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JEE Main & Advanced JEE Main Paper (Held On 19 April 2014)

  • question_answer
    A piece of wood from a recently cut tree shows 20 decays per minute. A wooden piece of same size placed in a museum (obtained from a tree cut many years back) shows 2 decays per minute. If half life of \[{{C}^{14}}\]is 5730 years, then age of the wooden piece placed in the museum is approximately:     JEE Main Online Paper (Held On 19 April 2016)

    A) 10439 years                       

    B) 13094 years

    C) 19039 years                       

    D) 39049 years

    Correct Answer: C

    Solution :

    Given: \[\frac{d{{N}_{0}}}{dt}=20\]decays/min \[\frac{dN}{dt}=2\] decays/min \[{{T}_{1/2}}=5730\] years As we know, \[N={{N}_{0}}{{e}^{-\lambda t}}\] \[Log\frac{{{N}_{0}}}{N}=\lambda t\] \[\therefore \]\[t=\frac{1}{\lambda }Log\frac{{{N}_{0}}}{N}\] \[=\frac{2.303\times {{T}_{1/2}}}{0.693}\times Lo{{g}_{10}}\frac{{{N}_{0}}}{N}\] But\[=\frac{\frac{d{{N}_{0}}}{dt}}{\frac{dN}{dt}}=\frac{{{N}_{0}}}{N}=\frac{20}{2}=10\] \[\therefore \]\[t=\frac{2.303\times 5730}{0.693}\times 1\] = 19039 years


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