JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Radioactivity and a,b rays

  • question_answer
    Calculate mass defect in the following reaction                 \[_{1}{{H}^{2}}{{+}_{1}}{{H}^{3}}{{\to }_{1}}H{{e}^{4}}{{+}_{0}}{{n}^{1}}\]                 (Given : mass \[{{H}^{2}}=2.014,\,{{H}^{3}}=3.016,\,He=4.004,\] \[n=1.008\,amu\])          [Kerala CET 2005]

    A)                 0.018 amu          

    B)                 0.18 amu

    C)                 0.0018 amu

    D)                 1.8 amu

    E)                 18 amu

    Correct Answer: A

    Solution :

                    Mass loss = mass of reactant ? mass of product.                                 \[=(2.014+3.016)-(4.004+1.008)\]                                 \[=5.030-5.012=0.018\,amu\]


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