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(A) Given: Work function $\phi_0 = 4.2 	ext{ eV}$,Wavelength $\lambda = 2000 	ext{ Å}$.Energy of incident photon $E = \frac{12400}{\lambda(	ext{in

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$8.4 	imes 10^5 	ext{ m/s}$

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(A) Given: Work function $\phi_0 = 4.2 	ext{ eV}$,Wavelength $\lambda = 2000 	ext{ Å}$.Energy of incident photon $E = \frac{12400}{\lambda(	ext{in Å})} 	ext{ eV} = \frac{12400}{2000} = 6.2 	ext{ eV}$.According to Einstein's photoelectric equation,the maximum kinetic energy $K_{\max} = E - \phi_0$.$K_{\max} = 6.2 	ext{ eV} - 4.2 	ext{ eV} = 2 	ext{ eV}$.Converting kinetic energy to Joules: $K_{\max} = 2 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-19} 	ext{ J}$.Using $K_{\max} = \frac{1}{2} m v_{\max}^2$,where mass of electron $m = 9.1 	imes 10^{-31} 	ext{ kg}$.$v_{\max}^2 = \frac{2 	imes K_{\max}}{m} = \frac{2 	imes 3.2 	imes 10^{-19}}{9.1 	imes 10^{-31}} \approx 0.703 	imes 10^{12} 	ext{ m}^2/	ext{s}^2$.$v_{\max} = \sqrt{0.703 	imes 10^{12}} \approx 8.38 	imes 10^5 	ext{ m/s} \approx 8.4 	imes 10^5 	ext{ m/s}$.

Energy required to remove an electron from an aluminum surface is $4.2 \text{ eV}$. If light of wavelength $2000 \text{ Å}$ falls on the surface,the velocity of the fastest electron ejected from the surface will be:

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