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

  • question_answer
    If potential energy between a proton and an electron is given by \[|U|=k{{e}^{2}}/2{{R}^{3}}\], where K is the charge of electron and R is the radius of atom, then radius of Bohr's orbit is given by (h = Planck's constant, k=constant)  

    A)  \[\frac{k{{e}^{2}}m}{{{h}^{2}}}\]

    B)  \[\frac{6{{\pi }^{2}}}{{{n}^{2}}}\frac{k{{e}^{2}}m}{{{h}^{2}}}\]

    C)  \[\frac{2\pi }{n}\frac{k{{e}^{2}}m}{{{h}^{2}}}\]

    D)  \[\frac{4{{\pi }^{2}}k{{e}^{2}}m}{{{n}^{2}}{{h}^{2}}}\]

    Correct Answer: B

    Solution :

    [b]  \[U=-\frac{k{{e}^{2}}}{2{{R}^{2}}},F=-\frac{dU}{dR}=\frac{3k{{e}^{2}}}{2{{R}^{4}}}\] But, \[F=\frac{m{{v}^{2}}}{R}\Rightarrow \frac{m{{v}^{2}}}{R}=\frac{3k{{e}^{2}}}{2{{R}^{4}}}\] Also, \[mvR=\frac{nh}{2\pi }\] Solve to get: \[R=\frac{6{{\pi }^{2}}k{{e}^{2}}m}{{{n}^{2}}{{h}^{2}}}\]


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