The natural logarithm of the activity $R$ of a radioactive sample varies with time $t$ as shown. At $t=0$, there are $N_0$ undecayed nuclei. Then, $N_0$ is equal to [Take $e^2=7.5$ ]

• A

  $7500$
• B

  $3500$
• Correct answer
  C

  $75000$
• D

  $150000$

Solution

From the diagram at $t=0$ $\log _e R_0=2$
$\Rightarrow \quad R_0=e^2=7.5$ Since, $R_0=\lambda N_0$ ....(i) Since, $\lambda=$ Slope of graph $=\frac{1}{10 \times 10^3}=10^{-4} \mathrm{~s}^{-1}$ $\therefore$ From Eq. (i), $N_0=\frac{R_0}{\lambda}=\frac{7.5}{10^{-4}}=75000$

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