J & K CET Engineering J and K - CET Engineering Solved Paper-2009

  • question_answer
    The mutual electrostatic potential energy between two protons which are at a distance of  \[9\times 10{{-}^{15}}\text{ }m,\] in \[_{92}{{U}^{235}}\] nucleus is

    A)  \[1.56\times {{10}^{-14}}J\]

    B)  \[5.5\times {{10}^{-14}}J\]

    C)  \[2.56\times {{10}^{-14}}J\]

    D)  \[4.56\times {{10}^{-14}}J\]

    Correct Answer: C

    Solution :

    Electrostatic potential energy \[U=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{r}\] Here, \[{{q}_{1}}={{q}_{2}}=1.6\times {{10}^{-19}}C\] and  \[r=9\times {{10}^{-15}}m\] \[\therefore \] \[U=\frac{9\times {{10}^{9}}\times 1.6\times {{10}^{-19}}\times 1.6\times {{10}^{-19}}}{9\times {{10}^{-15}}}\] \[=2.56\times {{10}^{-14}}J\]


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