JEE MAIN - Physics (2025 - 28th January Morning Shift - No. 8)
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.
$\frac{1}{c} \sqrt{\frac{E}{m_p}}$
$\frac{1}{\mathrm{c}} \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}$
$\frac{1}{\mathrm{2c}} \sqrt{\frac{ \mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}$
$\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}_{\mathrm{p}}}}$
Explanation
We know, $${E_{proton}} = {{hc} \over \lambda } = E$$
$${E_{proton}} = E$$ (given)
$$ \Rightarrow K{E_{proton}} = {1 \over 2}{m_P}{v^2} = E$$
So, $${{{\lambda _{proton}}} \over {{\lambda _{photon}}}} = {{h/p} \over {hc/E}} = {h \over p} \times {E \over {hc}}$$
$$ = {E \over {\sqrt {2{m_P}E} c}}$$ (As $$P = \sqrt {2m\,K.E.} $$)
$${{{\lambda _{proton}}} \over {{\lambda _{photon}}}} = {1 \over c}\sqrt {{E \over {2{m_P}}}} $$
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