Chapter 9.7, Problem 15PPS

a.

To determine

How much of a 0.2-gram sample remains after 72 years.

Answer to Problem 15PPS

  $0.00625\text{ grams}$

Explanation of Solution

Given:

The number of grams A of Plutonium-241 left after t years can be modeled by $A=p{\left(0.5\right)}^{\frac{t}{14.4}}$ , where p is the original amount of the element.

Calculation:

The number of grams A of Plutonium-241 left after t years can be modeled by $A=p{\left(0.5\right)}^{\frac{t}{14.4}}$ .

Here given that initially the sample amount was $p=0.2$ grams. So after $t=72$ years, the amount remaining can be calculated as

  $\begin{array}{c}A=p{\left(0.5\right)}^{\frac{t}{14.4}}\\ =0.2{\left(0.5\right)}^{\frac{72}{14.4}}\\ =0.2{\left(0.5\right)}^5\\ =0.00625\text{ grams}\end{array}$

Thus, the amount of Plutonium-241 left after 72 years was $0.00625\text{ grams}$ .

b.

To determine

How much of a 5.4-gram sample remains after 1095 days.

Answer to Problem 15PPS

  $4.6739\text{ grams}$

Explanation of Solution

Given:

The number of grams A of Plutonium-241 left after t years can be modeled by $A=p{\left(0.5\right)}^{\frac{t}{14.4}}$ , where p is the original amount of the element.

Calculation:

The number of grams A of Plutonium-241 left after t years can be modeled by $A=p{\left(0.5\right)}^{\frac{t}{14.4}}$ .

Here given that initially the sample amount was $p=5.4$ grams. So, after $t=\frac{1095}{365}=3$ years, the amount remaining can be calculated as

  $\begin{array}{c}A=p{\left(0.5\right)}^{\frac{t}{14.4}}\\ =5.4{\left(0.5\right)}^{\frac{3}{14.4}}\\ =4.6739\text{ grams}\end{array}$

Thus, the amount of Plutonium-241 left after 1095 days was $4.6739\text{ grams}$ .