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

  • question_answer
    If the electron revolving around the nucleus in a radius 'r' with orbital speed 'v' has magnetic moment evr/2. Hence, using Bohr's postulate of the quantization of angular momentum obtain the magnetic moment (M) of hydrogen atom in its ground state and current (I) due to revolution of electron.

    A)  \[M=\frac{eh}{4\pi m},I=\frac{eV}{2\pi r}\]

    B) \[M=\frac{2eh}{5\pi m},I=\frac{eV}{4\pi r}\]

    C) \[M=\frac{h}{\pi m},I=\frac{e}{\pi r}\]

    D) \[M=\frac{eh}{\pi m},I=\frac{eV}{\pi r}\]

    Correct Answer: A

    Solution :

    [a] \[I=\frac{e}{T}=\frac{eV}{2\pi r}\] so, \[M=\frac{ev}{2\pi r}\times \pi {{r}^{2}}=\frac{evr}{2}\] According to Bohr's theory angular momentum \[mvr=\frac{nh}{2\pi }\text{ or }vr=\frac{nh}{2\pi m}\text{ so, M=}\frac{neh}{4\pi m}\] For the ground state \[n=1,\text{ so, }M=\frac{eh}{4\pi m}\]


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