BCECE Engineering BCECE Engineering Solved Paper-2004

  • question_answer
    The total energy of an electron in the first excited state of hydrogen is about -3.4 eV. Its               kinetic energy in this state is:

    A)  - 3.4 eV                               

    B)         - 6.8 eV                               

    C)         6.8 eV                                 

    D)         3.4 eV

    Correct Answer: D

    Solution :

    Kinetic energy of electron \[K=\frac{Z{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}\] Potential energy of electron \[U=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Z{{e}^{2}}}{r}\] \[\therefore \] Total energy \[E=K+U=\frac{Z{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}-\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\] or            \[E=-\frac{Z{{e}^{2}}}{8\pi {{\varepsilon }_{0}}r}\] or            \[E=-K\] or            \[K=-E=-(3.4)\] or            \[=3.4\,eV\]


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