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AIIMS AIIMS Solved Paper-2014

  • question_answer
    Balmer gives an equation for wavelength of visible radiation of \[{{H}^{-}}\] spectrum as \[\lambda =\frac{k{{n}^{2}}}{{{n}^{2}}-4}\]. The value of k in terms of Rydberg's constant R A is

    A) R                                            

    B) 4R                  

    C) \[\frac{R}{4}\]                                  

    D) \[\frac{4}{R}\]                  

    Correct Answer: D

    Solution :

                     \[\frac{1}{\lambda }=R{{z}^{2}}\left( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right)\] For H-atom z = 1 For visible radiation, \[{{n}_{1}}=2\] \[\frac{1}{\lambda }=R\left( \frac{1}{{{2}^{2}}}-\frac{1}{{{n}^{2}}} \right)=\frac{R({{n}^{2}}-4)}{4{{n}^{2}}}\] but         \[{{\lambda }^{2}}=\frac{4{{X}^{2}}}{R({{n}^{2}}-4)}-\frac{k{{n}^{2}}}{{{n}^{2}}-4}\] \[K=\frac{4}{R}\]


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