At t = 0,the counting rate from a radioactive | vedclass

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(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$,where $N(t)$ is the counting rate at time $t$.Given $N(0) = N_0 = 1600 	ext{ co

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$200$

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(C) The radioactive decay law is given by $N(t) = N_0 e^{-\lambda t}$,where $N(t)$ is the counting rate at time $t$.Given $N(0) = N_0 = 1600 	ext{ counts/s}$ and $N(8) = 100 	ext{ counts/s}$.Substituting these values: $100 = 1600 e^{-\lambda(8)}$.$e^{-8\lambda} = \frac{100}{1600} = \frac{1}{16}$.Taking the natural logarithm on both sides: $-8\lambda = \ln(1/16) = -\ln(16) = -4\ln(2)$.So,$8\lambda = 4\ln(2)$,which gives $\lambda = \frac{\ln(2)}{2} 	ext{ s}^{-1}$.Now,we need to find the counting rate $N(6)$ at $t = 6 	ext{ s}$:$N(6) = N_0 e^{-\lambda(6)} = 1600 e^{-(\frac{\ln 2}{2}) 	imes 6}$.$N(6) = 1600 e^{-3\ln 2} = 1600 e^{\ln(2^{-3})} = 1600 	imes 2^{-3}$.$N(6) = 1600 	imes \frac{1}{8} = 200 	ext{ counts/s}$.

At $t = 0$,the counting rate from a radioactive source is $1600 \text{ counts/s}$,and at $t = 8 \text{ s}$,it is $100 \text{ counts/s}$. The counting rate at $t = 6 \text{ s}$ will be:

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