A proton of mass ‘mp ’ has same energy as that of a photon of wavelength ‘λ’. If the proton is moving

Question

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 de Broglie wavelength to the wavelength of photon is.

(1) \(\frac{1}{c} \sqrt{\frac{2 E}{m_{p}}}\)

(2) \(\frac{1}{c} \sqrt{\frac{E}{m_{p}}}\)

(3) \(\frac{1}{c} \sqrt{\frac{E}{2 m_{p}}}\)

(4) \(\frac{1}{2 c} \sqrt{\frac{E}{m_{p}}}\)

Answer 1

Correct option is : (3) \(\frac{1}{c} \sqrt{\frac{E}{2 m_{p}}}\) 

E is missing in the question but considering E as energy, the solution will be

\(\mathrm{E}_{\text {phooon }}=\frac{\mathrm{hc}}{\lambda}=\mathrm{E} ; \ \mathrm{E}_{\text {proton }}=\frac{1}{2} \mathrm{~m}_{\mathrm{p}} \mathrm{v}^{2}=\mathrm{E}\) 

\( \frac{\lambda_{\text {proton }}}{\lambda_{\text {photon }}}=\frac{\mathrm{h} / \mathrm{p}}{\mathrm{hc} / \mathrm{E}}=\frac{\mathrm{h} / \sqrt{2 \mathrm{~m}_{\mathrm{p}} \mathrm{E}}}{\mathrm{hc} / \mathrm{E}}\) 

\(=\frac{E}{c \sqrt{2 m_{p} E}}\) 

\(\frac{\lambda_{\text {proton }}}{\lambda_{\text {photon }}}=\frac{1}{c} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}_{\mathrm{p}}}}\)