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

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Aicharya · Accepted Answer

Energy of photon $=E=\frac{h c}{\lambda}$
$\Rightarrow$ Wavelength of photon $=\lambda=\frac{h c}{E}$
Energy of proton $=E=\frac{1}{2} m_p v^2=\frac{P^2}{2 m_p}$
$\Rightarrow$ Linear momentum of proton $=P=\sqrt{2 m_p E}$
Or de-Broglie wavelength of proton
$\begin{aligned}
=\lambda_p= & \frac{h}{P}=\frac{h}{\sqrt{2 m_p E}} \\ \text { Ratio } \frac{\lambda_p}{\lambda} & =\frac{h}{\sqrt{2 m_p E}} \times \frac{E}{h c} \\ & =\frac{1}{c} \sqrt{\frac{E}{2 m_p}}
\end{aligned}$

A proton of mass ' $m_{\mathrm{p}}$ ' has same energy as that of a photon of wavelength ' $\lambda$ '. If…

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