JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2025
- Medium
- MCQ
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 _______.
Pick an option to check your answer.
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
\(\begin{aligned}
=\lambda_p= & \frac{h}{P}=\frac{h}{\sqrt{2 m_p E}} \\
\text { Ratio } \frac{\lambda_p}{\lambda} & =\frac{h}{\sqrt{2 m_p E}} \times \frac{E}{h c} \\
& =\frac{1}{c} \sqrt{\frac{E}{2 m_p}}
\end{aligned}\)
\(\Rightarrow\) 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
\(\begin{aligned}
=\lambda_p= & \frac{h}{P}=\frac{h}{\sqrt{2 m_p E}} \\
\text { Ratio } \frac{\lambda_p}{\lambda} & =\frac{h}{\sqrt{2 m_p E}} \times \frac{E}{h c} \\
& =\frac{1}{c} \sqrt{\frac{E}{2 m_p}}
\end{aligned}\)
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