The work function of a metal is 2 text{ eV}. | 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}$$E = 6.6 	imes 10^{-19} 	ext{ J}$Now,$KE_{max} = E - \phi_0 = 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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