A) \[1.5\times {{10}^{9}}yr\]
B) \[2.25\times {{10}^{9}}yr\]
C) \[4.5\times {{10}^{8}}yr\]
D) \[9.0\times {{10}^{9}}yr\]
E) \[13.5\times {{10}^{9}}yr\]
Correct Answer: D
Solution :
All radioactive processes follow first order kinetics. Initially sample is free from lead. If its proportion is\[x,\]the amount of sample at time T is\[4x\]. \[\Rightarrow \] \[\frac{0.693}{4.5\times {{10}^{9}}yr}=\frac{2.303}{T}\log \frac{4x}{x}\] \[T=2\times 4.5\times {{10}^{9}}yr=9.0\times {{10}^{9}}yr\]
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