A) 40 years
B) 60 years
C) 80 years
D) 100 years
Correct Answer: A
Solution :
Here \[N=2.5g,\] \[{{N}_{0}}=10g\] \[n=\frac{N}{{{N}_{0}}}=\frac{2.5}{10}=\frac{1}{4}={{\left( \frac{1}{2} \right)}^{2}}\] Again \[{{\left( \frac{1}{2} \right)}^{2}}={{\left( \frac{1}{2} \right)}^{n}}\]so, \[n=2\] The year after which it will decay \[=2\times {{T}_{1/2}}=2\times 20=40\,years\]You need to login to perform this action.
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