Question

An activity of a radioactive sample decreases to $(1/3{)}^{rd}$ of its original value in $3$ days. Then, in $9$ days its activity will become:

• A. $\left(1/27\right)$ of the original value
• B. $\left(1/9\right)$ of the original value
• C. $\left(1/18\right)$ of the original value
• D. $\left(1/3\right)$ of the original value

Answer 1

Answer: (A)

$\left(1/27\right)$ of the original value

$\frac{R}{R_0}=e^{-\lambda t}$
Activity of a radioactive sample decreases to one third of its original value in 3 days.
$\frac{1}{3}=e^{-\lambda \times 3}=e^{-3\lambda }$ ....... (1)
Let activity in 9 days be R′. Then
$\frac{R^{\prime }}{R_0}=e^{-\lambda \times 9}$
$\frac{R^{\prime }}{R_0}=e^{-\lambda \times 3\times 3}$
$\frac{R^{\prime }}{R_0}=(e^{-3\lambda }{)}^3$
Substitute equation (1) in the above equation.
$\frac{R^{\prime }}{R_0}={\left(\frac{1}{3}\right)}^3$
$\frac{R^{\prime }}{R_0}=\frac{1}{27}$
$R^{\prime }=\frac{R_0}{27}$
Then, in 9 days its activity will become (1/27)${}^{\text{th}}$ of the original value.