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(A) The kinetic energy $K$ of an electron with de Broglie wavelength $\lambda_b$ is given by $\lambda_b = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.Rearrang

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$0.05$

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(A) The kinetic energy $K$ of an electron with de Broglie wavelength $\lambda_b$ is given by $\lambda_b = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$.Rearranging for $K$,we get $K = \frac{h^2}{2m\lambda_b^2}$.Using the photoelectric equation: $\frac{hc}{\lambda} = K + \phi$,where $\phi$ is the work function.Substituting the values: $\phi = \frac{hc}{\lambda} - \frac{h^2}{2m\lambda_b^2}$.Given $\lambda = 800 	imes 10^{-9} \ m$ and $\lambda_b = 1 	imes 10^{-9} \ m$:$\phi = \frac{(6.63 	imes 10^{-34} \ J \cdot s)(3 	imes 10^8 \ m/s)}{800 	imes 10^{-9} \ m} - \frac{(6.63 	imes 10^{-34} \ J \cdot s)^2}{2(9.11 	imes 10^{-31} \ kg)(1 	imes 10^{-9} \ m)^2}$.$\phi = (2.486 	imes 10^{-19} \ J) - (2.412 	imes 10^{-19} \ J) = 0.074 	imes 10^{-19} \ J$.Converting to electron-volts: $\phi = \frac{0.074 	imes 10^{-19} \ J}{1.6 	imes 10^{-19} \ J/eV} \approx 0.046 \ eV \approx 0.05 \ eV$.

In a photoelectric experiment,light of wavelength $800 \ nm$ produces photoelectrons with the smallest de Broglie wavelength of $1 \ nm$. Then the work function of the metal used in the experiment is nearly (in $eV$)

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