MHT CET · Physics · Dual Nature of Matter
An electron and photon are accelerated through the same potential difference. The ratio of the de-Broglie wavelength \(\lambda_{\mathrm{p}}\) to \(\lambda_{\mathrm{e}}\) is \(\left[\mathrm{m}_{\mathrm{e}}=\right.\) mass of electron, \(\mathrm{m}_{\mathrm{p}}=\) mass of proton \(]\)
- A \(\left(\frac{m_{p}}{m_{e}}\right)^{\frac{1}{2}}\)
- B \(\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}\)
- C \(\left(\frac{m_{e}}{m_{p}}\right)\)
- D \(\left(\frac{m p}{m}\right)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}\)
Step-by-step Solution
Detailed explanation
Both will have same kinetic energy \(\mathrm{k}\). Momentum of electron \(\quad P_{\mathrm{e}}=\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{k}}\) Momentum of proton \(P_{p}=\sqrt{2 m_{p} k}\)
\(\begin{array}{l}
\lambda_{p}=\frac{h}{p_{p}}, \lambda_{e}=\frac{h}{p_{e}} \\
\therefore \frac{\lambda_{p}}{\lambda_{e}}=\frac{P_{e}}{P_{p}}=\frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}
\end{array}\)
\(\begin{array}{l}
\lambda_{p}=\frac{h}{p_{p}}, \lambda_{e}=\frac{h}{p_{e}} \\
\therefore \frac{\lambda_{p}}{\lambda_{e}}=\frac{P_{e}}{P_{p}}=\frac{\sqrt{m_{e}}}{\sqrt{m_{p}}}=\left(\frac{m_{e}}{m_{p}}\right)^{\frac{1}{2}}
\end{array}\)
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