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 :
\[R={{R}_{0}}{{e}^{-\lambda t}}\] \[\Rightarrow \] \[\left( \frac{1}{3} \right)={{e}^{-\lambda \times 3}}={{e}^{-3\lambda }}\] 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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