A radioactive element has a rate of disintegr | vedclass

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(C) The rate of disintegration $R$ at time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.Here, $R_0 = 16,000 	ext{ disintegr

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$0.75 \log_e 2$

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(C) The rate of disintegration $R$ at time $t$ is given by the law of radioactive decay: $R = R_0 e^{-\lambda t}$.Here, $R_0 = 16,000 	ext{ disintegrations/minute}$, $R = 2000 	ext{ disintegrations/minute}$, and $t = 4 	ext{ minutes}$.Substituting these values into the equation: $2000 = 16,000 e^{-\lambda 	imes 4}$.Dividing both sides by $16,000$: $\frac{2000}{16,000} = e^{-4\lambda}$, which simplifies to $\frac{1}{8} = e^{-4\lambda}$.Taking the natural logarithm $(log_e)$ on both sides: $\log_e(1/8) = -4\lambda$.Since $\log_e(1/8) = \log_e(2^{-3}) = -3 \log_e 2$, we have $-3 \log_e 2 = -4\lambda$.Solving for $\lambda$: $\lambda = \frac{3}{4} \log_e 2 = 0.75 \log_e 2 	ext{ min}^{-1}$.

$A$ radioactive element has a rate of disintegration of $16,000 \text{ disintegrations per minute}$ at a particular instant. After $4 \text{ minutes}$, it becomes $2000 \text{ disintegrations per minute}$. The decay constant per minute is:

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