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Manipal Medical Manipal Medical Solved Paper-2008

  • question_answer
    The binding energies of the atoms of elements A and B are\[{{E}_{a}}\]and\[{{E}_{b}}\]respectively. Three atoms of the element B fuse to give one atom of element A. This fusion process is accompanied by release of energy e. Then\[{{E}_{a}},{{E}_{b}}\]and e are related to each other as

    A)  \[{{E}_{a}}+e=3{{E}_{b}}\]

    B)  \[{{E}_{a}}=3{{E}_{b}}\]

    C)  \[{{E}_{a}}-e=3{{E}_{b}}\]

    D)  \[{{E}_{a}}+3{{E}_{b}}+e=0\]

    Correct Answer: A

    Solution :

     \[Q=\Sigma {{B}_{r}}-\Sigma {{B}_{p}}\] Here, \[\Sigma {{B}_{r}}=3\Sigma {{B}_{b}},\Sigma {{B}_{p}}={{E}_{a}},Q=e\] \[\therefore \] \[e=3{{E}_{b}}-{{E}_{a}}\] \[\therefore \] \[{{E}_{a}}+e=3{{E}_{b}}\]


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