CBSE 2024 · Region 1 · Set 1 · Q20 · 2 marks
Consider a neutron (mass m ) of kinetic energy E and a photon of the same energy. Let $\displaystyle \lambda_{\mathrm{n}}$ and $\displaystyle \lambda_{\mathrm{p}}$ be the de Broglie wavelength of neutron and the wavelength of photon respectively. Obtain an expression for $\displaystyle \frac{\lambda_{\mathrm{n}}}{\lambda_{\mathrm{p}}}$.
Marking-scheme solution
Obtaining an expression for \(\displaystyle \lambda_{\mathrm{n}} / \lambda_{\mathrm{p}}\) \[\begin{aligned}
& \mathrm{E}=\frac{\mathrm{hc}}{\lambda \mathrm{p}} \Rightarrow \lambda \mathrm{p}=\frac{\mathrm{hc}}{\mathrm{E}} \\
& \lambda \mathrm{n}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{(2 \mathrm{mE})}} \\
& \frac{\lambda \mathrm{n}}{\lambda \mathrm{p}}=\frac{\mathrm{h}}{\sqrt{ }(2 \mathrm{mE})} \times \frac{\mathrm{E}}{\mathrm{hc}} \\
& \frac{\lambda \mathrm{n}}{\lambda \mathrm{p}}=\sqrt{ }\left(\frac{\mathrm{E}}{2 \mathrm{~m} c^{2}}\right)
\end{aligned}
\]
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CBSE Class 12 Physics past-paper question from the 2024board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.