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NEET Physics Atomic Physics NEET PYQ-Atomic Physics

  • question_answer
    The total energy of electron in the ground state of hydrogen atom is \[-13.6\text{ }eV\]. The kinetic energy of an electron in the first excited state is:                                                                                             [AIPMT (S) 2007]

    A)        3.4 eV               

    B)       6.8 eV

    C)  13.6 eV 

    D)  1.7 eV

    Correct Answer: A

    Solution :

    Key Idea: Total energy of electron in the orbit is equal to negative of its kinetic energy.
    The energy of hydrogen atom when the electron revolves in nth orbit is
                \[E=\frac{-13.6}{{{n}^{2}}}eV\]
    In the ground state; \[n=1\]
    \[\therefore \]      \[E=\frac{-13.6}{{{1}^{2}}}=-13.6\,eV\]
    For        \[n=2,\,E=\frac{-13.6}{{{2}^{2}}}=-3.4\,eV\] 
    So, kinetic energy of electron in the first excited state (i.e., for \[n=2\]) is
    \[K=-E\,=(-3.4)=3.4\,\,eV\]


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