Ionization energy of gaseous Na atoms is 499 .5 kJ $m o l^{- 1}$ . The lowest possible frequency of light that ionizes a sodium atom is $(G i v e n : h = 6.626 \times 10^{- 34} J s, N_{A} = 6.022 \times 10^{23} m o l^{- 1}$

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

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

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

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

answer is D.

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Detailed Solution

Ionization energy per gaseous atom is

$E = (\frac{495.5 \times 10^{3} J m o l^{- 1}}{6.022 \times 10^{23} m o l^{- 1}}) = 8.228 \times 10^{- 19} J; v = \frac{E}{h} = \frac{(8.228 \times 10^{- 19} J}{(6.626 \times 10^{- 34} J s)} = 1.24 \times 10^{15} s^{- 1}$

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