Chapter 7.4, Problem 26E

a.

To determine

To Find: The element’s half-life

Answer to Problem 26E

The element’s half-life $=138.63\text{days}$

Explanation of Solution

Given information: The number of radioactive atoms remaining after $t$ days in a sample of $\text{Polonium-210}$ that starts with $y_0$ radioactive atoms is $y=y_0{e}^{-0.005t}$

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:

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

  $y=y_0{e}^{-0.005t}$ $\text{equation1}$

Let, half-life $=t$

And,

  $y=\frac{y_0}{2}$

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

  $\begin{array}{l}\frac{y_0}{2}=y_0{e}^{-0.005t}\\ e^{-0.005t}=\frac{1}{2}\\ 0.005t=\ln 2\\ t=138.63\text{days}\end{array}$

b.

To determine

To Find: How many days after the sample arrives will be able to use the polonium

Answer to Problem 26E

The sample will be useful for $\approx 599.15\text{days}$

Explanation of Solution

Given information: Sample of $\text{Polonium-210}$ will not useful after $95\%$ of the radioactive nuclei present on the day the sample arrives have disintegrated

Formula used:

  $y=y_0{e}^{-kt}$

  $y_0=$ Number of bacteria present initially

  $y=$ Number of bacteria present after $t$ time

Where $k$ is a constant

Calculation:

Sample of $\text{Polonium-210}$ will not useful after $95\%$ of the radioactive nuclei present on the day the sample arrives have disintegrated

  $y=y_0{e}^{-0.005t}$ $\text{equation1}$

Here.

  $y=0.05y_0$

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

  $\begin{array}{l}0.05y_0=y_0{e}^{-0.005t}\\ e^{-0.005t}=0.05\\ -0.005t=\ln 0.05\\ t=\frac{-\ln 0.05}{0.005}\\ t\approx 599.15\text{days}\end{array}$