Questions: The radioactive substance cesium-137 has a half-life of 30 years. The amount A(t) (in grams) of a sample of cesium-137 remaining after t years is given by the following exponential function. A(t)=523(1/2)^(t/30) Find the initial amount in the sample and the amount remaining after 80 years. Round your answers to the nearest gram as necessary.

The radioactive substance cesium-137 has a half-life of 30 years. The amount A(t) (in grams) of a sample of cesium-137 remaining after t years is given by the following exponential function.

A(t)=523(1/2)^(t/30)

Find the initial amount in the sample and the amount remaining after 80 years.
Round your answers to the nearest gram as necessary.

Transcript text: The radioactive substance cesium-137 has a half-life of 30 years. The amount A(t) (in grams) of a sample of cesium-137 remaining after t years is given by the following exponential function. $A(t)=523(\frac{1}{2})^{\frac{t}{30}}$ Find the initial amount in the sample and the amount remaining after 80 years. Round your answers to the nearest gram as necessary. EXPLANATION First, we find the initial amount in the sample. It occurs when 0 years have passed. So, we let t = 0 and compute A(0). $A(0)=523(\frac{1}{2})^{\frac{0}{30}}=523(1)^0=523(1)=523$ The initial amount in the sample was 523 grams. Next, we find the amount remaining after 80 years.

Solution

Solution Steps

To find the initial amount of cesium-137, we evaluate the function $ A(t) $ at $ t = 0 $. This will give us the initial amount since no time has passed. For the amount remaining after 80 years, we substitute $ t = 80 $ into the function and calculate the result.

Step 1: Initial Amount Calculation

To find the initial amount of cesium-137, we evaluate the function $ A(t) $ at $ t = 0 $:

\[ A(0) = 523 \left( \frac{1}{2} \right)^{\frac{0}{30}} = 523 \cdot 1 = 523 \]

Thus, the initial amount in the sample is $ 523 $ grams.

Step 2: Amount Remaining After 80 Years

Next, we calculate the amount remaining after $ 80 $ years by substituting $ t = 80 $ into the function:

\[ A(80) = 523 \left( \frac{1}{2} \right)^{\frac{80}{30}} \approx 82.3673 \]

Rounding this to the nearest gram gives us $ 82 $ grams.