BHU PMT BHU PMT Solved Paper-2002

  • question_answer
    The disintegration constant of a radioactive isotope whose half-life is 3 h is:

    A)  1.57 per hour

    B)  1.92 per hour

    C)  1.04 per hour

    D)  0.231 per hour

    Correct Answer: D

    Solution :

    Key Idea: Use the following formula \[\lambda =\frac{0.693}{{{t}_{1/2}}}\] where   \[\lambda =?\]                    \[{{t}_{1/2}}=3h\] Substituting the values,                 \[\lambda =\frac{0.693}{3}=0.23/h\]


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