The counting rate observed from a radioactive | vedclass

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(C) The radioactive decay law is given by $A = A_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}$,where $A$ is the activity at time $t$,$A_0$ is the

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$200$

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(C) The radioactive decay law is given by $A = A_0 \left( \frac{1}{2} ight)^{\frac{t}{T_{1/2}}}$,where $A$ is the activity at time $t$,$A_0$ is the initial activity,and $T_{1/2}$ is the half-life.Given $A_0 = 1600$ counts/sec at $t = 0$ and $A = 100$ counts/sec at $t = 8$ seconds.Substituting these values: $100 = 1600 \left( \frac{1}{2} ight)^{\frac{8}{T_{1/2}}}$.$\frac{100}{1600} = \left( \frac{1}{2} ight)^{\frac{8}{T_{1/2}}} \implies \frac{1}{16} = \left( \frac{1}{2} ight)^{\frac{8}{T_{1/2}}}$.Since $\frac{1}{16} = (\frac{1}{2})^4$,we have $\frac{8}{T_{1/2}} = 4$,which gives $T_{1/2} = 2$ seconds.Now,to find the counting rate at $t = 6$ seconds:$A = 1600 \left( \frac{1}{2} ight)^{\frac{6}{2}} = 1600 \left( \frac{1}{2} ight)^3$.$A = 1600 	imes \frac{1}{8} = 200$ counts/sec.

The counting rate observed from a radioactive source at $t=0$ second was $1600$ counts per second and at $t=8$ seconds it was $100$ counts per second. The counting rate observed,as counts per second at $t=6$ seconds,will be

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