A) 1485 years
B) 1845 years
C) 530 years
D) 4767 years.
Correct Answer: B
Solution :
Radioactive decay follows first order kinetics, there-fore, |
Decay constant \[(\lambda )=\frac{0.693}{{{t}_{1}}/2}=\frac{0.693}{5730}\] |
Given, \[{{R}_{0}}=100\] \[\therefore \,\,\,R=80\] |
and \[t=\frac{2.303}{\lambda }\log \frac{{{[R]}_{0}}}{[R]}=\frac{2.303}{\left( \frac{0.693}{5730} \right)}\log \frac{100}{80}\] |
\[=\frac{2.303\times 5730}{0.693}\times 0.0969=1845\,years\] |
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