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{\frac{E}{2m_p}} $
• 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

Let $ E $ be the energy of both the photon and the proton.

Photon:
Let $ \lambda $ be the wavelength of the photon. The energy of the photon is given by:
$$ E = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E} $$

Proton:
The de Broglie wavelength of a proton is given by:
$$ \lambda_p = \frac{h}{p} $$
where $ p $ is the momentum of the proton.

The kinetic energy of the proton is:
$$ E = \frac{p^2}{2m_p} \Rightarrow p^2 = 2m_p E \Rightarrow p = \sqrt{2m_p E} $$

Substituting for $ p $, the de Broglie wavelength becomes:
$$ \lambda_p = \frac{h}{\sqrt{2m_p E}} $$

Finding the Ratio $ \frac{\lambda_p}{\lambda} $:

$$ \frac{\lambda_p}{\lambda} = \frac{\frac{h}{\sqrt{2m_p E}}}{\frac{hc}{E}} = \frac{h}{\sqrt{2m_p E}} \times \frac{E}{hc} = \frac{E}{c \sqrt{2m_p E}} = \frac{\sqrt{E}}{c \sqrt{2m_p}} = \frac{1}{c} \sqrt{\frac{E}{2m_p}} $$

Final Answer:
The final answer is $ \frac{1}{c} \sqrt{\frac{E}{2m_p}} $.