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

  • question_answer
    \[1.0g\] of a radioactive isotope was found to reduce to \[125\,mg\] after 24 hours. The half-life of the isotope is             [MP PET 1996]

    A)                 8 hours

    B)                 24 hours

    C)                 6 hours

    D)                 4 hours

    Correct Answer: A

    Solution :

           \[N={{\left[ \frac{1}{2} \right]}^{n}}\times {{N}_{o}}\]= 125 mg =\[{{\left( \frac{1}{2} \right)}^{n}}\times 1000\,mg\]                    \[{{\left( \frac{1}{2} \right)}^{n}}=\frac{125}{1000}=\frac{1}{8}\]                    \[{{\left( \frac{1}{2} \right)}^{n}}={{\left( \frac{1}{2} \right)}^{3}},n=3,\] so number, of \[{{t}_{1/2}}=3\]                                 Total time = 24 hours, Half-life time \[=\frac{24}{3}\]\[=8\,\text{hours}\].


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