Question

A proton of mass 'mp' has same energy as that of a photon of wavelength 'λ'. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

• A. \( \frac{1}{c\sqrt{2m_p}} \frac{E}{\lambda} \)
• B. \( \frac{1}{c\sqrt{m_p}} \frac{E}{\lambda} \)
• C. \( \frac{1}{2c\sqrt{m_p}} \frac{E}{\lambda} \)
• D. \( \frac{1}{c\sqrt{2m_p}} \frac{2E}{\lambda} \)

Hint:

When equating energies of different particle types, ensure unit consistency and correct formula application for each type's specific energy expression.

Answer 1

The Correct Option is A

First, calculate the energy of the photon: \( E = \frac{hc}{\lambda} \).
The energy of the proton is equal to \( E = \frac{1}{2} m_p v^2 \).
Equate the two and solve for \( v \), and then find the de Broglie wavelength of the proton: \( \lambda_p = \frac{h}{m_p v} \).
The ratio of \( \lambda_p \) to \( \lambda \) yields the expression \( \frac{1}{c\sqrt{2m_p}} \frac{E}{\lambda} \).