Particle Nature of Light: The Photon

Particle Nature Of Light: The Photon (Energy $ E = h\nu $, Momentum $ p = h/\lambda $)

The photoelectric effect provided strong evidence for the idea that light, in its interaction with matter, behaves as if its energy is concentrated in discrete packets or quanta. This quantum of light energy was later named the photon. The concept of the photon forms the basis of the particle theory of light.

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The Photon

Based on Einstein's explanation of the photoelectric effect, a photon is a discrete packet or bundle of energy associated with electromagnetic radiation. It is the fundamental quantum of light.

Key properties of a photon:

1. Energy: The energy ($E$) of a single photon is directly proportional to the frequency ($\nu$) of the electromagnetic radiation.

   $ E = h\nu $

   Where $h$ is Planck's constant ($h \approx 6.626 \times 10^{-34} \, J \cdot s$). This relation was originally proposed by Planck for the energy of radiation emitted or absorbed by oscillators in a blackbody, but Einstein extended it to the energy of the light itself. Using the relation between frequency and wavelength ($c = \nu\lambda$), where $c$ is the speed of light in vacuum:

   $ \nu = c/\lambda $

   The energy of a photon can also be expressed in terms of its wavelength:

   $ E = \frac{hc}{\lambda} $

   The energy of a photon is typically very small, so it is often expressed in electron volts (eV).
2. Momentum: Although a photon has zero rest mass, it carries momentum. The magnitude of the momentum ($p$) of a photon is given by:

   $ p = \frac{E}{c} $

   Substituting $E = h\nu$:

   $ p = \frac{h\nu}{c} $

   Using $\nu/c = 1/\lambda$:

   $ p = \frac{h}{\lambda} $

   The direction of the photon's momentum is the same as the direction of propagation of the electromagnetic wave. The pressure exerted by light (radiation pressure) is evidence of photons carrying momentum.
3. Speed: Photons always travel at the speed of light ($c$) in vacuum, regardless of the source's motion. In a medium, they travel at the speed of light in that medium ($v < c$).
4. Charge and Rest Mass: A photon is electrically neutral ($q=0$) and has zero rest mass ($m_0 = 0$).
5. Interaction: Photons interact with matter by being absorbed or emitted by charged particles (like electrons). In interactions like photoelectric effect or Compton scattering, energy and momentum are conserved in collisions between photons and particles, similar to collisions between classical particles.
6. Quantum Nature: Photons are bosons, obeying Bose-Einstein statistics. They are quanta of the electromagnetic field.

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Wave-Particle Duality of Light

The successful explanation of the photoelectric effect and other phenomena involving the interaction of light with matter (like Compton scattering) using the photon concept highlights the particle nature of light. However, phenomena like interference and diffraction can only be explained by treating light as a wave. This leads to the concept of wave-particle duality: light exhibits both wave-like properties and particle-like properties, depending on the experiment or phenomenon being observed. Neither the wave model alone nor the particle model alone can fully explain all aspects of light's behaviour.

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Answer:

Given:

Wavelength of yellow light, $\lambda = 580 \, nm = 580 \times 10^{-9} \, m = 5.80 \times 10^{-7} \, m$

Planck's constant, $h = 6.63 \times 10^{-34} \, J \cdot s$

Speed of light, $c = 3 \times 10^8 \, m/s$

First, calculate the frequency ($\nu$) of the light using $c = \nu\lambda$:

$ \nu = \frac{c}{\lambda} = \frac{3 \times 10^8 \, m/s}{5.80 \times 10^{-7} \, m} $

$ \nu = \frac{3}{5.80} \times 10^{15} \, Hz \approx 0.5172 \times 10^{15} \, Hz = 5.172 \times 10^{14} \, Hz $

Now, calculate the energy ($E$) of the photon using $E = h\nu$:

$ E = (6.63 \times 10^{-34} \, J \cdot s) \times (5.172 \times 10^{14} \, Hz) $

$ E = (6.63 \times 5.172) \times 10^{-20} \, J \approx 34.38 \times 10^{-20} \, J = 3.438 \times 10^{-19} \, J $

Alternatively, using $E = hc/\lambda$ directly:

$ E = \frac{(6.63 \times 10^{-34} \, J \cdot s) \times (3 \times 10^8 \, m/s)}{5.80 \times 10^{-7} \, m} $

$ E = \frac{19.89 \times 10^{-26}}{5.80 \times 10^{-7}} \, J = \frac{19.89}{5.80} \times 10^{-19} \, J \approx 3.43 \times 10^{-19} \, J $ (Slight difference due to rounding $\nu$)

The energy of the yellow light photon is approximately $3.43 \times 10^{-19}$ Joules.

Now, calculate the momentum ($p$) of the photon using $p = h/\lambda$:

$ p = \frac{6.63 \times 10^{-34} \, J \cdot s}{5.80 \times 10^{-7} \, m} $

$ p = \frac{6.63}{5.80} \times 10^{-27} \, kg \cdot m/s $ (Unit check: $J \cdot s/m = (N \cdot m) \cdot s / m = N \cdot s = (kg \cdot m/s^2) \cdot s = kg \cdot m/s$)

$ p \approx 1.143 \times 10^{-27} \, kg \cdot m/s $

Alternatively, using $p = E/c$ (with the calculated energy):

$ p = \frac{3.43 \times 10^{-19} \, J}{3 \times 10^8 \, m/s} \approx 1.143 \times 10^{-27} \, kg \cdot m/s $ (Unit check: $J / (m/s) = (N \cdot m) / (m/s) = N \cdot s = kg \cdot m/s$)

The momentum of the yellow light photon is approximately $1.143 \times 10^{-27} \, kg \cdot m/s$.