Ionization energy of gaseous Na atoms is 495.5 kJ mol−1. The lowest possible frequency of light that ionizes a sodium atom is (Given h=6.626×10−34Js,NA=6.022×1023mol−1

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Ionization energy of gaseous Na atoms is $495.5 kJ {mol}^{- 1}$ . The lowest possible frequency of light that ionizes a sodium atom is (Given $h = 6 .626 \times 10^{- 34} Js, N_{A} = 6 .022 \times 10^{23} {mol}^{- 1})$

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$3.15 \times 10^{15} s^{- 1}$

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

$7.50 \times 10^{4} s^{- 1}$

$4.76 \times 10^{14} s^{- 1}$

answer is D.

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Detailed Solution

Ionization energy per gaseous atom is

$E = (\frac{495 .5 \times 10^{3} {Jmol}^{- 1}}{6 .022 \times 10^{23} {mol}^{- 1}}) = 8 .228 \times 10^{- 19} J; v = \frac{E}{h} = \frac{(8 .228 \times 10^{- 19} J)}{(6 .626 \times 10^{- 34} Js)} = 1 .24 \times 10^{15} s^{- 1}$