What is the work function of the metal if the | vedclass

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{4000 	imes 10^{-10}

Vedclass · Accepted Answer

$2.1$

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(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{4000 	imes 10^{-10}} = 4.9695 	imes 10^{-19} \, J$.The kinetic energy of the emitted photoelectron is $K.E. = \frac{1}{2}mv^2$.$K.E. = \frac{1}{2} 	imes 9 	imes 10^{-31} 	imes (6 	imes 10^5)^2 = 1.62 	imes 10^{-19} \, J$.According to Einstein's photoelectric equation,$E = \phi + K.E.$,where $\phi$ is the work function.$\phi = E - K.E. = 4.9695 	imes 10^{-19} - 1.62 	imes 10^{-19} = 3.3495 	imes 10^{-19} \, J$.To convert the work function into $eV$,divide by the charge of an electron $(1.6 	imes 10^{-19} \, C)$:$\phi = \frac{3.3495 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 2.093 \, eV \approx 2.1 \, eV$.

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