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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 31

  • question_answer
    The Binding energy of \[_{17}^{35}Cl\] nucleus is 298 MeV. Find its atomic mass. \[\left[ {{m}_{p}}=1.007825\text{ }amu,\text{ }{{\text{m}}_{n}}=1.008665\text{ }amu \right]\]

    A) 24.9 amu       

    B)        34.9 amu

    C) 54.9 amu       

    D)        35.289 amu

    Correct Answer: B

    Solution :

    [b] \[m=17{{m}_{p}}+18{{m}_{n}}\] \[=17\times 1.007825+18\times 1.008665\] = 35.289 \[\Delta m=\frac{298MeV}{931.2MeV/amu}=0.3200\,amu\] \[\therefore \text{ }Atomic\text{ }mass=m-\Delta m\] \[=35.289-0.32\] = 34.969 amu


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