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(B) Energy incident $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3.0 	imes 10^{8}}{248 	imes 10^{-9}} \, J$Converting to $eV$: $E = \

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$9$

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(B) Energy incident $E = \frac{hc}{\lambda} = \frac{6.63 	imes 10^{-34} 	imes 3.0 	imes 10^{8}}{248 	imes 10^{-9}} \, J$Converting to $eV$: $E = \frac{6.63 	imes 10^{-34} 	imes 3.0 	imes 10^{8}}{248 	imes 10^{-9} 	imes 1.6 	imes 10^{-19}} \approx 5.0 \, eV$Using Einstein's photoelectric equation: $E = \phi + K.E.$$5.0 \, eV = 3.0 \, eV + K.E. \implies K.E. = 2.0 \, eV$$K.E. = 2.0 	imes 1.6 	imes 10^{-19} \, J = 3.2 	imes 10^{-19} \, J$De-Broglie wavelength $\lambda = \frac{h}{\sqrt{2mK.E.}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 9.1 	imes 10^{-31} 	imes 3.2 	imes 10^{-19}}}$$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{58.24 	imes 10^{-50}}} = \frac{6.63 	imes 10^{-34}}{7.63 	imes 10^{-25}} \approx 0.868 	imes 10^{-9} \, m$$\lambda \approx 8.68 	imes 10^{-10} \, m = 8.68 \, \mathring{A}$Rounding to the nearest integer,we get $9 \, \mathring{A}$.

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