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J & K CET Medical J & K - CET Medical Solved Paper-2004

  • question_answer
    If the mass defect of\[_{4}^{9}X\]is 0.090 amu, then binding energy per nucleon is: (1 amu =931.5 MeV)

    A)  9.315 MeV        

    B)  931.5 MeV

    C)  83.0 MeV                           

    D)  8.38 MeV

    Correct Answer: A

    Solution :

                    \[\Delta m=0.090\text{ }a\text{ }mu\] Number of nucleon = 9 \[\therefore \] binding energy per nucleon \[=\frac{total\text{ }binding\text{ }energy}{number\text{ }of\text{ }nucleon}\]               \[=\frac{\Delta M\times 931.5}{number\text{ }of\text{ }nucleon}\]                        \[=\frac{0.090\times 931.5}{9}=9.315\,MeV\]


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