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

1 answer

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}