Chapter 7.4, Problem 37E

To determine

To Find: The half-life of a substance.

Answer to Problem 37E

The half-life of a substance $=3.15\text{ years}$ .

Explanation of Solution

Given information: A substance that decay to $\frac{1}{3}$ of its original radioactive in $5$ years.

Formula used:

  $y=y_0{e}^{-kt}$

  $y_0=$ Number of radioactive atoms initially

  $y=$ Number of radioactive atoms remaining after $t$ time

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:

A substance that decay to $\frac{1}{3}$ of its original radioactive in $5$ years

  $y=y_0{e}^{-kt}$

  $y_0=$ Number of radioactive atoms initially

  $y=$ Number of radioactive atoms remaining after $t$ time

Where $k$ is a constant

  $\begin{array}{l}y=y_0{e}^{-kt}\\ \frac{y_0}{3}=y_0{e}^{-5k}\\ e^{-5k}=\frac{1}{3}\\ k\approx 0.2197\end{array}$

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

  $y=y_0{e}^{-0.2197t}$

  $\text{equation1}$

Let, half-life $=t$

And,

  $y=\frac{y_0}{2}$

Putting the value of $y$ in $\text{equation1}$

  $\begin{array}{l}y=y_0{e}^{-0.2197t}\\ \frac{y_0}{2}=y_0{e}^{-0.2197t}\\ e^{-0.2197t}=\frac{1}{2}\\ 0.2197t=\ln 2\\ t=3.15\text{ years}\end{array}$