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(B) The activity $R$ of a radioactive sample at time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.Rearranging the formula,we

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$0.4\, \log_e 2$

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(B) The activity $R$ of a radioactive sample at time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.Rearranging the formula,we get: $\frac{R_0}{R} = e^{\lambda t}$.Taking the natural logarithm on both sides: $\lambda t = \log_e \left( \frac{R_0}{R} ight)$.Therefore,the decay constant $\lambda$ is: $\lambda = \frac{1}{t} \log_e \left( \frac{R_0}{R} ight)$.Given values are: $R_0 = 5000 	ext{ disintegrations/min}$,$R = 1250 	ext{ disintegrations/min}$,and $t = 5 	ext{ min}$.Substituting these values into the equation:$\lambda = \frac{1}{5} \log_e \left( \frac{5000}{1250} ight) = \frac{1}{5} \log_e (4)$.Since $4 = 2^2$,we have:$\lambda = \frac{1}{5} \log_e (2^2) = \frac{2}{5} \log_e 2$.Calculating the fraction: $\frac{2}{5} = 0.4$.Thus,the decay constant is $\lambda = 0.4 \log_e 2 	ext{ min}^{-1}$.

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

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