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13.Nuclei
medium
Activity of radioactive element decreased to one third of original activity ${R_0}$ in $9$ years. After further $9$ years, its activity will be
A
${R_0}$
B
$\frac{2}{3}{R_0}$
C
${R_0}/9$
D
${R_0}/6$
Solution
(c) 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 $(ii)$ $R' = \frac{{{R_0}}}{9}$.
Standard 12
Physics
