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−0.0230t
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=15(0.9770)t

Answers

GPT-4o mini
To model the decay of cesium-137 using an exponential equation, we start with the recursive relationship given:

\[ B(x) \approx B(x-1) \cdot 0.977 \]

This indicates that the activity decreases to 97.7% of its previous value each year.

To convert this recursive relationship into an exponential function, we can express it in the form:

\[ B(t) = B(0) \cdot (0.977)^t \]

Where \( B(0) \) is the initial activity. Given \( B(0) = 15 \) mCi, the equation becomes:

\[ B(t) = 15 \cdot (0.977)^t \]

To further convert to an exponential form, we can use the natural exponential function. The equivalency can be established by recognizing that \( (0.977)^t = e^{t \ln(0.977)} \).

Calculating the natural logarithm of 0.977:

\[
\ln(0.977) \approx -0.0230 \quad \text{(rounded to four decimal places)}
\]

Thus:

\[
B(t) = 15 \cdot e^{-0.0230t}
\]

Now, we can compare this with the provided options to identify the correct equation:

1. \( B = 15e^{-0.0233t} \)
2. \( B = 15e^{-0.0230t} \)
3. \( B = 15e^{-0.0221t} \)
4. \( B = 15(0.9770)^t \)

We determine that the correct equation that models the decay of cesium-137 is:

**\( B = 15e^{-0.0230t} \)**.