Question

A proton and a photon have the same energy. Then the de-Broglie wavelength of proton \( \lambda_p \) and wavelength of photon \( \lambda_0 \) are related by:

• A. \( \lambda_0 \propto \frac{1}{\sqrt{\lambda_p}} \)
• B. \( \lambda_0 \propto \sqrt{\lambda_p} \)
• C. \( \lambda_0 \propto \lambda_p \)
• D. \( \lambda_0 \propto \lambda_p^2 \)
• E. \( \lambda_0 \propto \frac{1}{\lambda_p} \)

Hint:

The de-Broglie wavelength is inversely proportional to the momentum. For a photon, \( p = \frac{E}{c} \) and for a proton, \( p = mv \).

Answer 1

The Correct Option is D

Step 1: The de-Broglie wavelength \( \lambda \) for a particle is given by: \[ \lambda = \frac{h}{p} \] where \( p \) is the momentum of the particle.
Step 2: For a proton, \( p = mv \), where \( m \) is the mass and \( v \) is the velocity of the proton. For a photon, \( p = \frac{E}{c} \), where \( E \) is the energy and \( c \) is the speed of light. 
Step 3: Since both the proton and the photon have the same energy, we can relate their wavelengths using their respective momenta. For the photon, the wavelength is inversely proportional to the momentum, while for the proton, the momentum is proportional to its velocity. 
Thus, the de-Broglie wavelength of the proton and photon are related by: \[ \lambda_0 \propto \lambda_p^2 \]