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

  • question_answer
    The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of 108.5 nm. The ground state energy of an electron of this ion will be

    A) 3.4 eV

    B) 13.6 eV

    C) 54.4 eV

    D) 122.4 eV

    Correct Answer: C

    Solution :

    (c)           For third line of Balmer series \[{{n}_{1}}=2\], \[{{n}_{2}}=5\]
    \ \[\frac{1}{\lambda }=R{{Z}^{2}}\left[ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\] gives \[{{Z}^{2}}=\frac{n_{1}^{2}n_{2}^{2}}{(n_{2}^{2}-n_{1}^{2})\lambda R}\]
    On putting values Z = 2
    From \[E=-\frac{13.6{{Z}^{2}}}{{{n}^{2}}}=\frac{-13.6{{(2)}^{2}}}{{{(1)}^{2}}}=-54.4\,eV\]
     


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