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J & K CET Engineering J and K - CET Engineering Solved Paper-2010

  • question_answer
    10 g of a radioisotope is reduced to 1.25 g of . active material after 12 yr. Therefore, the half-life of the isotope (in yr) is

    A)  24

    B)  4

    C)  2

    D)  8

    Correct Answer: B

    Solution :

     Since, all radioactive process follows first order kinetics, \[k\text{ }=\text{ }\frac{2.303}{t}\log \frac{{{N}_{0}}}{N}\] \[=\frac{2.303}{12}\log \,8\] \[k=0.1733\] Half-life,\[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{0.1733}\] \[=3.99\] \[=4yr\]


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