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NEET Sample Paper NEET Sample Test Paper-85

  • question_answer
    The ionisation energy of hydrogen atom is 13.6 eV. Following Bohr's theory, the energy corresponding to a transition between the 3rd and the 4th orbit is

    A) \[3.40\,eV\]       

    B)        \[1.51\,eV\]

    C) \[0.85\,eV\]       

    D)        \[0.66\,eV\]

    Correct Answer: D

    Solution :

    [d]        \[{{E}_{3}}=\frac{-13.6}{{{3}^{2}}}=-1.5\,\,leV\] and       \[{{E}_{4}}=\frac{-13.6}{{{4}^{2}}}=-0.85\,\,eV\] \[\therefore \]      \[{{E}_{4}}-{{E}_{3}}=0.66\,\,eV\]


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