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12th Class Chemistry Nuclear Chemistry / नाभिकीय रसायन Question Bank Critical Thinking Nuclear chemistry

  • question_answer
    The activity of carbon-14 in a piece of an ancient wood is only 12.5%. If the half-life period of carbon-14 is 5760 years, the age of the piece of wood will be \[(\log \,2=0.3010)\] [MP PMT 1999]

    A)            \[17.281\times {{10}^{2}}\] years                                   

    B)            \[172.81\times {{10}^{2}}\] years

    C)            \[1.7281\times {{10}^{2}}\] years                                   

    D)            \[1728.1\times {{10}^{2}}\] years

    Correct Answer: B

    Solution :

           \[{{t}_{1/2}}\] of C?14 = 5760 year, \[\lambda =\frac{0.693}{5760},\]                    Now \[t=\frac{2.303}{\lambda }\log \frac{^{14}C\,\text{original}}{^{14}C\,\text{after}\,\text{time}\,\,t}\]                    \[=\frac{2.303\times 5760}{0.693}\log \frac{100}{12.5}=\frac{2.303\times 5760\times 0.9030}{0.693}\]                    = 17281= 172.81 × \[{{10}^{2}}\] years.  


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