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BVP Medical BVP Medical Solved Paper-2002

  • question_answer
    Mass of an a-particle is 4.0028 amu, its binding energy per nucleon will be (given, \[{{m}_{p}}=1.0073\] and  \[{{m}_{n}}=1.0087\]):

    A)  \[2.86MeV\]    

    B)  \[6.79MeV\]

    C)  \[4.46MeV\]                    

    D)  \[8.24MeV\]

    Correct Answer: B

    Solution :

                    Mass of \[\alpha -\]-particle \[{{(}_{2}}H{{e}^{4}})=\] = mass of 2 protons + mass of 2 neutrons                 \[=2\times 1.0073+2\times 1.0087\]                 \[=4.032amu\] Mass decay \[=4.032-4.0028\]                                 \[=0.02992amu\] Binding energy = mass defect \[\times \text{ }931MeV\]                 \[=0.0292\times 931\]                 \[=27.1852MeV\] Binding energy per nucleon \[=\frac{27.1852}{4}=6.7963MeV\]


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