Question

The radioactive isotope of lead, $\mathbf{Pb}\mathbf{-}\mathbf{209}$, decays at a rateproportional to the amount present at time tand has a half-life of 3.3hours. If 1gram of this isotope is present initially, howlong will it take for 90% of the lead to decay?

Short answer

The time at which the ninety percentage of lead is decayed is $10.96\mathrm{hrs}$.

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 population of a community as x be,

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

And with the conditions,$\mathrm{t}(\mathrm{x}=1\mathrm{grams})=0$and $\mathrm{t}(\mathrm{x}=0.5\mathrm{grams})=3.3\mathrm{hrs}$. As the equation (1) is linear and separable, so integrate the equation and separate the variables.

$\begin{array}{rcl}& & \frac{\mathrm{dx}}{\mathrm{x}}=\; -\mathrm{kdt}\\ & & \int \frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx}=\; -\mathrm{k}\int \mathrm{dt}\\ & & \mathrm{lnx}=\; -\mathrm{kt}+{\mathrm{c}}_{\mathrm{1}}\\ & & {\mathrm{e}}^{\mathrm{lnx}}={\mathrm{e}}^{\left(-\mathrm{kt}+{\mathrm{c}}_{\mathrm{1}}\right)}\end{array}$

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$\text{(x,t) = (1,0)}$in the equation (2), then

$$\begin{gathered} {\text{1 = ce}}^0 \\ \text{c = 1} \end{gathered}$$

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

role="math" localid="1663848729022" $\mathrm{x}={\mathrm{e}}^{-\mathrm{kt}}$… (3)

Again, apply the other point$\text{(x,t) =}\left(\text{0.5,3.3}\right.\text{)}$in the equation (3).

$\begin{array}{rcl}0.5& =& {\mathrm{e}}^{-3.3\mathrm{k}}\\ \mathrm{k}& =& \frac{\ln 0.5}{-3.3}\\ & =& \frac{-0.693}{-3.3}\\ & =& 0.21\end{array}$

Substitute the value of in the equation (3).

$\mathrm{x}={\mathrm{e}}^{-0.21\mathrm{t}}$… (4)

Obtain the time at which the ninety percentage of lead is decayed.

Substitute the value$\mathrm{x}=0.1$ into the equation (4).

$\begin{array}{rcl}0.1& =& {\mathrm{e}}^{-0.21\mathrm{t}}\\ \ln 0.1& =& -0.21\mathrm{t}\\ \mathrm{t}& =& \frac{\ln 0.1}{-0.21}\\ & =& \frac{-2.3}{-0.21}\\ & =& 10.96\mathrm{hrs}\end{array}$