BCECE Medical BCECE Medical Solved Papers-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) -3.4eV        

    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                 \[K=K+U=\frac{Z{{e}^{2}}}{8\pi {{\varepsilon }_{0}}}-\frac{Z{{e}^{2}}}{4\pi {{\varepsilon }_{0}}r}\] 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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