Question
Cesium-137 is a radioactive isotope with a variety of industrial and medical applications, known for its relatively long half-life of about 30 years. It has been observed that the activity of the sample decreases by 2.3% each year. This decay can be represented by the recursive equation B(x)≈B(x−1)⋅0.977
, in which B(x)
is the activity in millicuries after x years, and B(0)=15
mCi. Create an exponential equation that models the decay of cesium-137 in the sample over time. Round your answer to four decimal places. Which of the following is the correct equation?(1 point)
Responses
B=15e−0.0233t
upper B equals 15 e superscript negative 0.0233 t baseline
B=15e−0.0221t
upper B equals 15 e superscript negative 0.0221 t baseline
B=15e−0.0230t
upper B equals 15 e superscript negative 0.0230 t baseline
B=15(0.9770)t
, in which B(x)
is the activity in millicuries after x years, and B(0)=15
mCi. Create an exponential equation that models the decay of cesium-137 in the sample over time. Round your answer to four decimal places. Which of the following is the correct equation?(1 point)
Responses
B=15e−0.0233t
upper B equals 15 e superscript negative 0.0233 t baseline
B=15e−0.0221t
upper B equals 15 e superscript negative 0.0221 t baseline
B=15e−0.0230t
upper B equals 15 e superscript negative 0.0230 t baseline
B=15(0.9770)t
Answers
GPT-4o mini
To find the exponential decay model for the activity of cesium-137, we can start with the information given in the problem. We know the decay rate per year is approximately 2.3%, which corresponds to the decay factor 0.977 (i.e., \(1 - 0.023 = 0.977\)).
We can model the decay using the exponential decay formula:
\[
B(t) = B(0) \cdot e^{-kt}
\]
where:
- \(B(t)\) is the activity after \(t\) years,
- \(B(0) = 15\) mCi (the initial amount),
- \(k\) is the decay constant,
- \(t\) is the time in years.
An equivalent approach is to use the decay factor formula, starting from the initial activity:
\[
B(t) = B(0) \cdot (decay\ factor)^t
\]
Given that the decay factor is 0.977 (which describes the decrease by 2.3% per year), we have:
\[
B(t) = 15 \cdot (0.977)^t
\]
Now, to relate this to the exponential form, we note that \(0.977\) can be expressed as \(e^{-k}\) where:
\[
0.977 = e^{-k}
\]
Taking the natural logarithm (ln) of both sides to solve for \(k\):
\[
-k = \ln(0.977)
\]
Calculating \(k\):
\[
k \approx -\ln(0.977) \approx 0.0230
\]
Thus, the decay model using the exponential function is:
\[
B(t) = 15 \cdot e^{-0.0230t}
\]
From the options given:
- **B = 15e−0.0230t** is the correctly rounded equation for the exponential model.
Final answer:
**B = 15e−0.0230t**
We can model the decay using the exponential decay formula:
\[
B(t) = B(0) \cdot e^{-kt}
\]
where:
- \(B(t)\) is the activity after \(t\) years,
- \(B(0) = 15\) mCi (the initial amount),
- \(k\) is the decay constant,
- \(t\) is the time in years.
An equivalent approach is to use the decay factor formula, starting from the initial activity:
\[
B(t) = B(0) \cdot (decay\ factor)^t
\]
Given that the decay factor is 0.977 (which describes the decrease by 2.3% per year), we have:
\[
B(t) = 15 \cdot (0.977)^t
\]
Now, to relate this to the exponential form, we note that \(0.977\) can be expressed as \(e^{-k}\) where:
\[
0.977 = e^{-k}
\]
Taking the natural logarithm (ln) of both sides to solve for \(k\):
\[
-k = \ln(0.977)
\]
Calculating \(k\):
\[
k \approx -\ln(0.977) \approx 0.0230
\]
Thus, the decay model using the exponential function is:
\[
B(t) = 15 \cdot e^{-0.0230t}
\]
From the options given:
- **B = 15e−0.0230t** is the correctly rounded equation for the exponential model.
Final answer:
**B = 15e−0.0230t**