Question

a toxic radioactive substance with a density of 5 milligrams per square centimeter is detected in the ventilating ducts of a nuclear processing building that was used 38 years ago. if the half - life of the substance is 10 years, what was the density of the substance when it was deposited 38 years ago? round to two decimal places.

a. 6.00 mg/cm²
b. 174.11 mg/cm²
c. 69.64 mg/cm²
d. 0.36 mg/cm²

Explanation:

Step1: Identify the half - life formula

The formula for radioactive decay is $A = A_0(\frac{1}{2})^{\frac{t}{h}}$, where $A$ is the final amount (density in this case), $A_0$ is the initial amount (the density we want to find), $t$ is the time elapsed, and $h$ is the half - life.
We know that $A = 5$ mg/cm², $t = 38$ years, and $h = 10$ years.

Step2: Rearrange the formula to solve for $A_0$

Starting from $A = A_0(\frac{1}{2})^{\frac{t}{h}}$, we can solve for $A_0$:
$A_0=\frac{A}{(\frac{1}{2})^{\frac{t}{h}}}=A\times2^{\frac{t}{h}}$.

Step3: Substitute the values into the formula

Substitute $A = 5$, $t = 38$, and $h = 10$ into the formula:
$A_0=5\times2^{\frac{38}{10}}=5\times2^{3.8}$.
We know that $2^{3.8}=2^{3}\times2^{0.8}$.
$2^{3}=8$ and $2^{0.8}\approx1.7411$. So $2^{3.8}\approx8\times1.7411 = 13.9288$.
Then $A_0 = 5\times13.9288=69.644\approx69.64$ mg/cm².

Answer:

C. $69.64$ mg/cm²