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Strontium-90 has a half-life of 28 days.

(a) A sample has an initial mass of [tex]$50 \text{ mg}$[/tex]. Find a formula for the mass remaining after [tex]t[/tex] days.
[tex]\[ y(t) = \square \][/tex]

(b) Find the mass (in [tex]\text{mg}[/tex]) remaining after 20 days. (Round your answer to one decimal place.)
[tex]\[ \square \text{ mg} \][/tex]

(c) How long does it take (in days) for the sample to decay to a mass of [tex]2 \text{ mg}$[/tex]? (Round your answer to one decimal place.)
[tex]\[ \square \text{ days} \][/tex]


Sagot :

Sure! Let's go through this problem step by step:

### Part (a):
We need to find a formula for the mass remaining after \( t \) days.

Given:
- Initial mass, \( y(0) = 50 \) mg
- Half-life of Strontium-90, \( t_{1/2} = 28 \) days

The mass remaining after \( t \) days can be modeled using the exponential decay formula:
[tex]\[ y(t) = y(0) \times (0.5)^{\frac{t}{t_{1/2}}} \][/tex]

Plugging in the given values:
[tex]\[ y(t) = 50 \times (0.5)^{\frac{t}{28}} \][/tex]

So, the formula for the mass remaining after \( t \) days is:
[tex]\[ y(t) = 50 \times (0.5)^{\frac{t}{28}} \][/tex]

### Part (b):
We need to find the mass remaining after 20 days.

Using our formula:
[tex]\[ y(t) = 50 \times (0.5)^{\frac{t}{28}} \][/tex]

Plug in \( t = 20 \):
[tex]\[ y(20) = 50 \times (0.5)^{\frac{20}{28}} \][/tex]

After performing the calculation, the mass remaining after 20 days is approximately:
[tex]\[ y(20) \approx 30.5 \text{ mg} \][/tex]

### Part (c):
We need to find the time it takes for the sample to decay to 2 mg.

Given:
- Final mass, \( y(t) = 2 \) mg

Using our formula:
[tex]\[ 2 = 50 \times (0.5)^{\frac{t}{28}} \][/tex]

We need to solve for \( t \).

1. Divide both sides by 50:
[tex]\[ \frac{2}{50} = (0.5)^{\frac{t}{28}} \][/tex]

2. Simplify the fraction:
[tex]\[ 0.04 = (0.5)^{\frac{t}{28}} \][/tex]

3. Take the natural logarithm of both sides:
[tex]\[ \ln(0.04) = \ln\left((0.5)^{\frac{t}{28}}\right) \][/tex]

4. Use the logarithm power rule:
[tex]\[ \ln(0.04) = \frac{t}{28} \ln(0.5) \][/tex]

5. Solve for \( t \):
[tex]\[ t = \frac{28 \times \ln(0.04)}{\ln(0.5)} \][/tex]

After performing the calculation, the time it takes for the sample to decay to 2 mg is approximately:
[tex]\[ t \approx 130.0 \text{ days} \][/tex]

In summary:
(a) The formula for the mass remaining after \( t \) days is:
[tex]\[ y(t) = 50 \times (0.5)^{\frac{t}{28}} \][/tex]
(b) The mass remaining after 20 days is approximately:
[tex]\[ 30.5 \text{ mg} \][/tex]
(c) The time it takes for the sample to decay to 2 mg is approximately:
[tex]\[ 130.0 \text{ days} \][/tex]

I hope this explanation helps you to understand the solution!