Dual Nature of Matter

JEE Main level, single correct, medium

12 of 200

Question 12 of 200

A photo-emissive substance is illuminated with a radiation of wavelength λi\lambda_i so that it releases electrons with de-Broglie wavelength λe\lambda_e. The longest wavelength of radiation that can emit photoelectron is λ0\lambda_0. Expression for de-Broglie wavelength is given by :
( m:\mathrm{m}: mass of the electron, h:\mathrm{h}: Planck's constant and cc : speed of light)

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Answer: (A) λe=h2mc(1λi−1λ0)\lambda_{\mathrm{e}}=\sqrt{\frac{\mathrm{h}}{2 \mathrm{mc}\left(\frac{1}{\lambda_{\mathrm{i}}}-\frac{1}{\lambda_0}\right)}}

K.E=E−Wλe=h2mK⋅E,E=hcλi,W=hcλ0 h22 mλe2=hcλi−hcλ0λe=h2mc(1λi−1λ0)\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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