A) \[(1/27)\] of the original value
B) \[(1/9)\] of the original value
C) \[(1/18)\] of the original value
D) \[(1/3)\]of the original value
Correct Answer: A
Solution :
\[R={{R}_{0}}{{e}^{-\lambda t}}\] \[\Rightarrow \] \[\left( \frac{1}{3} \right)={{e}^{-\lambda \times 3}}={{e}^{-3\lambda }}\] ? (i) Again, \[\frac{R'}{{{R}_{0}}}={{e}^{-\lambda \times 9}}={{e}^{-9\lambda }}={{({{e}^{-3\lambda }})}^{3}}\] \[={{\left( \frac{1}{3} \right)}^{3}}\] [from Eq.(i)] \[=\frac{1}{27}\] \[\Rightarrow \] \[R'=\frac{{{R}_{0}}}{27}\] Hence, in \[9\] days activity will become\[\left( \frac{1}{27} \right)\]of the original value.
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