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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    The binding energy of the deuterium is L 2.23 MeV. The mass defect in amu is

    A)  0.0024                                 

    B)  0.0012

    C)  2.23                      

    D)         0.024

    Correct Answer: A

    Solution :

    1 amu=931 MeV Here: binding energy \[\Delta \]E=2.23 MeV \[\therefore \]mass defect                 \[\Delta \,m=\frac{2.23\,Me\,V}{931\,MeV}=0.0024\,amu\]


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