Ionization energy of gaseous Na atoms is 495. | vedclass

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Question

(D) The energy required to ionize one atom of $Na$ is given by $E = \frac{IE}{N_A}$.$E = \frac{495.5 	imes 10^3 \ J \ mol^{-1}}{6.022 	imes 10^{23}

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$1.24 	imes 10^{15} \ s^{-1}$

Vedclass · Answer

(D) The energy required to ionize one atom of $Na$ is given by $E = \frac{IE}{N_A}$.$E = \frac{495.5 	imes 10^3 \ J \ mol^{-1}}{6.022 	imes 10^{23} \ mol^{-1}} = 8.228 	imes 10^{-19} \ J$.Using the relation $E = h
u$,the frequency $
u$ is $
u = \frac{E}{h}$.$
u = \frac{8.228 	imes 10^{-19} \ J}{6.626 	imes 10^{-34} \ J \ s} = 1.2417 	imes 10^{15} \ s^{-1}$.Thus,the lowest frequency is approximately $1.24 	imes 10^{15} \ s^{-1}$.

Ionization energy of gaseous $Na$ atoms is $495.5 \ kJ \ mol^{-1}$. The lowest possible frequency of light that ionizes a sodium atom is $(h = 6.626 \times 10^{-34} \ J \ s, N_A = 6.022 \times 10^{23} \ mol^{-1})$.

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