Chapter 7.4, Problem 22E

To determine

To Find: The decay constant $k$

Answer to Problem 22E

The decay constant $\text{=0}\text{.01066}$

Explanation of Solution

Given information: An isotope of neptunium has a half-life of 65 minutes. If the decay of $\text{Np-240}$ is modelled by the differential equation $\frac{dy}{dt}=-ky$ , where t is measured in minutes

Concept used: If $y$ is the mass remaining from an initial mass $y_0$ of the substance after time $t$ , then the mass decays exponentially

  $y=y_0{e}^{-kt}$

Where $k$ is a constant

Half-life is defined as the time it takes for one-half of the atoms of a radioactive material to disintegrate

Calculation: the differential equation $\frac{dy}{dt}=-ky$

Dividing by $y$ both sides

  $\frac{1}{y}\frac{dy}{dt}=-k$

Multiply by $dt$ both sides

  $\frac{dy}{y}=-kdt$

Initial mass $=y_0$

After time $t$ the mass remaining from an initial mass $=y$

Integrate both sides

  $\begin{array}{l}{\displaystyle \int \frac{dy}{y}={\displaystyle \int -kdt}}\\ \ln y=-kt+c\end{array}$

When $t=0$

the initial mass is $y_0$

  $\ln y_0=c$

Substituting value of $c$ into equation $\ln y=-kt+c$

  $\begin{array}{l}\ln y=-kt+\ln y_0\\ \ln y-\ln y_0=-kt\\ \ln \frac{y}{y_0}=-kt\\ \frac{y}{y_0}=e^{-kt}\\ y=y_0{e}^{-kt}\end{array}$

Half-life is defined as the time it takes for one-half of the atoms of a radioactive material to disintegrate

Half-life $=t$

  $=65\text{minutes}$

And $y=\frac{y_0}{2}$

Substituting value of Half-life and $y$ into equation $y=y_0{e}^{-kt}$

  $\begin{array}{l}\frac{y_0}{2}=y_0{e}^{-65k}\\ \frac{1}{2}=e^{-65k}\\ \text{k=0}\text{.01066}\end{array}$

The decay constant $\text{=0}\text{.01066}$