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(B) Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{310 	imes 10^{-9}} \approx 6.416 	imes 10^{

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$4$

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(B) Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{310 	imes 10^{-9}} \approx 6.416 	imes 10^{-19} \ J$. Converting to $eV$: $E = \frac{6.416 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 4.01 \ eV$. Kinetic energy $K.E. = E - \Phi = 4.01 \ eV - 3.55 \ eV = 0.46 \ eV$. $K.E. = 0.46 	imes 1.6 	imes 10^{-19} \ J = 0.736 	imes 10^{-19} \ J$. Using $K.E. = \frac{1}{2} m_e v^2$: $0.736 	imes 10^{-19} = \frac{1}{2} 	imes 9 	imes 10^{-31} 	imes v^2$. $v^2 = \frac{1.472 	imes 10^{-19}}{9 	imes 10^{-31}} \approx 0.1635 	imes 10^{12} = 16.35 	imes 10^{10}$. $v \approx 4.04 	imes 10^5 \ ms^{-1}$. Thus,$x \approx 4$.

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