A proton of mass $$'m_{p}'$$has same energy as that of a photon of wavelength $$'\lambda'$$. If the proton is moving at non- relativistic speed, then ratio of its deBroglie wavelength to the wavelength of photon is.

A proton of mass $$m_p$$ has the same energy $$E$$ as a photon of wavelength $$\lambda$$, and we seek the ratio of the proton’s de Broglie wavelength to the photon’s wavelength.

To begin, we express the photon energy in terms of its wavelength. For a photon, $$E = \frac{hc}{\lambda}$$, which implies $$\lambda = \frac{hc}{E}$$.

Next, we determine the proton’s de Broglie wavelength by first relating its kinetic energy to its momentum. In the non-relativistic regime, the proton’s energy is $$E = \frac{p^2}{2m_p}$$, from which the momentum follows as

$$p = \sqrt{2m_p E}$$

and hence its de Broglie wavelength is

$$\lambda_p = \frac{h}{p} = \frac{h}{\sqrt{2m_p E}}$$

Combining these expressions to form the desired ratio gives

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

The correct answer is Option D) $$\frac{1}{c}\sqrt{\frac{E}{2m_p}}$$.