Chapter 3, Problem 22RE

a.

To determine

the initial quantity when $t=0$

Answer to Problem 22RE

The initial quantity when $t=0$ is $50grams$

Explanation of Solution

Given information : The question is

Let $Q$ represent the mass, in grams of a quantity of plutonium $241\left({}^{241}Pu\right)$ , whose half-life is about $14$ years. The quantity of plutonium present after $t$ years is given by $Q=50{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$14$}\right.}$

Calculation:

  $\begin{array}{l}t=0\\ Q=50{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$14$}\right.}\\ Q=50grams\end{array}$

b.

To determine

the quantity present after $10$ years

Answer to Problem 22RE

The quantity present after $10$ years is $30.48grams$

Explanation of Solution

Given information : The question is

Let $Q$ represent the mass, in grams of a quantity of plutonium $241\left({}^{241}Pu\right)$ , whose half-life is about $14$ years. The quantity of plutonium present after $t$ years is given by $Q=50{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$14$}\right.}$

Calculation :

  $\begin{array}{l}t=10\\ Q=50{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$10$}\!\left/ \!\raisebox{-1ex}{$14$}\right.}\\ Q=30.48grams\end{array}$

c.

To determine

To use : the graphing utility to graph the function over the interval $t=0$ to $t=50$

Explanation of Solution

Given information : The question is

Let $Q$ represent the mass, in grams of a quantity of plutonium $241\left({}^{241}Pu\right)$ , whose half-life is about $14$ years. The quantity of plutonium present after $t$ years is given by $Q=50{\left(\frac{1}{2}\right)}^{\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$14$}\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

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

The graph is obtained as: