed that the wavelength and intensity of light have a certain impact on the ejected electrons. In this article, we will dive deeper into the topic and discuss what actually is the particle nature of light, what are its characteristics, and other relevant sub-topics.

Key Terms: Photons, Dual nature of Light, Particle Nature of Light, Photoelectric Effect, Matter, Quanta, Energy


What is Particle Nature of Light?

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In addition to the wave nature of light, the photoelectric effect experiment gave rise to another strange fact. It suggested that light when interacts with matter, behaves as it is made up of energy packets or quanta. These energy packets or quanta are what we call photons today. This very experiment led to the arrival of another theory known as the Particle Nature of Light.

Particle Nature of Light
Particle Nature of Light

Photon

Photons, also known as light quantum or energy packets, are elementary particles i.e., constituents of energy. In the simplest terms, a photon is a particle of light. 

Photon
Photon

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Characteristics of Photons

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  1. Photons are the basic constituent of light which is considered as the smallest quantum of radiation.
  2. Photons travel at the speed of 2.99 × 108 m/s in a vacuum and never stay at the state of rest.
  3. The energy of a photon is represented as E = hv, where h = Planck's Constant = 6.626 × 10-34 m2kg/s and v = frequency. You can further represent it in the form E = hc/λ, where c = speed of light and λ = wavelength of light.
  4. The formula for the momentum of a photon is represented as p = hv/c.
  5. Photons are massless.
  6. The number of photons per second gradually increases on increasing the intensity of the light of any given wavelength.
  7. They are electrically neutral.
  8. Photons are stable particles.
  9. They never decay on their own.
  10. Photons are not deflected by an electric or magnetic field.
  11. When photons interact with an electron or other subatomic particles, the collision that takes place is known as Compton Effect.
  12. In a photon-electron (or photo - other subatomic particles) collision, the complete energy and momentum are conserved.
  13. Photons transfer energy during a collision or interaction with other particles.

Things to Remember

  • The particle nature of light came into the picture when Albert Einstein looked up to the experiment conducted by Max Planck and observed that the wavelength and intensity of light have a certain impact on the ejected electrons.
  • In addition to the wave nature of light, the photoelectric effect experiment gave rise to another strange fact.
  • Photons, also known as light quantum or energy packets, are elementary particles i.e., constituents of energy.
  • In the simplest terms, a photon is a particle of light.
  • Photons are the basic constituent of light which is considered as the smallest quantum of radiation.

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Previous Year Questions

  1. If the X-ray tube is working at 20 kV then the minimum wavelength of X-rays will be….
  2. The initial momentum of electron will be...[NEET 2012]
  3. For photoelectric emission from certain metal the cutoff frequency is v. If radiation of frequency 2v… [NEET 2013]
  4. Electrons of mass mm with de-Broglie wavelength λ fall on the target in….[NEET 2016]
  5. An electron is accelerated under a potential difference of...[BITSAT 2011]

Sample Questions

Ques. The monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10-3 W.
a) What is the energy of a photon in the light beam?
b) How many photons per second, on average are emitted by the source? (3 marks)

Ans.

  1. Energy of a photon, E = hv = (6.63 × 10-34 ) (6.0 × 1014 ) = 3.98 × 10-19 J
  2. Considering 'N' as the number of photons emitted per second, the power P transmitted in the beam is equal to N times the energy per photon. 

i.e., P = NE

N = P/E = 2.0 × 10-3 W / 3.98 × 10-19 J

=> N = 5.0 × 1015 Photons per second

Ques. The wavelength of the light in the visible region is about 390 nm for violet color, about 550 nm (average wavelength) for yellow-green color, and about 760 nm for red color. 
a)Find the energies of photons in (eV) for the following cases-
1) Violet end
2) The average wavelength, yellow-green color and
3) Red end of the visible spectrum
b) From which of the photosensitive materials with work functions of Cs, K, and Na, and for the results of 1, 2, 3 of (a), can you build a photoelectric device that operates with visible light? (5 marks)

Ans.

(a) Energy of Photons = E = hv

=> E = (6.63 × 10-34 Js) (3 × 108 m/s)

= 1.989 × 10-25 Jm/λ

  1. For violet light of wavelength 390 nm,

E1 = 1.989 × 10-25 Jm/ 390 × 10-9 m = 5.10 × 10-19 J

=> E1 = 5.10 × 10-19 J/ 1.6 × 10-19 J/ eV = 3.19 eV

  1. For yellow-green light of wavelength 550 nm,

E2 = 1.989 × 10-25 Jm/ 550 × 10-9 m = 3.62 × 10-19 J = 2.26 eV

  1. For red light of wavelength 760 nm,

E3 = 1.989 × 10-25 Jm/ 760 × 10-9 m = 2.62 × 10-19 J = 1.64 eV

(b) For a photoelectric device to operate, we require incident light energy E to be equal to or greater than the work function of the material. So, a photoelectric device will operate with the violet light (with E = 3.19 eV) photosensitive material: Na, K, and Cs. 

It will also operate with yellow-green light (with E = 2.26 eV) for Cs only. But in the case of the red light (with E = 1.64 eV) it will not operate for any of these materials. 

Ques. Which experiment helped in the discovery of the 'Particle Nature of Light'? (1 mark)

Ans. The photoelectric effect experiment helped in the discovery of the particle nature of light.

Ques. What is the true nature of light? Is it a particle or wave? (1 mark)

Ans. Light possesses dual nature, which means it shows both particles as well as wave nature. 

Ques. Which effect takes place during the collision of a photon with an electron? (1 mark)

Ans. When a photon collides with an electron, the effect that takes place is called the Compton Effect. 

Ques. Why are photons not affected by the magnetic field? (1 mark)

Ans. Since photons are electrically neutral, i.e., they do not contain any charge in themselves, they only interact with other charged particles and not with one another. This is why they are not affected by the magnetic field.

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CBSE CLASS XII Related Questions

  • 1.
    In the figure, curved lines represent equipotential surfaces. A charge \( Q \) is moved along different paths A, B, C, and D. The work done on the charge will be maximum along the path:
     curved lines represent equipotential surfaces

      • A
      • B
      • C
      • D

    • 2.
      Write two characteristics of equipotential surfaces. A uniform electric field of 50 NC\(^{-1}\) is set up in a region along the \( x \)-axis. If the potential at the origin \( (0, 0) \) is 220 V, find the potential at a point \( (4m, 3m) \).


        • 3.

          Two slits 0.1 mm apart are arranged 1.20 m from a screen. Light of wavelength 600 nm from a distant source is incident on the slits. How far apart will adjacent bright interference fringes be on the screen? 


            • 4.
              The electric field (\( \vec{E} \)) and electric potential (\( V \)) at a point inside a charged hollow metallic sphere are respectively:

                • \( E = 0, \quad V = 0 \)
                • \( E = 0, \quad V = V_0 \text{ (a constant)} \)
                • \( E \ne 0, \quad V \ne 0 \)
                • \( E = E_0 \text{ (a constant)}, \quad V = 0 \)

              • 5.

                A current-carrying coil is placed in an external uniform magnetic field. The coil is free to turn in the magnetic field. What is the net force acting on the coil? Obtain the orientation of the coil in stable equilibrium. Show that in this orientation the flux of the total field (field produced by the loop + external field) through the coil is maximum. 


                  • 6.
                    A current flows through a cylindrical conductor of radius \( R \). The current density at a point in the conductor is \( j = \alpha r \) (along its axis), where \( \alpha \) is a constant and \( r \) is the distance from the axis of the conductor. The current flowing through the portion of the conductor from \( r = 0 \) to \( r = \frac{R}{2} \) is proportional to:

                      • \( R \)
                      • \( R^2 \)
                      • \( R^3 \)
                      • \( R^4 \)
                    CBSE CLASS XII Previous Year Papers

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