Chapter 9.7, Problem 14PPS

a)

To determine

To evaluate: the amount of 0.2 − gram sample remains after 72 years.

Answer to Problem 14PPS

The amount of 0.2−gram sample remains after 72 years is $6.25\times {10}^{-3}gm$ .

Explanation of Solution

Given:

A radioactive element’s half life is the time it takes for one half of the element’s quanity to decay. The half-life of Plutonium-241 is 14.4years.

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.

Model is given as,

  $\begin{array}{c}A=p{\left(0.5\right)}^{\frac{t}{14.4}}\\ =\left(0.2\right){\left(0.5\right)}^{\frac{72}{14.4}}\\ =\left(0.2\right){\left(0.5\right)}^5\\ =6.25\times {10}^{-3}gm\end{array}$

So,

The amount of 0.2−gram sample remains after 72 years is $6.25\times {10}^{-3}gm$ .

b)

To determine

To evaluate: the amount of 5.4 − gram sample remains after 1095 days.

Answer to Problem 14PPS

The amount of 5.4−gram sample remains after 72 years is $6.68\times {10}^{-2}gm$ .

Explanation of Solution

Given:

A radioactive element’s half life is the time it takes for one half of the element’s quanity to decay. The half-life of Plutonium-241 is 14.4years.

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.

The 1095 days has 91.25 years.

Model is given as,

  $\begin{array}{c}A=p{\left(0.5\right)}^{\frac{t}{14.4}}\\ =\left(5.4\right){\left(0.5\right)}^{\frac{91.25}{14.4}}\\ =\left(5.4\right){\left(0.5\right)}^{6.336}\\ =6.68\times {10}^{-2}gm\end{array}$

So,

The amount of 5.4−gram sample remains after 72 years is $6.68\times {10}^{-2}gm$ .