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(C) The rate of disintegration $R$ at any time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.Given $R_0 = 8000$ disintegratio

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$0.5$

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(C) The rate of disintegration $R$ at any time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.Given $R_0 = 8000$ disintegrations per minute at $t = 0$.At $t = 4$ minutes,$R = 2000$ disintegrations per minute.Substituting these values into the equation: $2000 = 8000 e^{-\lambda (4)}$.Dividing both sides by $8000$: $\frac{2000}{8000} = e^{-4\lambda}$.$\frac{1}{4} = e^{-4\lambda}$.Taking the natural logarithm on both sides: $\ln(1/4) = -4\lambda$.$-\ln(4) = -4\lambda$.$\ln(2^2) = 4\lambda$.$2 \ln(2) = 4\lambda$.$\lambda = \frac{2 \ln(2)}{4} = 0.5 \ln(2)$ per minute.Therefore,the decay constant is $0.5 \log _e 2$ per minute.

$A$ radioactive element has a rate of disintegration of $8000$ disintegrations per minute at a particular instant. After $4$ minutes,it becomes $2000$ disintegrations per minute. The decay constant per minute is: (in $log _e 2$)

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