• # question_answer Ionization energy of gaseous Na atoms is $495.5\,\text{kJ}\,\text{mo}{{\text{l}}^{-1}}$ The lowest possible frequency of light that ionizes a sodium atom is $(h=6.626\times {{10}^{-34}}Js,{{N}_{A}}=6.022\times {{10}^{23}}mo{{l}^{-1}})$     JEE Main Online Paper (Held On 19 April 2016) 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) $1.24\times {{10}^{15}}{{s}^{-1}}$

Energy $={{N}_{A}}hv$ $495.5=6.023\times {{10}^{23}}\times 6.6\times {{10}^{-34}}\times v$ $v=\frac{495.5\times {{10}^{3}}J}{6.023\times {{10}^{23}}\times 6.6\times {{10}^{-34}}}=12.4\times {{10}^{14}}$ $=1.24\times {{10}^{15}}{{s}^{-1}}.$