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NEET Sample Paper NEET Sample Test Paper-71

  • question_answer
    The fossil bone has a \[^{14}C{{:}^{12}}C\] ratio, which is \[\left[ \frac{1}{16} \right]\] of that in a living animal bone. If the half-life of \[^{14}C\] is 5730 years,  then the age  of the fossil bone is                                                

    A) 11460 years                  

    B) 17190 years  

    C) 22920 years                  

    D) 45840 years     

    Correct Answer: C

    Solution :

    After n half-lives (i.e., at \[t=nT\]) the number of nuclides left undecayed, \[N={{N}_{0}}{{\left( \frac{1}{2} \right)}^{n}}\] \[Given\,\,=\,\,\frac{N}{{{N}_{0}}}=\frac{1}{16}\] \[\therefore \,\,\,\,\,\,\,\,\,\frac{1}{16}={{\left( \frac{1}{2} \right)}^{n}}\] \[{{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{n}}\] Equating the powers, we obtain \[n=4\] That is, \[\frac{t}{T}=4\] or \[t=4T\] or \[t= 4 \times  5730 = 22920 years\] \[\left( \because ~\,\,\,\,T= 5730 years \right)\]


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