Question

If \( \lambda \) be the wavelength of any electromagnetic radiation, the de-Broglie wavelength of its quantum (photon) is

• A. \( \frac{\lambda}{4} \)
• B. \( \lambda \)
• C. \( \frac{\lambda}{2} \)
• D. \( 2\lambda \)
• E. \( \frac{3\lambda}{4} \)

Hint:

Photon already follows de-Broglie relation.

Answer 1

The Correct Option is B

Concept: For photon: \[ \lambda = \frac{h}{p} \]

Step 1: Conclusion.
de-Broglie wavelength equals electromagnetic wavelength. \[ \lambda_{de\text{-}Broglie} = \lambda \]