A) \[{{R}_{0}}\]
B) \[\frac{2}{3}{{R}_{0}}\]
C) \[{{R}_{0}}/9\]
D) \[{{R}_{0}}/6\]
Correct Answer: C
Solution :
Activity \[R={{R}_{0}}{{e}^{-\lambda t}}\] \[\frac{{{R}_{0}}}{3}={{R}_{0}}{{e}^{-\lambda \times 9}}\Rightarrow {{e}^{-9\lambda }}=\frac{1}{3}\] ...(i) After further 9 years \[{R}'=R\,{{e}^{-\lambda t}}=\frac{{{R}_{0}}}{3}\times {{e}^{-\lambda \times 9}}\] ...(ii) From equation (i) and (ii) \[{R}'=\frac{{{R}_{0}}}{9}\].
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