BHU PMT BHU PMT (Screening) Solved Paper-2008

  • question_answer
    If the series limit wavelength of the Lyman series for hydrogen atom is \[912\overset{\text{o}}{\mathop{\text{A}}}\,\] then the series limit wavelength for the Balmer series for the hydrogen atom is

    A) \[912\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) \[912\times 2\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[912\times 4\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D) \[\frac{912}{2}\,\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

                     For series limit of Balmer series \[{{n}_{2}}=2,{{n}_{1}}=\infty \] \[\frac{1}{\lambda }=R\left[ \frac{1}{n_{2}^{2}}-\frac{1}{n_{1}^{2}} \right]\]                 \[=\left[ \frac{1}{{{(2)}^{2}}}-\frac{1}{{{(\infty )}^{2}}} \right]=\frac{R}{4}\] \[\therefore \]  \[\lambda =\frac{4}{R}\]                 \[=\frac{4}{10967800}m\]                 \[=4\times 912\times {{10}^{-10}}m\]                 \[=4\times 912{AA}\]


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