A) \[1.64\times {{10}^{10}}\text{ }yrs\]
B) \[7.14\times {{10}^{9}}\text{ }yrs\]
C) \[2.25\times {{10}^{9}}\text{ }yrs\]
D) None of these
Correct Answer: B
Solution :
\[t=\frac{1}{\lambda }\ln \frac{{{n}_{U}}+{{n}_{Pb}}}{{{n}_{U}}};{{n}_{U}}=no.\]of mole of U238 \[t=\frac{1}{\lambda }\ln \left( \frac{0.15}{0.05} \right);{{n}_{Pb}}=no.\] of mole of Pb206 \[t=\frac{1}{\lambda }\ln (3)=\frac{1.1}{0.693}\times 4.5\times {{10}^{9}}=7.14\times {{10}^{9}}\]years
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