Chapter 7.4, Problem 8E

a.

To determine

To calculate : the formula for the mass remaining after $t$ days

Answer to Problem 8E

The formula for the mass remaining after $t$ days is $50{\left(0.5\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$28$}\right.}$

Explanation of Solution

Given information : Strontium- $\text{90}$ has a half life of $\text{28}$ days

A sample has a mass of $\text{50}$ mg initially

Calculation : let $\text{m}\left(t\right)=m_0{e}^{kt}$ be the mass of Strontium- $\text{90}$ left at $t$ days

The half life is $\text{28}$ days so let $m\left(28\right)=0.5$ and initial amount $m_0=1$ , then solve for k

  $\begin{array}{l}0.5=\left(1\right)e^{k\left(28\right)}\\ \ln \left(0.5\right)=\ln e^{28k}\\ \ln \left(0.5\right)=28k\\ k=\frac{\ln \left(0.5\right)}{28k}\approx -0.02475526\end{array}$

The general formula is then

  $\text{m}\left(t\right)=m_0{e}^{\frac{\ln \left(0.5\right)}{\left(\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$28$}\right.\right)}}$

If $\text{50}$ mg then it becomes

  $\text{m}\left(t\right)=50e^{\frac{\ln \left(0.5\right)}{\left(\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$28$}\right.\right)}}=50{\left(0.5\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$28$}\right.}$

b.

To determine

To calculate : the mass remaining after $\text{40}$ days

Answer to Problem 8E

The mass remaining after $\text{40}$ days is $18.57mg$

Explanation of Solution

Given information : Strontium- $\text{90}$ has a half life of $\text{28}$ days

Calculation :

  $\begin{array}{l}m\left(r\right)=50e^{\frac{-\ln \left(2\right)}{28}\times r}\\ m\left(r\right)=50e^{\frac{-\ln \left(2\right)}{28}\times 40}\\ m\left(r\right)=18.57mg\end{array}$

c.

To determine

How long does it take the sample to decay to the mass of $2$ mg

Answer to Problem 8E

In $130$ days the sample to decay to the mass of $2$ mg

Explanation of Solution

Given information : Strontium- $\text{90}$ has a half life of $\text{28}$ days.

Calculation :

  $\begin{array}{l}m\left(r\right)=50e^{\frac{-\ln \left(2\right)}{28}\times r}\\ 2=50e^{\frac{-\ln \left(2\right)}{28}\times r}\\ \frac{2}{50}=e^{\frac{-\ln \left(2\right)}{28}\times r}\\ \frac{1}{25}=e^{\frac{-\ln \left(2\right)}{28}\times r}\\ \ln \left(\frac{1}{25}\right)=\frac{-\ln \left(2\right)}{28}\times r\\ r=130days\end{array}$

d.

To determine

To sketch : the graph of the mass function

Explanation of Solution

Given information : the equation is $\text{m}\left(t\right)=50{\left(0.5\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$28$}\right.}$

Graph : Sketch the graph using graphing utility.

Step 1: Press WINDOW button to access the Window editor.

Step 2: Press $\boxed{\text{Y=}}$ button.

Step 3: Enter the expression $50{\left(0.5\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$28$}\right.}$ which is required to graph.

Step 4: Press $\boxed{\text{GRAPH}}$ button to graph the function.

The graph is obtained as:

  

Interpretation : from the above graph it is observed that the point where half of the initial amount remaining is shown