Question

The relation between the wavelength of electromagnetic radiation ($\lambda$) and de Broglie wavelength of its quantum (photon) ($\lambda'$) is ______.

• A. $\lambda'>\lambda$
• B. $\lambda' = \lambda$
• C. $\lambda'<\lambda$
• D. $\lambda' = \frac{\lambda}{2}$

Hint:

For light, the "wave" wavelength and the "particle" de Broglie wavelength are identical. This shows the perfect consistency of the wave-particle duality.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Electromagnetic radiation can be treated as a wave with wavelength $\lambda$ or as a stream of photons with momentum $p$.
Step 2: Formula Application:
For a photon, energy $E = h\nu = hc/\lambda$. Also, momentum $p = E/c = h/\lambda$.
Step 3: Explanation:
According to de Broglie's hypothesis, the wavelength of any particle is $\lambda' = h/p$. Substituting $p = h/\lambda$ into this equation: $\lambda' = h / (h/\lambda) = \lambda$.
Step 4: Final Answer:
The wavelengths are equal, $\lambda' = \lambda$.