A proton of mass $^{'} m_{p}^{'}$ 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.

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$\frac{1}{c} \sqrt{\frac{E}{2 m_{p}}}$

$\frac{1}{2 c} \sqrt{\frac{E}{m_{p}}}$

$\frac{1}{c} \sqrt{\frac{2 E}{m_{p}}}$

$\frac{1}{c} \sqrt{\frac{E}{m_{p}}}$

answer is C.

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Detailed Solution

E is missing in the question but considering E as energy, the solution will be
$\begin{aligned} E_{photon} = \frac{hc}{λ} = E; E_{proton} = \frac{1}{2} m_{p} v^{2} = E \\ \frac{λ_{proton}}{λ_{photon}} = \frac{h / p}{hc / E} = \frac{h / \sqrt{2 m_{p} E}}{hc / E} \\ = \frac{E}{c \sqrt{2 m_{p} E}} \end{aligned}$
$\frac{λ_{proton}}{λ_{photon}} = \frac{1}{c} \sqrt{\frac{E}{2 m_{p}}}$