Math Problem Statement

A specific radioactive substance follows a continuous exponential decay model. It has a half-life of 4 hours. At the start of the experiment, 89.8g is present. How much will be present in 21 hours? Do not round any intermediate computations, and round your answer to the nearest tenth.

Solution

To solve this problem, we need to use the exponential decay formula, which is given by:

A(t)=A0ektA(t) = A_0 \cdot e^{kt}

where:

  • A(t)A(t) is the amount of substance remaining at time tt.
  • A0A_0 is the initial amount of the substance.
  • kk is the decay constant.
  • tt is the time elapsed.

Step 1: Determine the decay constant kk

The decay constant kk is related to the half-life (t1/2t_{1/2}) by the formula:

k=ln(2)t1/2k = \frac{\ln(2)}{t_{1/2}}

Given the half-life t1/2=4t_{1/2} = 4 hours, we can calculate kk:

k=ln(2)4=0.69314740.1732867hours1k = \frac{\ln(2)}{4} = \frac{0.693147}{4} \approx 0.1732867 \, \text{hours}^{-1}

Step 2: Apply the exponential decay formula

Now that we have kk, we can substitute the values into the exponential decay formula to find the amount of substance remaining after 21 hours.

A(21)=89.8e(0.1732867×21)A(21) = 89.8 \cdot e^{(0.1732867 \times -21)}

Step 3: Compute the exponent and the remaining amount

First, compute the exponent:

0.1732867×21=3.63802070.1732867 \times -21 = -3.6380207

Then, calculate e3.6380207e^{-3.6380207}:

e3.63802070.026211e^{-3.6380207} \approx 0.026211

Finally, multiply by the initial amount A0=89.8A_0 = 89.8:

A(21)89.8×0.0262112.353208A(21) \approx 89.8 \times 0.026211 \approx 2.353208

Step 4: Round the result

Rounding to the nearest tenth:

A(21)2.4gA(21) \approx 2.4 \, \text{g}

Thus, after 21 hours, approximately 2.4 grams of the substance will be present.

Would you like any further details or have any questions? Here are some related questions you might consider:

  1. How would the remaining amount change if the half-life was 2 hours instead of 4?
  2. What would happen if the experiment lasted for 30 hours instead of 21?
  3. How can we find the time required for the substance to decay to a specific amount?
  4. What does the decay constant kk represent in practical terms?
  5. How would the equation change if the substance followed a different decay model?
  6. What is the importance of not rounding intermediate steps in this kind of calculation?
  7. How would the calculations differ if the initial amount were different, say 100g?
  8. Can you derive the formula for the half-life from the exponential decay equation?

Tip: When dealing with exponential decay, always ensure to handle the decay constant carefully, as small errors in kk can lead to significant differences in the final result.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Half-life

Formulas

Exponential decay formula A(t) = A0 * e^(kt)
Decay constant formula k = ln(2) / t_half

Theorems

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Suitable Grade Level

Grades 11-12