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(D) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - \phi_0$Given: $\phi_0 = 1 \ eV$, $\lambda = 3000 \ \mathring{A} =

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$1 	imes 10^6 \ m/s$

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(D) According to Einstein's photoelectric equation: $K_{\max} = \frac{hc}{\lambda} - \phi_0$Given: $\phi_0 = 1 \ eV$, $\lambda = 3000 \ \mathring{A} = 3 	imes 10^{-7} \ m$.The energy of the incident photon is $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{3 	imes 10^{-7}} \ J = 6.63 	imes 10^{-19} \ J$.Converting to $eV$: $E = \frac{6.63 	imes 10^{-19}}{1.6 	imes 10^{-19}} \ eV \approx 4.14 \ eV$.$K_{\max} = 4.14 \ eV - 1 \ eV = 3.14 \ eV = 3.14 	imes 1.6 	imes 10^{-19} \ J = 5.024 	imes 10^{-19} \ J$.Using $K_{\max} = \frac{1}{2} m v_{\max}^2$, where $m = 9.1 	imes 10^{-31} \ kg$:$v_{\max} = \sqrt{\frac{2 K_{\max}}{m}} = \sqrt{\frac{2 	imes 5.024 	imes 10^{-19}}{9.1 	imes 10^{-31}}} \approx \sqrt{1.104 	imes 10^{12}} \approx 1.05 	imes 10^6 \ m/s$.Thus, the closest option is $1 	imes 10^6 \ m/s$.

The work function of a metal is $1 \ eV$. If light of wavelength $3000 \ \mathring{A}$ is incident on the surface of this metal, the maximum velocity of the emitted photoelectrons is ........

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