JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • 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) \[1.24\times {{10}^{15}}{{s}^{-1}}\]

    Correct Answer: D

    Solution :

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


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