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Particle Nature of Light: The Photon

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Estimated time: 10 minutes
CBSE: Class 12

Definition: Photons

The photoelectric effect demonstrates that light behaves as if it consists of energy packets called quanta or photons.

CBSE: Class 12

Formula: Photons

E = hν

where:

  • E = energy of one photon
  • h = Planck’s constant = 6.626 × 10-34 J s
  • ν = frequency of radiation
CBSE: Class 12

Key Properties of Photons

  • A photon has energy equal to hν.
  • A photon moves with the speed of light in vacuum, that is, c.
  • Photons are electrically neutral, so they are not deflected by electric or magnetic fields.
  • All photons of the same frequency have the same energy.
  • Increasing intensity increases the number of photons, not the energy of each photon.

Momentum of a Photon

A photon also carries momentum:

p = \[\frac {E}{c}\] = \[\frac {hν}{c}\] = \[\frac {h}{λ}\]

where:

  • p = momentum of the photon
  • λ = wavelength of radiation
CBSE: Class 12

Importance of Photon

The photon model is important because it explains the photoelectric effect, where electrons are emitted from a metal surface when light of suitable frequency falls on it.

Cause-and-Effect Logic

  • Each photon interacts with one electron.
  • If photon energy is less than the work function, no electron is emitted.
  • If the photon energy is greater than the work function, the electron is emitted with kinetic energy.

Photoelectric Equation

Kmax = hν − ϕ

where:

  • Kmax​ = maximum kinetic energy of emitted electron
  • ϕ = work function of the metal
CBSE: Class 12

Example 1

  • The laser emits monochromatic light of frequency 6.0 × 1014 Hz with power 2.0 × 10−3 W.
  • Part (a) finds the energy of one photon using E = hν, giving E = 3.98 × 10−19 J.
  • Part (b) finds the number of photons per second using P = NE ⇒ N = P/E, giving N ≈ 5.0 × 1015 photons per second.

So this example shows that, given the light’s frequency and power, you can calculate the energy of each photon and the number of photons emitted per second.

CBSE: Class 12

Example 2

  • The work function of caesium is 2.14 eV; part (a) uses ϕ = hν0​ to get the threshold frequency ν0 = 5.16 × 1014 Hz.
  • This means: below this frequency, no photoelectrons are emitted regardless of intensity.
  • Part (b) uses Einstein’s equation eV0 = hν − ϕ = hc/λ − ϕ with a stopping potential of 0.60 V to find the wavelength of the incident light, which comes out as 454 nm.

So this example shows how work function, threshold frequency, stopping potential, and wavelength are linked in photoelectric effect numericals.

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