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−0.0230t
upper B equals 15 e superscript negative 0.0230 t baseline
B=15e−0.0221t
upper B equals 15 e superscript negative 0.0221 t baseline
B=15e−0.0233t
upper B equals 15 e superscript negative 0.0233 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−0.0230t
upper B equals 15 e superscript negative 0.0230 t baseline
B=15e−0.0221t
upper B equals 15 e superscript negative 0.0221 t baseline
B=15e−0.0233t
upper B equals 15 e superscript negative 0.0233 t baseline
B=15(0.9770)t
Answers
GPT-4o mini
To derive the exponential decay equation for cesium-137 based on the information provided, we know that the recursive formula you've given is:
\[ B(x) \approx B(x-1) \cdot 0.977 \]
This indicates a decay factor of \(0.977\) per year. To convert this into an exponential function, we can express it in the form:
\[ B(t) = B(0) \cdot e^{-kt} \]
where \(k\) is the decay constant. We can relate the decay factor to the exponential function. Since the activity decreases by 2.3% each year, the decay factor directly implies:
\[ 0.977 = e^{-k} \]
To find \(k\), we take the natural logarithm of both sides:
\[ \ln(0.977) = -k \]
Calculating \( \ln(0.977) \):
\[ k = -\ln(0.977) \]
\[ k \approx 0.0230 \] (when rounded to four decimal places)
Now we can write the function:
\[ B(t) = 15e^{-0.0230t} \]
Since \(B(0) = 15\) mCi, this confirms that the initial value at \(t = 0\) is correct.
The correct exponential equation modeling the decay of cesium-137 in the sample over time is:
\[ B = 15e^{-0.0230t} \]
Thus, the answer is:
**B = 15e^{-0.0230t}**
\[ B(x) \approx B(x-1) \cdot 0.977 \]
This indicates a decay factor of \(0.977\) per year. To convert this into an exponential function, we can express it in the form:
\[ B(t) = B(0) \cdot e^{-kt} \]
where \(k\) is the decay constant. We can relate the decay factor to the exponential function. Since the activity decreases by 2.3% each year, the decay factor directly implies:
\[ 0.977 = e^{-k} \]
To find \(k\), we take the natural logarithm of both sides:
\[ \ln(0.977) = -k \]
Calculating \( \ln(0.977) \):
\[ k = -\ln(0.977) \]
\[ k \approx 0.0230 \] (when rounded to four decimal places)
Now we can write the function:
\[ B(t) = 15e^{-0.0230t} \]
Since \(B(0) = 15\) mCi, this confirms that the initial value at \(t = 0\) is correct.
The correct exponential equation modeling the decay of cesium-137 in the sample over time is:
\[ B = 15e^{-0.0230t} \]
Thus, the answer is:
**B = 15e^{-0.0230t}**