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AMU Medical AMU Solved Paper-2012

  • question_answer
    The electron of a hydrogen atom revolves round the proton in a circular n^ orbit of radius\[{{r}_{n}}={{\varepsilon }_{0}}{{n}^{2}}{{h}^{2}}/(\pi m{{e}^{2}})\] with a speed \[{{v}_{n}}=\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}nh}\]. The current due to the circulating charge is proportional to

    A)                  \[{{e}^{2}}\]                                     

    B)                  \[{{e}^{3}}\]

    C)                  \[{{e}^{5}}\]                                     

    D)                  \[{{e}^{6}}\]

    Correct Answer: C

    Solution :

                     Time period of electron, \[T=\frac{2\pi r}{v}\]                                 \[=\frac{\frac{2\pi \times {{\varepsilon }_{0}}\,\,{{n}^{2}}{{h}^{2}}}{\pi m{{e}^{2}}}}{\frac{{{e}^{2}}}{2{{\varepsilon }_{0}}nh}}\]                                 \[T=\frac{4\varepsilon _{0}^{2}{{n}^{2}}{{h}^{3}}}{m{{e}^{4}}}\] Current, \[i=\frac{e}{T}=\frac{e}{\frac{4\varepsilon _{0}^{2}{{n}^{2}}{{h}^{2}}}{m{{e}^{4}}}}\]                 \[i=\frac{m{{e}^{5}}}{4\varepsilon _{0}^{2}{{n}^{2}}{{h}^{5}}}\] i.e.,        \[i\propto {{e}^{5}}\]


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