Question

What are de Broglie waves? Show that the wavelength of the electromagnetic radiation is equal to the de Broglie wavelength of its quantum (photon).

Hint:

For a photon, the mass is zero, so we cannot use \(p = mv\). We must always use the relativistic relation \(p = E/c\).

Answer 1

Step 1: Understanding the Concept:
De Broglie proposed that matter, like radiation, exhibits dual nature. "Matter waves" or de Broglie waves are associated with moving particles. For a photon, the particle and wave descriptions must yield the same wavelength.

Step 2: Key Formula or Approach:
1. De Broglie Wavelength: \(\lambda = \frac{h}{p}\). 2. Photon Energy: \(E = hf = \frac{hc}{\lambda_{em}}\). 3. Einstein’s mass-energy relation for a photon: \(E = pc\).

Step 3: Detailed Explanation:
Definition: De Broglie waves are waves associated with moving material particles. The wavelength depends on the particle's momentum. Derivation for Photon: For a photon of electromagnetic radiation: From Planck's theory: \(E = \frac{hc}{\lambda_{em}}\) ...(1) From Einstein’s theory: \(E = pc \implies p = \frac{E}{c}\) ...(2) Substitute \(E\) from (1) into (2): \[ p = \frac{hc/\lambda_{em}}{c} = \frac{h}{\lambda_{em}} \] Rearranging for wavelength: \[ \lambda_{em} = \frac{h}{p} \] According to de Broglie, the wavelength of any "quantum" (particle) is \(\lambda_{dB} = \frac{h}{p}\). Comparing the two, we see \(\lambda_{em} = \lambda_{dB}\).

Step 4: Final Answer:
The wavelength of electromagnetic radiation (\(\lambda = c/f\)) is mathematically identical to the de Broglie wavelength of a photon (\(\lambda = h/p\)).