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AMU Medical AMU Solved Paper-2014

  • question_answer
    For the following fission reaction\[^{235}U+n{{\xrightarrow{{}}}^{140}}Ce{{+}^{94}}Zr+2n\],Find the disintegration energy. \[({{M}_{U}}=23502\,u,\,{{M}_{n}}=1.0\,u,\,\,{{M}_{Ce}}=139.9\,\,u\],\[{{M}_{Zr}}=93.9\,\,u)\]

    A)  205 MeV            

    B)  198 MeV

    C)  123 MeV            

    D)  89 MeV

    Correct Answer: A

    Solution :

                     The reaction is, \[^{235}U+n{{\xrightarrow{{}}}^{140}}Ce+\,{{\,}^{94}}Zr+2n\] The mass lost in this reaction (mass defect) is                 \[\Delta m={{M}_{U}}+{{M}_{n}}-({{M}_{Ce}}+{{M}_{Zr}}+2{{M}_{n}})\]                 \[\Delta m=[235.02+1.0-(139.9+93.9+2)]\,u\]                 \[\Delta m=(236.02-235.8)\,u\]                 \[\Delta m=0.22\,u\]                      [\[\because \,1amu=931\,MeV\]] \[\therefore \] The disintegration energy                 \[\Delta E=0.22\times 931=204.82\,MeV\]                 \[\cong 205.00\,MeV\]


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