Question

(a) Consider the initial-value problem $\mathit{d}\mathit{A}\mathbf{/}\mathit{d}\mathit{t}\mathbf{=}\mathit{k}\mathit{A}\mathbf{,}\mathit{A}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}{\mathit{A}}_{\mathbf{0}}$as the model for the decay of a radioactive substance.Show that, in general, the half-life Tof the substance is $\mathit{T}\mathbf{=}\mathbf{-}\mathbf{\left(}\mathit{l}\mathit{n}\mathbf{2}\mathbf{\right)}\mathbf{/}\mathit{k}$.

(b) Show that the solution of the initial-value problem in part(a) can be written localid="1663839717694" $\mathit{A}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}{\mathit{A}}_{\mathbf{0}}{\mathbf{2}}^{\mathbf{-}\mathbf{t}\mathbf{/}\mathbf{T}}$

(c) If a radioactive substance has the half-life Tgiven in part (a),how long will it take an initial amount ${\mathit{A}}_{\mathbf{0}}$of the substance todecay to how long will it take an initial amount $\frac{\mathbf{1}}{\mathbf{8}}{\mathit{A}}_{\mathbf{0}}$?

Short answer

(a) Hence showed that in general, the half-life T of the substance is $T=\; -\frac{ln(2)}{k}$

(b) Hence showed that the solution of the initial value problem is $A=A_0{2}^{-\frac{1}{T}}$

(c) The time required to decay is$\text{3T}$.

Step-by-step solution

Define growth and decay.

The initial-value problem, $\frac{\text{dx}}{\text{dt}}{\text{= kx, x(t}}_{\text{0}}{\text{) =x}}_{\text{0}}$where k is a constant of proportionality, serves as a model for diverse phenomena involving either growth or decay. This is in the form of a first-order reaction (i.e.) a reaction whose rate, or velocity, $\text{dx/dt}$is directly proportional to the amount x of a substance that is unconverted or remaining at time t.

Solve for first order growth and decay equation.

(a)

Let the linear equation with the amount of radioactive substance as A be,

$\frac{\text{dA}}{\text{dt}}\text{= kA}$… (1)

And with the conditions,${\text{A(t = 0) =A}}_0$ and ${\text{A(t = T) = 0.5A}}_0$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}& & \frac{\text{dx}}{x}\text{= - kdt}\\ & & \int \frac{\text{1}}{A}\text{dA = k}\int \text{dt}\\ & & {\text{lnA = kt +c}}_{\text{1}}\\ & & {\text{e}}^{\text{ln(A)}}{\text{=e}}^{\left({\text{kt +c}}_{\text{1}}\right)}\end{array}$

Then, the equation becomes,

$\begin{array}{rcl}& & \mathrm{A}={\mathrm{e}}^{\mathrm{kt}}{\mathrm{e}}^{\mathrm{C}}\\ & & ={\mathrm{he}}^{\mathrm{kt}}\end{array}$… (2)

Obtain the values of constants.

To find the values of constants, apply the point ${\text{(A,t) = (A}}_0\text{,0)}$in the equation (2), then

localid="1663839736130" $$\begin{gathered} {\text{A}}_0{\text{= he}}^0 \\ {\text{h =A}}_0 \end{gathered}$$

Substitute the value of in the equation (2).

$A=A_0{e}^{kt}$… (3)

Again, apply the other point${\text{(A,t) = (0.5A}}_{\text{0}}\text{,t)}$in the equation (3).

$$\begin{gathered} 0.5A_0=A_0{e}^{kT} \\ 0.5=e^{kT} \\ ln(0.5)=ln{e}^{kT} \\ -ln\left(2\right)=kT \\ T=-\frac{\left(ln2\right)}{k} \end{gathered}$$

(b)

The value of k is $k=-\frac{\left(ln2\right)}{T}$. Substitute it in the equation (3).

$\begin{array}{rcl}A& =& A_0{e}^{\left(-\frac{\left(ln2\right)}{T}t\right)}\\ & =& A_0{e}^{{\left(ln2\right)}^{\frac{1}{T}}}\end{array}$

$A=A_0{2}^{-\frac{1}{T}}$… (4)

Obtain therequired time.

(c)

Substitute the value $A=\frac{1}{8}{A}_0$and $\text{T = -}\frac{\text{ln(2)}}{\text{k}}$into the equation (4).

$$\begin{gathered} \frac{\text{1}}{\text{2}}{\text{A}}_{\text{0}}\Rightarrow \text{t = -}\frac{\text{(ln2)}}{\text{k}}\text{= T} \\ \frac{\text{1}}{\text{4}}{\text{A}}_{\text{0}}\text{=}\frac{\text{1}}{\text{2}}\left(\frac{\text{1}}{\text{2}}{\text{A}}_{\text{0}}\right)\Rightarrow \text{t = - 2}\frac{\text{(ln2)}}{\text{k}}\text{= 2T} \\ \frac{\text{1}}{\text{8}}{\text{A}}_{\text{0}}\text{=}\frac{\text{1}}{\text{2}}\left(\frac{\text{1}}{\text{4}}{\text{A}}_{\text{0}}\right)\Rightarrow \text{t = - 3}\frac{\text{(ln2)}}{\text{k}}\text{= 3T} \end{gathered}$$

Hence proved.