An electromagnetic radiation of wavelength 33 | vedclass

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(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Converting wavelength to meters: $\lambda = 331.5 	imes 10^{-9} \ m$.Energ

Vedclass · Accepted Answer

$2.5$

Vedclass · Answer

(D) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Converting wavelength to meters: $\lambda = 331.5 	imes 10^{-9} \ m$.Energy per photon: $E = \frac{6.63 	imes 10^{-34} 	imes 3 	imes 10^8}{331.5 	imes 10^{-9}} = 6 	imes 10^{-19} \ J$.Energy per mole: $E_{mol} = E 	imes N_A = 6 	imes 10^{-19} 	imes 6 	imes 10^{23} = 3.6 	imes 10^5 \ J \ mol^{-1}$.Using the photoelectric equation: $E_{mol} = W + KE_{mol}$, where $W$ is the work function.$W = 3.6 	imes 10^5 - 1.2 	imes 10^5 = 2.4 	imes 10^5 \ J \ mol^{-1}$.To convert to $eV$ per atom: $W_{eV} = \frac{2.4 	imes 10^5}{6 	imes 10^{23} 	imes 1.6 	imes 10^{-19}} = \frac{2.4 	imes 10^5}{9.6 	imes 10^4} = 2.5 \ eV$.

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