A) 10439 years
B) 13094 years
C) 19039 years
D) 39049 years
Correct Answer: C
Solution :
Given: \[\frac{d{{N}_{0}}}{dt}=20\]decays/min \[\frac{dN}{dt}=2\] decays/min \[{{T}_{1/2}}=5730\] years As we know, \[N={{N}_{0}}{{e}^{-\lambda t}}\] \[Log\frac{{{N}_{0}}}{N}=\lambda t\] \[\therefore \]\[t=\frac{1}{\lambda }Log\frac{{{N}_{0}}}{N}\] \[=\frac{2.303\times {{T}_{1/2}}}{0.693}\times Lo{{g}_{10}}\frac{{{N}_{0}}}{N}\] But\[=\frac{\frac{d{{N}_{0}}}{dt}}{\frac{dN}{dt}}=\frac{{{N}_{0}}}{N}=\frac{20}{2}=10\] \[\therefore \]\[t=\frac{2.303\times 5730}{0.693}\times 1\] = 19039 yearsYou need to login to perform this action.
You will be redirected in
3 sec