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CMC Medical CMC-Medical Ludhiana Solved Paper-2011

  • question_answer
    An electron revolving in an orbit of radius 0.5 A in a hydrogen atom executes \[{{10}^{16}}\]rev / s. The magnetic moment of electron due to its orbital motion will be

    A)  \[1256\times {{10}^{-26}}\,A-{{m}^{2}}\]            

    B)  \[6.53\times {{10}^{-26}}\,A-{{m}^{2}}\]

    C)  zero                                     

    D)  \[256\times {{10}^{-26}}\,A-{{m}^{2}}\]

    Correct Answer: A

    Solution :

                    \[M=I\,A=(eV)\,\pi {{r}^{2}}\] \[=1.6\times {{10}^{-19}}\times {{10}^{16}}\times \frac{22}{7}\times {{(0.5\times {{10}^{-10}})}^{2}}\] \[=1256\times {{10}^{-26}}A{{m}^{2}}\]


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