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J & K CET Medical J & K - CET Medical 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, \[k=\frac{2.303}{t}\log \frac{{{N}_{0}}}{N}\] \[=\frac{2.303}{12}\log \frac{10}{1.25}\] \[=\frac{2.303}{12}\log 8\] \[k=0.1733\] Half-life, \[{{t}_{1/2}}=\frac{0.693}{k}=\frac{0.693}{0.1733}\]                 \[=3.99\]                 \[\approx 4yr\]


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