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(A) The activity $A$ is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.The decay constant $\lambda$ is

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$3.45 	imes 10^{10}$

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(A) The activity $A$ is given by $A = \lambda N$,where $\lambda$ is the decay constant and $N$ is the number of nuclei.The decay constant $\lambda$ is related to the half-life $T_{1/2}$ by $\lambda = \frac{\ln 2}{T_{1/2}}$.Given $T_{1/2} = 138.6 	ext{ days} = 138.6 	imes 24 	imes 3600 	ext{ seconds} = 1.1975 	imes 10^7 	ext{ s}$.Given activity $A = 2000 	ext{ disintegrations/sec}$.Substituting the values into $N = \frac{A}{\lambda} = \frac{A 	imes T_{1/2}}{\ln 2}$:$N = \frac{2000 	imes 138.6 	imes 24 	imes 3600}{0.693} = \frac{2000 	imes 1.1975 	imes 10^7}{0.693} \approx 3.45 	imes 10^{10}$.

$A$ radioactive sample is an $\alpha$-emitter with a half-life of $138.6$ days. $A$ student observes its activity to be $2000$ disintegrations per second. The number of radioactive nuclei for this given activity is:

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