the count rate of a radioactive source at t=0 was 1600 count/s and at t=8s it was 100 counts/s. The count rate (in counts) at t =6s was ----??

Dear Student,

Please find below the solution to the asked query:

Given that the initial count rate is N₀ = 1600 count/sec. Let the number of counts at any time is N. Then, According to the law of radioactivity,

$N=N_0 e^{-\lambda t}$
Where λ = disintegration constant.

It is also given that, after a time t = 8 sec. the count rate is 100 counts/sec.

$$\begin{gathered} N = N_0 e^{-\lambda t} \Rightarrow 100 = 1600 e^{-8\lambda } \Rightarrow e^{-8\lambda } = \frac{1}{16} \Rightarrow 8\lambda = 4 \ln \left(2\right) \\ \Rightarrow T_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{\ln \left(2\right)}{\lambda }=2 sec. \end{gathered}$$

Therefore, after t = 6 sec. The count rate is,

$N = 200 \raisebox{1ex}{$counts$}\!\left/ \!\raisebox{-1ex}{$sec$}\right.$

 

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards