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Math  /  Algebra

QuestionA specific radioactive substance follows a continuous exponential decay model. It has a half-life of 21 hours. At the start of the experiment, 99.8 g is present. (a) Let tt be the time (in hours) since the start of the experiment, and let yy be the amount of the substance at time tt.
Write a formula relating yy to tt. Use exact expressions to fill in the missing parts of the formula. Do not use approximations. y=□e(□)ty=\square e^{(\square) t} (b) How much will be present in 12 hours?
Do not round any intermediate computations, and round your answer to the nearest tenth.

Studdy Solution

STEP 1

1. The decay of the substance follows a continuous exponential decay model.
2. The half-life of the substance is 21 hours.
3. The initial amount of the substance is 99.8 grams.
4. We need to find the formula relating y y to t t and calculate the amount present after 12 hours.

STEP 2

1. Determine the decay constant using the half-life.
2. Write the formula relating y y to t t .
3. Calculate the amount of substance after 12 hours.

STEP 3

Determine the decay constant using the half-life.
The formula for the decay constant k k in terms of the half-life T1/2 T_{1/2} is:
k=ln⁡(0.5)T1/2 k = \frac{\ln(0.5)}{T_{1/2}}
Substitute the given half-life:
k=ln⁡(0.5)21 k = \frac{\ln(0.5)}{21}

STEP 4

Write the formula relating y y to t t .
The general formula for exponential decay is:
y=y0ekt y = y_0 e^{kt}
Where y0 y_0 is the initial amount. Substitute y0=99.8 y_0 = 99.8 and k=ln⁡(0.5)21 k = \frac{\ln(0.5)}{21} :
y=99.8e(ln⁡(0.5)21)t y = 99.8 e^{\left(\frac{\ln(0.5)}{21}\right)t}

STEP 5

Calculate the amount of substance after 12 hours.
Substitute t=12 t = 12 into the formula:
y=99.8e(ln⁡(0.5)21)×12 y = 99.8 e^{\left(\frac{\ln(0.5)}{21}\right) \times 12}
Calculate the exponent:
ln⁡(0.5)21×12=12ln⁡(0.5)21 \frac{\ln(0.5)}{21} \times 12 = \frac{12 \ln(0.5)}{21}
Now calculate y y :
y=99.8e(12ln⁡(0.5)21) y = 99.8 e^{\left(\frac{12 \ln(0.5)}{21}\right)}
This simplifies to:
y≈99.8×e−0.396 y \approx 99.8 \times e^{-0.396}
Calculate the final value:
y≈99.8×0.672 y \approx 99.8 \times 0.672
y≈67.1 y \approx 67.1
The amount of substance present after 12 hours is approximately 67.1 67.1 grams.

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