The work function of a metal is 2 eV. If a r | vedclass

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{max})$ is given by:$KE_{max} = \frac{hc}{\lambda} - \phi_0$Given:W

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$3.4 	imes 10^{-19} \ 	ext{J}$

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(C) According to Einstein's photoelectric equation,the maximum kinetic energy $(KE_{max})$ is given by:$KE_{max} = \frac{hc}{\lambda} - \phi_0$Given:Work function $\phi_0 = 2 \ 	ext{eV} = 2 	imes 1.6 	imes 10^{-19} \ 	ext{J} = 3.2 	imes 10^{-19} \ 	ext{J}$Wavelength $\lambda = 3000 \ 	ext{Å} = 3000 	imes 10^{-10} \ 	ext{m} = 3 	imes 10^{-7} \ 	ext{m}$Energy of incident photon $E = \frac{hc}{\lambda} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{3 	imes 10^{-7}} \ 	ext{J} = 6.6 	imes 10^{-19} \ 	ext{J}$Now,calculate $KE_{max}$:$KE_{max} = 6.6 	imes 10^{-19} \ 	ext{J} - 3.2 	imes 10^{-19} \ 	ext{J} = 3.4 	imes 10^{-19} \ 	ext{J}$

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