Question

Potassium-40 Decay The chemical element potassium is a soft metal that can be found extensively throughout the Earth's crust and oceans. Although potassium occurs naturally in the form of three isotopes, only the isotope potassium-40 (K-40)is radioactive. This isotope is also unusual in that it decays by two different nuclear reactions. Over time, by emitting beta particles a great percentage of an initial amount ${\mathbf{K}}_{\mathbf{0}}$of K-40 decays into the stable isotope calcium-40 (Ca40), whereas by electron capture a smaller percentage of decays into the stable isotope argon-40 (Ar-40). Because the rates at which the amounts C(t)of Ca-40 and A(t)of Ar-40increase are proportional to the amount K(t)of potassium present, and the rate at which K(t)decays is also proportional to K(t), we obtain the system of linear first-order equations

$\frac{\mathbf{dC}}{\mathbf{dt}}\mathbf{=}{\mathbf{\lambda }}_{\mathbf{1}}\mathbf{K}$

$\frac{\mathbf{dA}}{\mathbf{dt}}\mathbf{=}{\mathbf{\lambda }}_{\mathbf{2}}\mathbf{K}$

$\frac{\mathbf{dK}}{\mathbf{dt}}\mathbf{=}\mathbf{-}\left({\mathbf{\lambda }}_{\mathbf{1}}\mathbf{+}{\mathbf{\lambda }}_{\mathbf{2}}\right)\mathbf{K}$

where ${\mathbf{\text{λ}}}_{\mathbf{1}}$and ${\mathbf{\text{λ}}}_2$are positive constants of proportionality. By proceeding as in Problem 1 we can solve the foregoing mathematical model.

(a) From the last equation in the given system of differential equations find K(t)if $\mathbf{K}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathbf{K}}_{\mathbf{0}}$. Then useto findandfrom the first and second equations. Assume that C(0)=0and A(0)=0.

(b) It is known that ${\mathbf{\lambda }}_{\mathbf{1}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{7526}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$and ${\mathbf{\lambda }}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5874}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{10}}$. Find the half-life of K-40.

(c) Use C(t)and A(t)found in part (a) to determine the percentage of an initial amount${\mathbf{K}}_{\mathbf{0}}$of K-40that decays into Ca-40and the percentage that decays into Ar-40over a very long period of time.

Short answer

(a)So, the value of$K=K_0{e}^{-\left({\lambda }_1+{\lambda }_2\right)}$.

(b) So, the half- life of K – 40 is $t=0.1298\times {10}^{10}$.

(c) Percentage of an initial amount is 89.

Step-by-step solution

Definition of linear first order equation

A first order differential equation is an equation of the form. A solution of a first order differential equation is a functionthat makesfor every value of .

System of linear first order equation

Consider the system of linear first-order equations

$$\begin{gathered} \frac{dC}{dt}={\lambda }_{1}k \\ \frac{dA}{dt}={\lambda }_{2}k \\ \frac{dK}{dt}=-\left({\lambda }_1+{\lambda }_2\right)k \end{gathered}$$

(a)

FindK(t).

Consider $\frac{dK}{dt}=-\left({\lambda }_1+{\lambda }_2\right)k$.

$\frac{1}{K}dK=-\left({\lambda }_1+{\lambda }_2\right)dt$

Integrate both sides.

$\int \frac{1}{K}dK=\int -\left({\lambda }_1+{\lambda }_2\right)dt$

$\ln K=-\left({\lambda }_1+{\lambda }_2\right)t+c_1$

role="math" localid="1663844302991" $$\begin{gathered} K=e^{-({\lambda }_1+{\lambda }_2)t+c1} \\ =e^{-({\lambda }_1+{\lambda }_2)}{e}^{c1} \\ K=c_2{e}^{-\left({\lambda }_1+{\lambda }_2\right)}.\left[\text{replace}{e}^{c_1\text{by}}{c}_2\right] \end{gathered}$$

It is given that $K\left(0\right)=K_0$.

$K_0=K\left(0\right)=c_2{e}^0$

$c_2=K_0$

$\text{So,}K=K_0{e}^{-\left({\lambda }_1+{\lambda }_2\right)}$

Find A(t)

Consider $\frac{dA}{dt}={\lambda }_{2}K$.

$\frac{dA}{dt}={\lambda }_2{K}_0{e}^{-\left({\lambda }_1+{\lambda }_2\right)}$

$\int dA=\int {\lambda }_2{K}_0{e}^{-\left({\lambda }_1+{\lambda }_2\right)}dt$

$A=-\frac{{\lambda }_2{K}_0}{\left({\lambda }_1+{\lambda }_2\right)}{e}^{-\left({\lambda }_1+{\lambda }_2\right)}+c_3$

$0=A\left(0\right)=-\frac{{\lambda }_2{K}_0}{\left({\lambda }_1+{\lambda }_2\right)}{e}^0+c_3$

$c_3=\frac{{\lambda }_2{K}_0}{{\lambda }_1+{\lambda }_2}$

So,$A=-\frac{{\lambda }_2{K}_0}{\left({\lambda }_1+{\lambda }_2\right)}{e}^{-\left({\lambda }_1+{\lambda }_2\right)}+\frac{{\lambda }_2{K}_0}{{\lambda }_1+{\lambda }_2}$

Find C(t)

Consider

$\begin{array}{rcl}\frac{dC}{dt}& =& {\lambda }_1{K}_0{e}^{-\left({\lambda }_1+{\lambda }_2\right)t}\\ \int dC& =& \int {\lambda }_1{K}_0{e}^{-\left({\lambda }_1+{\lambda }_2\right)}dt\\ C& =& -\frac{{\lambda }_1{K}_0}{{\lambda }_1+{\lambda }_2}{e}^{-\left({\lambda }_1+{\lambda }_2\right)}+c_4\\ 0& =& C\left(0\right)=-\frac{{\lambda }_1{K}_0}{{\lambda }_1+{\lambda }_2}{e}^0+c_4\\ c_4& =& \frac{{\lambda }_1{K}_0}{{\lambda }_1+{\lambda }_2}\\ \text{So,}C& =& -\frac{{\lambda }_1{K}_0}{{\lambda }_1+{\lambda }_2}{e}^{\left({\lambda }_1+{\lambda }_2\right)}+\frac{{\lambda }_1{K}_0}{{\lambda }_1+{\lambda }_2}\end{array}$

(b) Step 1: Find the half-life of K-40.

It is given that${\lambda }_1=4.7526\times {10}^{-10}$and${\lambda }_2=0.5874\times {10}^{-10}$.

Consider .

$\begin{array}{rcl}\frac{1}{2}{K}_0& =& K_0{e}^{-\left(4.7526\times {10}^{-10}+0.5874\times {10}^{-10}\right)t}\\ \frac{1}{2}& =& e^{-5.34\times {10}^{-10}t}\\ \ln \frac{1}{2}& =& -5.34\times {10}^{-10}t\\ -0.6931& =& -5.34\times {10}^{-10}t\\ t& =& 0.1298\times {10}^{10}\end{array}$

(c) Step 1: Find the percentage of an initial amount

Determine the percentage of an initial amountofthat decays intoand the percentage that decays intoover a long period of time.

$\underset{t\to \infty }{\lim }A\left(t\right)=\underset{t\to \infty }{\lim }\left[-\frac{{\lambda }_2{K}_0}{\left({\lambda }_1+{\lambda }_2\right)}{e}^{-\left({\lambda }_1+{\lambda }_2\right)}+\frac{{\lambda }_2{K}_0}{{\lambda }_1+{\lambda }_2}\right]$

$=\frac{{\lambda }_2{K}_0}{{\lambda }_1+{\lambda }_2}$

Percentage that decays intois

$$\begin{gathered} \frac{\left(\frac{{\lambda }_2{K}_0}{{\lambda }_1+{\lambda }_2}\right)}{K_o}\times 100=\frac{0.5874\times {10}^{-10}}{4.7526\times {10}^{-10}+0.5874\times {10}^{-10}}\times 100 \\ =11\% \end{gathered}$$

$\underset{t\to \infty }{\lim }C\left(t\right)=\underset{t\to \infty }{\lim }\left[-\frac{{\lambda }_2{K}_0}{\left({\lambda }_1+{\lambda }_2\right)}{e}^{-\left({\lambda }_1+{\lambda }_2\right)}+\frac{{\lambda }_2{K}_0}{{\lambda }_1+{\lambda }_2}\right]$

Percentage that decays into Ca-40 is

$$\begin{gathered} \frac{\left(\frac{{\lambda }_1{K}_0}{{\lambda }_1+{\lambda }_2}\right)}{K_o}\times 100=\frac{4.7526\times {10}^{-10}}{4.7526\times {10}^{-10}+0.5874\times {10}^{-10}}\times 100 \\ =89\% \end{gathered}$$