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JEE Main & Advanced Physics Atomic Physics Question Bank Topic Test - Atomic Physics

  • question_answer
    Every series of hydrogen spectrum has an upper and lower limit in wavelength. The spectral series which has an upper limit of wavelength equal to 18752 \[\overset{\text{o}}{\mathop{\text{A}}}\,\] is      

    A) Balmer series

    B) Lyman series

    C) Paschen series

    D) Pfund series

    Correct Answer: C

    Solution :

    (c)           \[\frac{1}{\lambda }=R\,\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] \[\Rightarrow \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}=\frac{1}{R\lambda }\]
    \[=\frac{1}{1.097\times {{10}^{7}}\times 18752\times {{10}^{-10}}}\]\[=0.0486=\frac{7}{144}.\] But
    \[\frac{1}{{{3}^{2}}}-\frac{1}{{{4}^{2}}}=\frac{7}{144}\Rightarrow {{n}_{1}}=3\] and n2 = 4          (Paschen series)
     


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