Question

Determine the half-life of the radioactive substance described in Problem 6.

Short answer

The half-life of the radioactive substance is 135.9 years.

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.

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

$\frac{\text{dx}}{\text{dt}}\text{= - kx}$… (1)

And with the conditions,$\text{t(x = 100 milligrams) = 0}$ and $\text{t(x = 97 milligrams) = 6 hrs}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$$\begin{gathered} \frac{\text{dx}}{x}\text{= - kdt} \\ \int \frac{\text{1}}{x}\text{dx = - k}\int \text{dt} \\ {\text{lnx = - kt +c}}_{\text{1}} \\ {\text{e}}^{\text{lnx}}{\text{=e}}^{\left({\text{- kt +c}}_{\text{1}}\right)} \end{gathered}$$

Then, the equation becomes,

$$\begin{gathered} {\text{x =e}}^{\text{- kt}}{\text{e}}^{{\text{c}}_{\text{1}}} \\ {\text{= ce}}^{\text{- kt}} \end{gathered}$$… (2)

Obtain the values of constants.

To find the values of constants, apply the point$\text{(x,t) = (100,0)}$in the equation (2), then

$$\begin{gathered} {\text{100 = ce}}^0 \\ \text{c = 100} \end{gathered}$$

Substitute the value of c in the equation (2).

$\mathrm{x}=100{\mathrm{e}}^{-\mathrm{kt}}$… (3)

Again, apply the other point$\text{(x,t) = (97,6)}$in the equation (3).

$\begin{array}{rcl}97& =& 100{\mathrm{e}}^{-6\mathrm{k}}\\ 0.97& =& {\mathrm{e}}^{-6\mathrm{k}}\\ \mathrm{k}& =& \frac{\ln 0.97}{-6}\\ & =& \frac{-0.0305}{-6}\\ & =& 0.0051\end{array}$

Substitute the value of in the equation (3).

$\mathrm{x}=100{\mathrm{e}}^{-0.0051\mathrm{t}}$… (4)

Substitute the valueinto the equation (4).

$\begin{array}{rcl}0.5{\mathrm{x}}_0& =& 100{\mathrm{e}}^{-0.0051\mathrm{t}}\\ 0.5\times 100& =& 100{\mathrm{e}}^{-0.0051\mathrm{t}}\\ 0.5& =& {\mathrm{e}}^{-0.0051\mathrm{t}}\\ \ln 0.5& =& -0.0051\mathrm{t}\\ \mathrm{t}& =& \frac{\ln 0.5}{-0.0051}\\ & =& \frac{-0.693}{-0.0051}\\ & =& 135.9\mathrm{hrs}\end{array}$