Chapter 5.8, Problem 3AYU

Radioactive Decay Strontium-90 is a radioactive material that decays according to the function $A(t)={\text{A}}_0{e}^{-0.0244t}$, where $A_0$ is the initial amount present and $A$ is the amount present at time $t$ (in years). Assume that a scientist has a sample of 500 grams of strontium-90.

(a) What is the decay rate of strontium-90?

(b) Graph the function using a graphing utility.

(c) How much strontium-90 is left after 10 years?

(d) When will 400 grams of strontium-90 be left?

(e) What is the half-life of strontium-90?

To determine

To find:

a. What is the decay rate of strontium-90?

Answer to Problem 3AYU

Solution:

a. $-2.44\%$

Explanation of Solution

Given:

$A\left(t\right)=A_0{e}^{kt}$

Calculation:

a. Compare $A\left(t\right)=A_0{e}^{kt}$ where $k$ is a negative number that represents the rate of decay. We will get the values of $k$ as $-0.0244$.

On converting the percentage,

Notation $k=-2.44\%$.

Therefore the decay rate of strontium-90 is $-2.44\%$.

To determine

To find:

b. Graph the function using a graphing utility.

Answer to Problem 3AYU

Solution:

b.

Explanation of Solution

Given:

$A\left(t\right)=A_0{e}^{kt}$

Calculation:

b. Graph:

To determine

To find:

c. How much strontium-90 is left after 10 years?

Answer to Problem 3AYU

Solution:

c. $391.7$

Explanation of Solution

Given:

$A\left(t\right)=A_0{e}^{kt}$

Calculation:

c. Substitute 10 for $t$ and 500 for $A_0$ in the given function.

$A\left(t\right)=500 e^{-0.0244\left(10\right)}$

$A\left(t\right)= 500\left(0.7835\right)$

$A\left(t\right)=391.7$

Therefore the amount of strontium 90 left after 10 years is about $391.7$ grams.

To determine

To find:

d. When will 400 grams of strontium-90 be left?

Answer to Problem 3AYU

Solution:

d. $9.14$

Explanation of Solution

Given:

$A\left(t\right)=A_0{e}^{kt}$

Calculation:

d. Substitute 400 for $A\left(t\right)$ in the given function.

$400=500 e^{-0.0244t}$

$0.8 = e^{-0.0244t}$

Taking log on both sides we get,

$\text{ln}0.8=\text{ln}{e}^{-0.0244t}$

$t=\frac{-0.2231}{-0.0244}=9.14$

Therefore, 400 grams of strontium 90 will be left after about $9.1$ years.

To determine

To find:

e. What is the half-life of strontium-90?

Answer to Problem 3AYU

Solution:

e. $28.4$

Explanation of Solution

Given:

$A\left(t\right)=A_0{e}^{kt}$

Calculation:

e. Half life is the time required for half of the radioactive substance to decay. Since $A_0$ is the initial amount of strontium 90.

Replace $A\left(t\right)=\frac{A_0}{2}$ in the given function.

$\frac{A_0}{2}= A_0 e^{-0.0244t}$

$\frac{1}{2}=e^{-0.0244t}$

Taking log on both sides we get,

$\text{ln}\frac{1}{2}= \text{ln}{e}^{-0.0244t}$

$t=\frac{-0.6931}{-0.0244}=28.4$

Therefore, the half life of strontium 90 is $28.4$ years.