A radioactive element has a rate of disintegr | vedclass

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(A) The rate of disintegration $R$ at any time $t$ is given by the radioactive decay law: $R = R_0 e^{-\lambda t}$.Given:Initial rate $R_0 = 9000 	ex

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$0.5 \log _e 3$

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(A) The rate of disintegration $R$ at any time $t$ is given by the radioactive decay law: $R = R_0 e^{-\lambda t}$.Given:Initial rate $R_0 = 9000 	ext{ disintegrations/minute}$.Rate after time $t = 2 	ext{ minutes}$ is $R = 3000 	ext{ disintegrations/minute}$.Substituting these values into the equation:$3000 = 9000 e^{-\lambda 	imes 2}$Divide both sides by $9000$:$\frac{3000}{9000} = e^{-2\lambda}$$\frac{1}{3} = e^{-2\lambda}$Taking the natural logarithm $(\log_e)$ on both sides:$\log_e(\frac{1}{3}) = -2\lambda$$-\log_e 3 = -2\lambda$$\lambda = \frac{\log_e 3}{2} = 0.5 \log_e 3$.Thus, the decay constant is $0.5 \log_e 3 	ext{ min}^{-1}$.

$A$ radioactive element has a rate of disintegration of $9000$ disintegrations per minute at a particular instant. After $2$ minutes, it becomes $3000$ disintegrations per minute. The decay constant per minute is:

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