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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2009

  • question_answer
    Energy released when one atom of uranium undergoes nuclear fission according to the following reaction is  (atomic mass of U= 235.060; n= 1.009; Ba = 143.881 and Kr= 89.947) about \[_{92}{{U}^{235}}{{+}_{0}}{{n}^{1}}{{\to }_{56}}B{{a}^{144}}{{+}_{36}}K{{r}^{90}}+{{2}_{0}}{{n}^{1}}\]

    A) 235 MeV             

    B) 208 MeV

    C) 931.5 MeV       

    D)        \[5.33\times {{10}^{23}}MeV\]

    E) 20.8 MeV

    Correct Answer: B

    Solution :

    Mass defect,\[\Delta m=Z\]mass of reactants\[-\Sigma \] mass of products \[\Delta m=235.060-(143.881+89.947+1.009)\] \[=0.223\] Binding energy or the energy released \[=0.223\times 931\,MeV\] \[=207.6MeV\] \[=208\text{ }MeV\]


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