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(D) The energy of one photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{(6.626 	imes 10^{-34} \ J \ s)(3 	imes 10^8 \

Vedclass · Accepted Answer

$2.31 	imes 10^5 \ J \ mol^{-1}$

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(D) The energy of one photon is given by $E = \frac{hc}{\lambda}$.Substituting the values: $E = \frac{(6.626 	imes 10^{-34} \ J \ s)(3 	imes 10^8 \ m \ s^{-1})}{300 	imes 10^{-9} \ m} = 6.626 	imes 10^{-19} \ J$.For $1 \ mol$ of photons,the total energy is $E_{total} = E 	imes N_A = 6.626 	imes 10^{-19} 	imes 6.022 	imes 10^{23} \approx 3.99 	imes 10^5 \ J \ mol^{-1}$.According to the photoelectric effect equation,$E_{total} = \Phi + K.E.$,where $\Phi$ is the work function (minimum energy).$\Phi = E_{total} - K.E. = 3.99 	imes 10^5 \ J \ mol^{-1} - 1.68 	imes 10^5 \ J \ mol^{-1} = 2.31 	imes 10^5 \ J \ mol^{-1}$.

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