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NEET AIPMT SOLVED PAPER 1999

  • question_answer
                The half-life of \[_{6}{{C}^{14}},\] (\[\lambda =2.31\,\times {{10}^{4}}\]per year) is:                                                                               

    A)              \[2\times {{10}^{2}}\] year             

    B)              \[3\times {{10}^{3}}\]  year            

    C)              \[3.3\times {{10}^{4}}\] year 

    D)              \[4\times {{10}^{3}}\] year

    Correct Answer: B

    Solution :

                              Radioactive decay is first order reaction.                 So,          \[\lambda =\frac{0.693}{{{T}_{1/2}}}\]                         \[\therefore {{T}_{1/2}}=\frac{0.693}{2.31\times {{10}^{-4}}}\,year\]             \[=0.3\times {{10}^{4}}=3\times {{10}^{3}}\,year\]


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