Answer :
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!
### 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!
