A radioactive nucleus with a decay constant l | vedclass

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Question

(B) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate:$\frac{dN}{dt} = P - \lambda N$Given $P = 100$

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$2\ln \left( \frac{4}{3} \right)s$

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(B) The rate of change of the number of nuclei $N$ is given by the production rate minus the decay rate:$\frac{dN}{dt} = P - \lambda N$Given $P = 100$ and $\lambda = 0.5$,we have $\frac{dN}{dt} = 100 - 0.5N$.Rearranging and integrating from $t=0$ $(N=0)$ to $t$ $(N=50)$:$\int_0^N \frac{dN}{100 - 0.5N} = \int_0^t dt$$-\frac{1}{0.5} [\ln(100 - 0.5N)]_0^N = t$$-2 [\ln(100 - 0.5N) - \ln(100)] = t$$\ln \left( \frac{100 - 0.5N}{100} \right) = -0.5t$$1 - \frac{0.5N}{100} = e^{-0.5t}$$N = \frac{100}{0.5} (1 - e^{-0.5t}) = 200(1 - e^{-0.5t})$.Setting $N = 50$:$50 = 200(1 - e^{-0.5t})$$0.25 = 1 - e^{-0.5t}$$e^{-0.5t} = 0.75 = \frac{3}{4}$$-0.5t = \ln(3/4) = -\ln(4/3)$$t = \frac{\ln(4/3)}{0.5} = 2 \ln \left( \frac{4}{3} \right) s$.

$A$ radioactive nucleus with a decay constant $\lambda = 0.5/s$ is being produced at a constant rate of $P = 100\, nuclei/s$. If at $t = 0$ there were no nuclei,the time when there are $N = 50\, nuclei$ is:

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