A) \[4000years\]
B) \[5700years\]
C) \[8100years\]
D) \[6000years\]
Correct Answer: C
Solution :
We know that, \[\frac{dN}{dt}=\lambda .N\] \[\therefore \] \[1.18\times {{10}^{13}}=\frac{1}{T}\times \frac{1}{12}\times 6.023\times {{10}^{23}}\] or \[T=\frac{6.023\times {{10}^{23}}}{1.18\times {{10}^{13}}\times 12}\text{minutes}\] \[T=\frac{6.23\times {{10}^{23}}}{1.18\times {{10}^{13}}\times 12\times 60\times 24\times 365}years\] \[=8100years\]
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