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(C) Let $R$ be the rate of production of nuclei,$R = 10 	ext{ nuclei/s}$.Let $\lambda$ be the decay constant,$\lambda = 0.5 	ext{ s}^{-1}$.Let $N(t)

Vedclass · Accepted Answer

$1.386 	ext{ s}$

Vedclass · Answer

(C) Let $R$ be the rate of production of nuclei,$R = 10 	ext{ nuclei/s}$.Let $\lambda$ be the decay constant,$\lambda = 0.5 	ext{ s}^{-1}$.Let $N(t)$ be the number of nuclei at time $t$.The rate of change of the number of nuclei is given by the differential equation: $\frac{dN}{dt} = R - \lambda N$.Rearranging the terms,we get $\frac{dN}{R - \lambda N} = dt$.Integrating both sides from $t=0$ (where $N=0$) to $t=t$ (where $N=N$):$\int_{0}^{N} \frac{dN}{R - \lambda N} = \int_{0}^{t} dt$.This yields $-\frac{1}{\lambda} \ln(R - \lambda N) \Big|_{0}^{N} = t$.$-\frac{1}{\lambda} [\ln(R - \lambda N) - \ln(R)] = t$.$\ln\left(\frac{R - \lambda N}{R}ight) = -\lambda t$.$1 - \frac{\lambda N}{R} = e^{-\lambda t}$.$N(t) = \frac{R}{\lambda} (1 - e^{-\lambda t})$.Given $N = 10$,$R = 10$,and $\lambda = 0.5$,we substitute these values:$10 = \frac{10}{0.5} (1 - e^{-0.5t})$.$10 = 20 (1 - e^{-0.5t})$.$0.5 = 1 - e^{-0.5t}$.$e^{-0.5t} = 0.5$.Taking the natural logarithm on both sides:$-0.5t = \ln(0.5) = -\ln(2)$.$0.5t = \ln(2)$.$t = \frac{\ln(2)}{0.5} = 2 \ln(2) \approx 2 	imes 0.693 = 1.386 	ext{ s}$.

$A$ radioactive substance is being produced at a constant rate of $10 \text{ nuclei/s}$. The decay constant of the substance is $0.5 \text{ s}^{-1}$. After what time will the number of radioactive nuclei become $10$? Initially,there are no nuclei present. Assume the decay law holds for the sample.

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