JEE Main & Advanced Physics Atomic Physics Question Bank Atomic Structure

  • question_answer
    The wavelength of Lyman series is                          [BHU 1997]

    A)            \[\frac{4}{3\times 10967}cm\]                                         

    B)            \[\frac{3}{4\times 10967}cm\]

    C)            \[\frac{4\times 10967}{3}cm\]                                         

    D)            \[\frac{3}{4}\times 10967\ cm\]

    Correct Answer: A

    Solution :

               \[\frac{1}{\lambda }={{R}_{H}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right].\] For Lyman series n1=1 and n2=2, 3, 4, When n2=2, we get \[\lambda =\frac{4}{3{{R}_{H}}}=\frac{4}{3\times 10967}cm\]


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