JEE Main & Advanced Chemistry Surface & Nuclear Chemistry / भूतल और नाभिकीय रसायन Question Bank Rate of decay and Half-life

  • question_answer
    The radioactive decay of \[_{35}{{X}^{88}}\] by a beta emission produces an unstable nucleus which spontaneously emits a neutron. The final product is       [MNR 1995; CBSE 2001]

    A)                 \[_{37}{{X}^{88}}\]         

    B)                 \[_{35}{{Y}^{89}}\]

    C)                 \[_{34}{{Z}^{88}}\]         

    D)                 \[_{36}{{W}^{87}}\]

    Correct Answer: D

    Solution :

              \[_{35}{{X}^{88}}\xrightarrow{-\beta }{{\,}_{36}}{{W}^{88}}\to {{\,}_{36}}{{W}^{87}}+{{\,}_{o}}{{n}^{1}}\]


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