Wave-Particle Duality of Light

Electromagnetic radiation behaves as both a particle and a wave known as Wave-Particle Duality. The waves associated the particles are called Matter Waves. In order for the electron to maintain moment, the perimeter of its orbit must be a multiple of its wavelength.

The energy of a photon (no rest mass) is given by the following formula:

\(E = hf\)

Where:

  • \(E\) is the energy
  • \(h\) is Planck's constant
  • \(f\) is the frequency

NOTE: Planck's constant can be represented as either \(6.626 \cdot 10^{-34} [\text{m}^2 \cdot \text{kg/s}]\) or \(4.136 \cdot 10^{-15} [\text{eVs}]\).


The wavelength is given by the following formula:

\(\lambda = \cfrac{hc}{E}\)

Where:

  • \(\lambda\) is the wavelength
  • \(h\) is Planck's constant
  • \(c\) is the speed of light (\(3.0 \times 10^8 [\text{m/s}]\))
  • \(E\) is the energy

The momentum is given by the following formula:

\(p = \cfrac{E}{c}\)

Where:

  • \(p\) is the momentum
  • \(E\) is the energy
  • \(c\) is the speed of light (\(3.0 \times 10^8 [\text{m/s}]\))

Compute the wavelength of a \(1 \; [\text{MeV}]\) photon.

We can determine the wavelength using the following formula:

\(\lambda = \cfrac{hc}{E}\)

Next, we can substitute the appropriate values and solve:

\(\lambda = \cfrac{(4.136 \cdot 10^{-15} \; [\text{eVs}])(2.9979 \cdot 10^{8} \; [\text{m/s}])}{1*10^6 \; [\text{eV}]} \)

\(\lambda = 1.24 \cdot 10^{-12} \; [\text{m}] \)

Therefore, we can determine the wavelength of the photon is \(\boldsymbol{1.24 \cdot 10^{-12} \; [\textbf{m}]}\).


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