A) 1672 years
B) 2391 years
C) 3291 years
D) 4453 years
Correct Answer: B
Solution :
Given, for \[^{14}C\] \[{{A}_{0}}=16\] dis \[{{\min }^{-1}}{{g}^{-1}}\] A = 12 dis\[{{\min }^{-1}}{{g}^{-1}}\] \[{{t}_{1/2}}=5760\]years Now, \[\lambda =\frac{0.693}{{{t}_{1/2}}}\] \[\lambda =\frac{0.693}{5760}\]per year Then, from, \[t=\frac{2.303}{\lambda }{{\log }_{10}}\frac{{{A}_{0}}}{A}\] \[=\frac{2.303}{0.693}{{\log }_{10}}\frac{16}{12}\] \[=\frac{2.303\times 5760}{0.693}{{\log }_{10}}1.333\] \[=\frac{2.303\times 5760\times 0.1249}{0.693}=2390.81\approx 2391\]years.
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