Question

De Broglie wavelength of a quantum photon having energy \( E \)?

• A. \( \lambda = \frac{h}{E} \)
• B. \( \lambda = \frac{h}{\sqrt{E}} \)
• C. \( \lambda = \frac{E}{h} \)
• D. \( \lambda = \frac{h}{\sqrt{2E}} \)

Hint:

The De Broglie wavelength relates a particle’s wave-like behavior to its energy. For photons, \( \lambda = \frac{h}{E} \).

Answer 1

The Correct Option is A

De Broglie’s hypothesis states that every moving particle has a wave-like nature. The wavelength \( \lambda \) of a photon is related to its energy \( E \) by the following equation: \[ \lambda = \frac{h}{p} \] Where \( h \) is Planck's constant and \( p \) is the momentum. For a photon, the momentum can be related to its energy \( E \) by \( p = \frac{E}{c} \), where \( c \) is the speed of light. For a photon, the wavelength is given by: \[ \lambda = \frac{h}{E} \] Thus, the correct answer is \( \lambda = \frac{h}{E} \).