Question

Initially 100milligrams of a radioactive substance was present.Afterhours the mass had decreased by 3%. If the rate of decay is proportional to the amount of the substance present at time t, and the amount remaining after 24hours.

Short answer

The amount of remaining substance after twenty four hours is$\text{88.53 milligrams}$

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 as x be,

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

And with the conditions, $\mathrm{t}(\mathrm{x}=100\mathrm{milligrams})=0$and $\mathrm{t}(\mathrm{x}=97\mathrm{milligrams})=6\mathrm{hrs}$. 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 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)

Obtain the amount of substance.

Substitute the value$\mathrm{t}=24$ into the equation (4).

$\begin{array}{rcl}\mathrm{x}& =& 100{\mathrm{e}}^{-0.0051\times 24}\\ & =& 100{\mathrm{e}}^{-0.12184}\\ & =& 100\times 0.8853\\ & =& 88.53\mathrm{milligrams}\end{array}$