To determine the half-life of a radioactive e | vedclass

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(D) The rate of radioactive decay is given by $|dN/dt| = \lambda N = \lambda N_0 e^{-\lambda t}$.Taking the natural logarithm on both sides: $\ln|dN/d

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$8$

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(D) The rate of radioactive decay is given by $|dN/dt| = \lambda N = \lambda N_0 e^{-\lambda t}$.Taking the natural logarithm on both sides: $\ln|dN/dt| = \ln(\lambda N_0) - \lambda t$.Comparing this with the equation of a straight line $y = mx + c$,the slope of the graph is $-\lambda$.From the given graph (assuming slope $= -0.5$),we get $\lambda = 0.5 \ 	ext{year}^{-1}$.The half-life is $t_{1/2} = \ln(2) / \lambda \approx 0.693 / 0.5 = 1.386 \ 	ext{years}$.After $t = 4.16 \ 	ext{years}$,the number of nuclei remaining is $N = N_0 / p$.Since $4.16 / 1.386 = 3$,the time elapsed is $3 \ t_{1/2}$.Thus,$N = N_0 / 2^3 = N_0 / 8$.Therefore,$p = 8$.

To determine the half-life of a radioactive element,a student plots a graph of $\ln|dN(t)/dt|$ versus $t$. Here $dN(t)/dt$ is the rate of radioactive decay at time $t$. If the number of radioactive nuclei of this element decreases by a factor of $p$ after $4.16 \ \text{years}$,the value of $p$ is

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