Question

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 = 6.626x10^-34 Js, Na = 6.022 $\times$ 1023 mol^-1)

• A. $7.50 \times 10^4 s^{-1}$
• B. $4.76 \times 10^{14} s{-1}$
• C. $3.15 \times 10^{15} s{-1}$
• D. $1.24 \times 10^{15} s{-1}$

Answer 1

Answer: (D)

$\Delta E = hv $

$v =\frac{\Delta E }{ h }$

$v =\frac{495.5 \times 10^{3} Joule }{6.023 \times 10^{23}} \times \frac{1}{6.626 \times 10^{-34}}$

$v =1.24 \times 10^{15} \,sec ^{-1}$