The ionization energy of sodium in kJ mol^{- | vedclass

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Question

(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Using the formula $E = \frac{1240}{\lambda (nm)} \ eV$:$E = \frac{1240}{242} \ eV \app

Vedclass · Accepted Answer

$494$

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(A) The energy of a photon is given by $E = \frac{hc}{\lambda}$.Using the formula $E = \frac{1240}{\lambda (nm)} \ eV$:$E = \frac{1240}{242} \ eV \approx 5.124 \ eV$.To convert this to $J \ mol^{-1}$,we multiply by the conversion factor $1.602 	imes 10^{-19} \ J/eV$ and Avogadro's number $N_A = 6.022 	imes 10^{23} \ mol^{-1}$.$E = 5.124 	imes 1.602 	imes 10^{-19} 	imes 6.022 	imes 10^{23} \ J \ mol^{-1}$.$E \approx 494,300 \ J \ mol^{-1} = 494 \ kJ \ mol^{-1}$.

The ionization energy of sodium in $kJ \ mol^{-1}$,if electromagnetic radiation of wavelength $242 \ nm$ is just sufficient to ionize a sodium atom,is . . . . . . .

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