Manipal Medical Manipal Medical Solved Paper-2014

  • question_answer
    A tin nucleus has charge 50 eV. If the proton is at \[{{10}^{-12}}\]m from the nucleus. Then, the potential at this position will be

    A) \[7.2\times {{10}^{8}}V\]                                

    B) \[3.6\times {{10}^{4}}V\]

    C) \[14.4\times {{10}^{4}}V\]                              

    D) \[7.2\times {{10}^{4}}V\]

    Correct Answer: D

    Solution :

    Charge on the nucleus                           \[Q=50\,eV\]    \[=50\times 1.6\times {{10}^{-19}}=80\times {{10}^{-19}}\] Distance of proton from nucleus \[r={{10}^{-12}}m\] Potential of this position is given by \[V=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Q}{r}\]                    \[=9\times {{10}^{9}}\times \frac{80\times {{10}^{-19}}}{{{10}^{-12}}}\]                                 \[=7.2\times {{10}^{4}}V\]


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