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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted electron is $K_{\max} = \frac{hc}{\lambda} - \phi_0$,wher

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$\ge 2.8 	imes 10^{-9}\, m$

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(D) According to Einstein's photoelectric equation,the maximum kinetic energy of the emitted electron is $K_{\max} = \frac{hc}{\lambda} - \phi_0$,where $\lambda$ is the wavelength of incident light and $\phi_0$ is the work function.Given: $\lambda = 500\, nm$,$hc = 1240\, eV\, nm$,and $\phi_0 = 2.28\, eV$.$K_{\max} = \frac{1240\, eV\, nm}{500\, nm} - 2.28\, eV = 2.48\, eV - 2.28\, eV = 0.2\, eV$.The de Broglie wavelength $\lambda_e$ is related to kinetic energy $K$ by $\lambda_e = \frac{h}{\sqrt{2mK}}$. Since $K \le K_{\max}$,the minimum wavelength corresponds to the maximum kinetic energy: $\lambda_{\min} = \frac{h}{\sqrt{2mK_{\max}}}$.Using $h = 6.6 	imes 10^{-34}\, J\, s$,$m = 9.1 	imes 10^{-31}\, kg$,and $K_{\max} = 0.2 	imes 1.6 	imes 10^{-19}\, J$:$\lambda_{\min} = \frac{6.6 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 0.32 	imes 10^{-19}}} \approx 2.8 	imes 10^{-9}\, m$.Since $K \le K_{\max}$,it follows that $\lambda_e \ge \lambda_{\min}$. Thus,$\lambda_e \ge 2.8 	imes 10^{-9}\, m$.

Light of wavelength $500\, nm$ is incident on a metal with work function $2.28\, eV$. The de Broglie wavelength of the emitted electron is:

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