JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2025
- Medium
- MCQ
A photo-emissive substance is illuminated with a radiation of wavelength \(\lambda_i\) so that it releases electrons with de-Broglie wavelength \(\lambda_e\). The longest wavelength of radiation that can emit photoelectron is \(\lambda_0\). Expression for de-Broglie wavelength is given by :
( \(\mathrm{m}:\) mass of the electron, \(\mathrm{h}:\) Planck's constant and \(c\) : speed of light)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \mathrm{K}. \mathrm{E}=\mathrm{E}-\mathrm{W} \\ & \lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK} \cdot \mathrm{E}}}, \mathrm{E}=\frac{\mathrm{hc}}{\lambda_{\mathrm{i}}}, \mathrm{W}=\frac{\mathrm{hc}}{\lambda_0} \\ & \frac{\mathrm{~h}^2}{2 \mathrm{~m} \lambda_{\mathrm{e}}^2}=\frac{\mathrm{hc}}{\lambda_{\mathrm{i}}}-\frac{\mathrm{hc}}{\lambda_0} \\ & \lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}\end{aligned}\)
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