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(B) The disintegration rate $A$ at any time $t$ is given by $A = A_0 e^{-\lambda t}$.Given,$A_0 = 5000$ disintegrations per minute at $t = 0$.At $t =

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$0.4$

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(B) The disintegration rate $A$ at any time $t$ is given by $A = A_0 e^{-\lambda t}$.Given,$A_0 = 5000$ disintegrations per minute at $t = 0$.At $t = 5$ minutes,$A = 1250$ disintegrations per minute.Substituting these values into the equation: $1250 = 5000 e^{-\lambda (5)}$.Dividing both sides by $5000$: $\frac{1250}{5000} = e^{-5\lambda}$.$\frac{1}{4} = e^{-5\lambda}$.Taking the natural logarithm on both sides: $\ln(1/4) = -5\lambda$.$-\ln(4) = -5\lambda$.$\ln(2^2) = 5\lambda$.$2 \ln 2 = 5\lambda$.$\lambda = \frac{2}{5} \ln 2 = 0.4 \ln 2 	ext{ per minute}$.

$A$ radioactive sample at any instant has its disintegration rate $5000$ disintegrations per minute. After $5$ minutes,the rate is $1250$ disintegrations per minute. Then,the decay constant (per minute) is (in $, \ln 2$)

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