NEET Sample Paper NEET Sample Test Paper-14

  • question_answer In the hydrogen atom the electron moves around the proton with a speed of\[~2.0\times 106\text{ }m{{s}^{-1}}\]in a circular orbit of radius \[5.0\times {{10}^{11}}m.\]What is the equivalent dipole moment?

    A) \[2\times {{10}^{-24}}A{{m}^{2}}\]                  

    B) \[4\times {{10}^{-24}}A{{m}^{2}}\]

    C) \[8\times {{10}^{-24}}A{{m}^{2}}\]                  

    D) \[16\times {{10}^{-24}}A{{m}^{2}}\]  

    Correct Answer: C

    Solution :

    The time period of electron motion in circular orbit is \[T=\frac{2\pi a}{v}=\frac{2\pi \times 5.0\times {{10}^{-11}}}{2.0\times {{10}^{6}}}=5\pi \times {{10}^{-17}}\sec .\] Therefore, the equivalent current is \[I=\frac{\text{Charge}}{\text{Time}}=\frac{e}{T}=\frac{1.6\times {{10}^{-19}}}{5\pi \times {{10}^{-17}}}=\frac{1.6}{5\pi }\times {{10}^{-2}}A\] Equivalent magnetic dipole moment = current\[\times \]area of circular orbit. \[=I\times (\pi {{a}^{2}})=\frac{1.6\times {{10}^{-2}}}{5\pi }\times \pi {{(5\times {{10}^{-11}})}^{2}}=8\times {{10}^{-24}}\,A{{m}^{2}}.\] Hence, the correction option is [c].


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