The graph in the figure shows how the count-r | vedclass

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(D) The radioactive decay law is given by $A = A_0 e^{-\lambda t}$. Taking the natural logarithm on both sides,we get $\ln A = \ln A_0 - \lambda t$. T

Vedclass · Accepted Answer

$A = 12 \, e^{-0.1t}$

Vedclass · Answer

(D) The radioactive decay law is given by $A = A_0 e^{-\lambda t}$. Taking the natural logarithm on both sides,we get $\ln A = \ln A_0 - \lambda t$. This is a linear equation of the form $y = mx + c$,where $y = \ln A$,$x = t$,$c = \ln A_0$ (intercept),and $m = -\lambda$ (slope).From the graph,the intercept on the $y$-axis (at $t = 0$) is $c = 2.6$. Thus,$\ln A_0 = 2.6$. Given $\ln 12 = 2.6$,we have $A_0 = 12$.The slope $m$ of the line is calculated using two points on the line: $(0, 2.6)$ and $(26, 0)$.$m = \frac{0 - 2.6}{26 - 0} = \frac{-2.6}{26} = -0.1$.Since $m = -\lambda$,we have $\lambda = 0.1 \, 	ext{s}^{-1}$.Substituting these values into the decay equation,we get $\ln A = 2.6 - 0.1t$.Taking the exponential of both sides,$A = e^{2.6 - 0.1t} = e^{2.6} \cdot e^{-0.1t}$.Since $e^{2.6} = 12$,the relationship is $A = 12 \, e^{-0.1t}$.

The graph in the figure shows how the count-rate $A$ of a radioactive source as measured by a Geiger counter varies with time $t$. The relationship between $A$ and $t$ is: (Assume $\ln 12 = 2.6$)

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