MHT CET · Physics · Dual Nature of Matter
The ratio of the wavelength of a photon of energy ' \(E\) ' to that of the electron of same energy is ( \(\mathrm{m}=\) mass of an electron, \(\mathrm{c}=\) speed of light, \(\mathrm{h}=\) Planck's constant)
- A \(\sqrt{\frac{\mathrm{m}}{\mathrm{cE}}}\)
- B \(\sqrt{\frac{2 \mathrm{~m}}{\mathrm{cE}}}\)
- C \(\mathrm{c} \sqrt{\frac{\mathrm{m}}{\mathrm{E}}}\)
- D \(\mathrm{c} \sqrt{\frac{2 \mathrm{~m}}{\mathrm{E}}}\)
Answer & Solution
Correct Answer
(D) \(\mathrm{c} \sqrt{\frac{2 \mathrm{~m}}{\mathrm{E}}}\)
Step-by-step Solution
Detailed explanation
We know,
Energy of a photon \(E=\frac{h c}{\lambda_p}\)
\(\therefore \quad \lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}\)
Wavelength of an electron \(\lambda_e=\frac{h}{m v}\) We also know, \(E=\frac{1}{2} m v^2 \Rightarrow v=\sqrt{\frac{2 E}{m}}\)
\(\begin{array}{ll}
\therefore \quad & m v=\sqrt{2 E m} \\
\therefore \quad & \lambda_e=\frac{h}{\sqrt{2 E m}} \\
& \Rightarrow \frac{\lambda_p}{\lambda_e}=\frac{h c}{E} \times \frac{\sqrt{2 E m}}{h}=c \sqrt{\frac{2 m}{E}}
\end{array}\)
Energy of a photon \(E=\frac{h c}{\lambda_p}\)
\(\therefore \quad \lambda_{\mathrm{p}}=\frac{\mathrm{hc}}{\mathrm{E}}\)
Wavelength of an electron \(\lambda_e=\frac{h}{m v}\) We also know, \(E=\frac{1}{2} m v^2 \Rightarrow v=\sqrt{\frac{2 E}{m}}\)
\(\begin{array}{ll}
\therefore \quad & m v=\sqrt{2 E m} \\
\therefore \quad & \lambda_e=\frac{h}{\sqrt{2 E m}} \\
& \Rightarrow \frac{\lambda_p}{\lambda_e}=\frac{h c}{E} \times \frac{\sqrt{2 E m}}{h}=c \sqrt{\frac{2 m}{E}}
\end{array}\)
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