Question

The ionisation energy of gaseous Na atom is $ 495.8\text{ }kJ\text{ }mo{{l}^{-1}} $ . The lowest possible frequency of light that can ionise a Na atom is:

• A. $ 1.24\times {{10}^{12}}{{s}^{-1}} $
• B. $ 3.15\times {{10}^{14}}{{s}^{-1}} $
• C. $ 4.76\times {{10}^{14}}{{s}^{-1}} $
• D. $ 1.24\times {{10}^{15}}{{s}^{-1}} $

Answer 1

Answer: (D)

$ E=hv $ , hence, $ v=\frac{E}{h} $ $ =\frac{495.8\times {{10}^{3}}J/\text{atom}}{6.02\times {{10}^{23}}\times 6.62\times {{10}^{-34}}Js} $ $ =1.24\times {{10}^{15}}{{s}^{-1}} $