Question

Consider the following scenario: A sample of radioactive material has a decay constant of 0.04 per hour. If there are initially 500 grams of the material, how much will remain after 3 hours?(1 point)
Responses

y=500(0.96)3
y is equal to 500 times 0 point 9 6 cubed

y=3(0.96)500
y is equal to 3 times 0 point 9 6 to the 500th power

y=3(0.04)500
y is equal to 3 times 0 point 0 4 to the 500th power

y=500(0.04)3

Answer 1

To solve the problem of how much radioactive material remains after 3 hours, we can use the formula for exponential decay:

\[
y = y_0 e^{-\lambda t}
\]

where:
- \(y\) is the amount remaining after time \(t\),
- \(y_0\) is the initial amount (500 grams in this case),
- \(\lambda\) is the decay constant (0.04 per hour),
- \(t\) is the time in hours.

In this case:

\[
y = 500 e^{-0.04 \times 3}
\]

Calculating this:

1. Calculate \(-0.04 \times 3\):
\[
-0.04 \times 3 = -0.12
\]

2. Now substitute into the equation:
\[
y = 500 e^{-0.12}
\]

3. To find \(e^{-0.12}\), we compute approximately:
\[
e^{-0.12} \approx 0.8869
\]

4. Therefore:
\[
y \approx 500 \times 0.8869 \approx 443.45 \text{ grams}
\]

Now, looking at the provided options, none of them directly correspond to the calculation we performed. However, if we are examining the exponential decay in terms of discrete decay steps using a base, we can think about using \(0.96\) (which is \(e^{-0.04}\) rounded for decay over three hours):

Using a decay of \(0.96^3\), the decay calculation would appear as:

\[
y = 500(0.96)^3
\]

Calculating \(0.96^3\):

1. \(0.96 \times 0.96 = 0.9216\)
2. \(0.9216 \times 0.96 \approx 0.884736\)

So,

\[
y \approx 500 \times 0.884736 \approx 442.37 \text{ grams}
\]

Since this option is similar to the actual exponential decay computation, the correct response closest to our derivation is:

**y = 500(0.96)^3**.

This implies that after 3 hours, approximately 442-443 grams are remaining.