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JEE Main & Advanced Chemistry Structure of Atom / परमाणु संरचना Question Bank Atomic models and Planck's quantum theory

  • question_answer
    The frequency of radiation emitted when the electron falls from \[n=4\] to \[n=1\] in a hydrogen atom will be (Given ionization energy of H\[=2.18\times {{10}^{-18}}J\,\,\,\text{ato}{{\text{m}}^{-1}}\]and \[h=6.625\times {{10}^{-34}}Js\])                                                [CBSE PMT 2004]

    A)                 \[3.08\times {{10}^{15}}{{s}^{-1}}\]        

    B)                 \[2.00\times {{10}^{15}}{{s}^{-1}}\]

    C)                 \[1.54\times {{10}^{15}}{{s}^{-1}}\]        

    D)                 \[1.03\times {{10}^{15}}{{s}^{-1}}\]

    Correct Answer: A

    Solution :

               \[{{E}_{\text{ionisation}}}={{E}_{\infty }}-{{E}_{n}}=\frac{13.6Z_{eff}^{2}}{{{n}^{2}}}eV\]                                   = \[\left[ \frac{13.6{{Z}^{2}}}{n_{2}^{2}}-\frac{13.6{{Z}^{2}}}{n_{1}^{2}} \right]\]                    \[E=h\nu =\frac{13.6\times {{1}^{2}}}{{{(1)}^{2}}}-\frac{13.6\times {{1}^{2}}}{{{(4)}^{2}}}\]; \[h\nu =13.6-0.85\]                    \[\because \]  \[h=6.625\times {{10}^{-34}}\]                                 \[\nu =\frac{13.6-0.85}{6.625\times {{10}^{-34}}}\times 1.6\times {{10}^{-19}}\] = \[3.08\times {{10}^{15}}{{s}^{-1}}\].


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