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JEE Main & Advanced Physics Atomic Physics Question Bank Self Evaluation Test - Atoms

  • question_answer
    If the angular momentum of an electron in an orbit is J then the K.E. of the electron in that orbit is

    A)  \[\,\frac{{{J}^{2}}}{2m{{r}^{2}}}\]

    B)  \[\,\frac{Jv}{r}\]

    C)  \[\,\frac{{{J}^{2}}}{2m}\]

    D)  \[\,\frac{{{J}^{2}}}{2\pi }\]

    Correct Answer: A

    Solution :

    Angular momentum\[=mrv=J\text{    }\therefore v=\frac{{{J}^{2}}}{2m{{r}^{2}}}\] \[\text{K}\text{.E}\text{. of electron=}\frac{1}{2}m{{v}^{2}}=\frac{1}{2}m{{\left( \frac{J}{mr} \right)}^{2}}=\frac{{{J}^{2}}}{2m{{r}^{2}}}\]


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