The bones of a newly discovered dinosaur weigh 170 pounds and measure 9 feet, with a 6-inch claw on one toe of each hind foot The age of the dinosaur was estimated using a radioactive substance dating of rocks surrounding the bones. Complete parts a and b. a. The radioactive substance decays exponentially with a half-life of approximately 1.22 billion years. Use the fact that after 1.22 billion years a given amount of the radioactive substance will have decayed to half the original amount to show that the decay model for the radioactive substance is given by A = An e -0 56815t where t is in billions of years. oe To show that the decay model for the radioactive substance is A= An e-0 56815t find the decay rate k for a substance, Substitute the values of A and t in the exponential decay model, A= A, e A=A, ek Ag = A, e122k 2 Substitute D=e1.22k Divide both sides of the equation by Ag-

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**Educational Resource on Radioactive Decay:**

**Understanding Radioactive Decay in Dinosaur Bones**

The bones of a newly discovered dinosaur weigh 170 pounds and measure 9 feet, with a 6-inch claw on one toe of each hind foot. The age of the dinosaur was estimated using a radioactive substance dating of rocks surrounding the bones.

**Activity Overview:**

Complete parts a and b as outlined below.

**a. Exponential Decay Model of Radioactive Substances**

The radioactive substance decays exponentially with a half-life of approximately 1.22 billion years. Given that after 1.22 billion years a given amount of the radioactive substance will have decayed to half the original amount, show that the decay model for the radioactive substance is given by:

\[ A = A_0 e^{-0.5681t} \]

where \( t \) is in billions of years.

**Steps to Derive the Decay Model:**

1. Start with the exponential decay formula:
   \[ A = A_0 e^{kt} \]

2. Substitute the condition where the substance is reduced to half its original amount after 1.22 billion years:
   \[ \frac{A_0}{2} = A_0 e^{1.22k} \]

3. Divide both sides of the equation by \( A_0 \):
   \[ \frac{1}{2} = e^{1.22k} \]

4. Solve for \( k \) by taking the natural logarithm of both sides:
   \[ \ln\left(\frac{1}{2}\right) = 1.22k \]

5. Rearrange to find \( k \):
   \[ k = \frac{\ln(0.5)}{1.22} \]

This solves to \( k \approx -0.5681 \).

Thus, the exponential decay model is confirmed as:
\[ A = A_0 e^{-0.5681t} \]

This formula allows scientists to estimate the age of fossils by analyzing the remaining amount of a radioactive substance found in the surrounding rocks.
Transcribed Image Text:**Educational Resource on Radioactive Decay:** **Understanding Radioactive Decay in Dinosaur Bones** The bones of a newly discovered dinosaur weigh 170 pounds and measure 9 feet, with a 6-inch claw on one toe of each hind foot. The age of the dinosaur was estimated using a radioactive substance dating of rocks surrounding the bones. **Activity Overview:** Complete parts a and b as outlined below. **a. Exponential Decay Model of Radioactive Substances** The radioactive substance decays exponentially with a half-life of approximately 1.22 billion years. Given that after 1.22 billion years a given amount of the radioactive substance will have decayed to half the original amount, show that the decay model for the radioactive substance is given by: \[ A = A_0 e^{-0.5681t} \] where \( t \) is in billions of years. **Steps to Derive the Decay Model:** 1. Start with the exponential decay formula: \[ A = A_0 e^{kt} \] 2. Substitute the condition where the substance is reduced to half its original amount after 1.22 billion years: \[ \frac{A_0}{2} = A_0 e^{1.22k} \] 3. Divide both sides of the equation by \( A_0 \): \[ \frac{1}{2} = e^{1.22k} \] 4. Solve for \( k \) by taking the natural logarithm of both sides: \[ \ln\left(\frac{1}{2}\right) = 1.22k \] 5. Rearrange to find \( k \): \[ k = \frac{\ln(0.5)}{1.22} \] This solves to \( k \approx -0.5681 \). Thus, the exponential decay model is confirmed as: \[ A = A_0 e^{-0.5681t} \] This formula allows scientists to estimate the age of fossils by analyzing the remaining amount of a radioactive substance found in the surrounding rocks.
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