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:
A photon is a quantum of electromagnetic radiation. It carries momentum $p$ which is related to its wavelength $\lambda$.
Step 2: Detailed Explanation:
The momentum of a photon is given by $p = \frac{h}{\lambda}$. According to de Broglie's hypothesis, the wavelength $\lambda'$ of any particle (including a photon) is $\lambda' = \frac{h}{p}$. Substituting $p = h/\lambda$ into the de Broglie equation: $$\lambda' = \frac{h}{(h/\lambda)} = \lambda$$
Step 3: Final Answer:
The correct relation is $\lambda' = \lambda$.