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(D) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.For one mole of photons,the energy is $E_{mol} = \frac{hcN_A}{\lambda}$.$E_{mol

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The energy of a single photon is given by $E = \frac{hc}{\lambda}$.For one mole of photons,the energy is $E_{mol} = \frac{hcN_A}{\lambda}$.$E_{mol} = \frac{6.626 	imes 10^{-34} \, J \, s 	imes 3 	imes 10^8 \, m \, s^{-1} 	imes 6.022 	imes 10^{23} \, mol^{-1}}{300 	imes 10^{-9} \, m} = 3.99 	imes 10^5 \, J \, mol^{-1}$.According to the photoelectric effect equation,$E_{photon} = \Phi + KE$,where $\Phi$ is the work function (minimum energy to remove an electron).$\Phi = E_{mol} - KE = (3.99 	imes 10^5 - 1.68 	imes 10^5) \, J \, mol^{-1} = 2.31 	imes 10^5 \, J \, mol^{-1}$.

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