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A sample of a radioactive substance has an initial mass of 769.3 mg. This substance follows a continuous exponential decay model and has a half-life of 23 hours.

(a) Let [tex]t[/tex] be the time (in hours) since the start of the experiment, and let [tex]y[/tex] be the amount of the substance at time [tex]t[/tex]. Write a formula relating [tex]y[/tex] to [tex]t[/tex]. Use exact expressions to fill in the missing parts of the formula. Do not use approximations.

[tex]\[ y = \square e^{ D t} \][/tex]

(b) How much will be present in 21 hours? Do not round any intermediate computations, and round your answer to the nearest tenth.

[tex]\[ \square \text{ mg} \][/tex]

Sagot :

GinnyAnsweridn
Sure, let's tackle this problem step by step.

### Part (a)

We are given the initial mass of the radioactive substance and its half-life. To find the formula relating the remaining mass [tex]\( y \)[/tex] to time [tex]\( t \)[/tex] under the continuous exponential decay model, we start by defining the relationship through the exponential decay formula:

[tex]\[ y = y_0 e^{kt} \][/tex]

Here:

- [tex]\( y_0 \)[/tex] is the initial mass of the substance.
- [tex]\( k \)[/tex] is the decay constant.
- [tex]\( t \)[/tex] is the time in hours.

We need to determine the decay constant [tex]\( k \)[/tex]. Using the fact that the half-life of the substance is 23 hours, we know that:

[tex]\[ y \bigg|_{t=23} = \frac{y_0}{2} \][/tex]

Using the exponential decay formula when [tex]\( t = 23 \)[/tex]:

[tex]\[ \frac{y_0}{2} = y_0 e^{k \cdot 23} \][/tex]

We can cancel [tex]\( y_0 \)[/tex] on both sides of the equation:

[tex]\[ \frac{1}{2} = e^{k \cdot 23} \][/tex]

Taking the natural logarithm on both sides:

[tex]\[ \ln\left(\frac{1}{2}\right) = k \cdot 23 \][/tex]

Simplifying for [tex]\( k \)[/tex]:

[tex]\[ k = \frac{\ln\left(\frac{1}{2}\right)}{23} \][/tex]

[tex]\[ k = \frac{\ln(0.5)}{23} \][/tex]

The formula relating [tex]\( y \)[/tex] to [tex]\( t \)[/tex] is then:

[tex]\[ y = 769.3 e^{\left(\frac{\ln(0.5)}{23}\right) t} \][/tex]

So in the given formula [tex]\( y = \square e^{Dt} \)[/tex], [tex]\( \square \)[/tex] should be replaced by the initial mass 769.3, and [tex]\( D \)[/tex] should be replaced by [tex]\( \frac{\ln(0.5)}{23} \)[/tex].

Thus, the exact formula is:

[tex]\[ y = 769.3 e^{\left(\frac{\ln(0.5)}{23} \right) t} \][/tex]

### Part (b)

To find out how much substance will be present after 21 hours, we substitute [tex]\( t = 21 \)[/tex] into our formula:

[tex]\[ y = 769.3 e^{\left( \frac{\ln(0.5)}{23} \right) 21} \][/tex]

Evaluating the exponent:

[tex]\[ \frac{\ln(0.5)}{23} \approx -0.0301368 \][/tex]

So:

[tex]\[ y = 769.3 e^{-0.0301368 \cdot 21} \][/tex]

[tex]\[ y \approx 769.3 e^{-0.632873} \][/tex]

[tex]\[ y \approx 769.3 \times 0.5312 \][/tex]

[tex]\[ y \approx 408.5 \][/tex]

Therefore, the amount of the substance present after 21 hours will be approximately 408.5 mg, rounded to the nearest tenth.
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