Question

The counting rate observed from a radioactive source at $t = 0$ was $1600$ counts $s^{-1}$, and $t = 8 \,s$, it was $100$ counts $s^{-1}$. The counting rate observed as counts $s^{-1}$ at $t = 6$ s will be

• A. 250
• B. 400
• C. 300
• D. 200

Answer 1

Answer: (D)

200

Answer 2

As we know. $\left[\frac{N}{N_{0}}\right]\quad\quad\quad\quad\ldots\left(i\right)$ n = no. of half life N - no. of atoms left $N_{0}$- initial no. of atoms By radioactive decay law. k - disintegration constant $\therefore \frac{\frac{dN}{dt}}{\frac{dN_{0}}{dt}} = \frac{N}{N_{0}}\quad\quad\quad\ldots\left(ii\right)$ From $\left(i\right)$ and $\left(ii\right)$ we get $\frac{\frac{dN}{dt}}{\frac{dN_{0}}{dt}} = \left[\frac{1}{2}\right]^{n}$ or, $\left[\frac{100}{1600}\right] = \left[\frac{1}{2}\right]^{n}\quad \Rightarrow \left[\frac{1}{2}\right]^{4} = \left[\frac{1}{2}\right]^{n}$ $\therefore n = 4$, Therefore, in 8 seconds 4 half life had occurred in which counting rate reduces to 100 counts $s^{-1}$. $\therefore$ Half life, $T_{\frac{1}{2}} = 2\,sec$ In 6 sec, 3 half life will occur $\therefore \left[\frac{\frac{dN}{dt}}{1600}\right] = \left[\frac{1}{2}\right]^{3}$ $\Rightarrow \frac{dN}{dt} = 200$ counts $s^{-1}$