JEE Mains · Physics · STD 12 - 11. Dual nature of radiation and matter
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 _______.
- A \(\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}\)
- B \(\frac{1}{\mathrm{c}} \sqrt{\frac{2 \mathrm{E}}{\mathrm{m}_{\mathrm{p}}}}\)
- C \(\frac{1}{2 c} \sqrt{\frac{E}{m_p}}\)
- D \(\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}_{\mathrm{p}}}}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{\mathrm{c}} \sqrt{\frac{\mathrm{E}}{2 \mathrm{~m}_{\mathrm{p}}}}\)
Step-by-step Solution
Detailed explanation
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…
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