A certain metal has a work function of Phi = | vedclass

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(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 400 \ nm = 400 	imes 10^{-9} \ m$:$E_1 = \frac{6.626 	imes 10^{-34}

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Only $400 \ nm$ light leads to the ejection of photoelectrons.

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(C) The energy of a photon is given by $E = \frac{hc}{\lambda}$.For $\lambda_1 = 400 \ nm = 400 	imes 10^{-9} \ m$:$E_1 = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{400 	imes 10^{-9}} \ J \approx 4.97 	imes 10^{-19} \ J$.Converting to $eV$: $E_1 = \frac{4.97 	imes 10^{-19}}{1.602 	imes 10^{-19}} \ eV \approx 3.1 \ eV$.For $\lambda_2 = 800 \ nm = 800 	imes 10^{-9} \ m$:$E_2 = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{800 	imes 10^{-9}} \ J \approx 2.48 	imes 10^{-19} \ J$.Converting to $eV$: $E_2 = \frac{2.48 	imes 10^{-19}}{1.602 	imes 10^{-19}} \ eV \approx 1.55 \ eV$.Since the work function $\Phi = 2 \ eV$,photoelectrons are ejected only if the incident photon energy $E \ge \Phi$.Here,$E_1 (3.1 \ eV) > 2 \ eV$ and $E_2 (1.55 \ eV) Therefore,only $400 \ nm$ light leads to the ejection of photoelectrons.

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