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(C) Let the energy of both the proton and the photon be $E$.For the photon,the energy is given by $E = \frac{hc}{\lambda_{photon}}$,so $\lambda_{photo

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$\frac{1}{c} \sqrt{\frac{E}{2 m_p}}$

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(C) Let the energy of both the proton and the photon be $E$.For the photon,the energy is given by $E = \frac{hc}{\lambda_{photon}}$,so $\lambda_{photon} = \frac{hc}{E}$.For the proton,the kinetic energy is $E = \frac{1}{2} m_p v^2$. The momentum of the proton is $p = \sqrt{2 m_p E}$.The de Broglie wavelength of the proton is $\lambda_{proton} = \frac{h}{p} = \frac{h}{\sqrt{2 m_p E}}$.The ratio of the de Broglie wavelength of the proton to the wavelength of the photon is:$\frac{\lambda_{proton}}{\lambda_{photon}} = \frac{h / \sqrt{2 m_p E}}{hc / E} = \frac{h}{\sqrt{2 m_p E}} 	imes \frac{E}{hc} = \frac{E}{c \sqrt{2 m_p E}} = \frac{1}{c} \sqrt{\frac{E}{2 m_p}}$.

$A$ proton of mass $m_p$ has the same energy as that of a photon of wavelength $\lambda$. If the proton is moving at non-relativistic speed,then the ratio of its de Broglie wavelength to the wavelength of the photon is:

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