Q88 / 160 · JIPMER 2011 · Physics

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

  • Correct answer
    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

Solution

$R=R_0 e^{-\lambda t}$ $\Rightarrow \quad\left(\frac{1}{3}\right)=e^{-\lambda \times 3}=e^{-3 \lambda}$ ...(i) Again, $\frac{R^{\prime}}{R_0}=e^{-\lambda \times 9}=e^{-9 \lambda}=\left(e^{-3 \lambda}\right)^3$ $=\left(\frac{1}{3}\right)^3$ [From Eq. (i)] $=\frac{1}{27}$ $\therefore \quad R^{\prime}=\frac{R_0}{27}$ Hence, in 9 days activity will become $\left(\frac{1}{27}\right)$ of the original value.

Hence, the correct answer is option: 'A'.

Relevant Questions