) The half-life of plutonium-239 is 24,300 years. If a nuclear bomb released 8 kg of this isotope, how many years would pass before the amount is reduced to 1 kg? The half-life of radon-222 is 3.8 days. How much of a 100 gram sample is left after 15.2 days? 0705 Carbon-14 has a half-life of 5,730 vears. If a sample contained 90 mg originally, how much is left after 17.190 years? 6. The half-life of cobalt-60 is 5.26 years. If 50 grams are left after 15.78 years, how many grams were in the original sample? No 7. The half-life of 1-137 is 8.07 days. If 25 grams are left after 40.35 days, how many grams were in the original sample? 8. If 100 grams of Au-198 decays to 6.25 grams in 10.8 days, what is the half-life of Au-198?
The Correct Answer and Explanation is :
Let’s break down each of the problems one by one. The common formula we will use for all of them is the radioactive decay formula:
[
A = A_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T}}
]
Where:
- ( A ) is the remaining amount of the substance.
- ( A_0 ) is the initial amount of the substance.
- ( t ) is the time elapsed.
- ( T ) is the half-life of the substance.
1. Half-life of Plutonium-239 (24,300 years)
We are given:
- Initial amount, ( A_0 = 8 \, \text{kg} ).
- Remaining amount, ( A = 1 \, \text{kg} ).
- Half-life, ( T = 24,300 \, \text{years} ).
- We need to find ( t ), the time it takes to decay from 8 kg to 1 kg.
Using the decay formula:
[
1 = 8 \times \left(\frac{1}{2}\right)^{\frac{t}{24,300}}
]
Solving for ( t ):
[
\frac{1}{8} = \left(\frac{1}{2}\right)^{\frac{t}{24,300}}
]
Taking the natural logarithm of both sides:
[
\ln\left(\frac{1}{8}\right) = \frac{t}{24,300} \times \ln\left(\frac{1}{2}\right)
]
[
t = \frac{\ln\left(\frac{1}{8}\right)}{\ln\left(\frac{1}{2}\right)} \times 24,300
]
[
t \approx 73,000 \, \text{years}
]
So, it would take approximately 73,000 years for the amount of plutonium-239 to reduce to 1 kg.
2. Half-life of Radon-222 (3.8 days)
Given:
- Initial amount, ( A_0 = 100 \, \text{g} ).
- Half-life, ( T = 3.8 \, \text{days} ).
- Time, ( t = 15.2 \, \text{days} ).
Using the decay formula:
[
A = 100 \times \left(\frac{1}{2}\right)^{\frac{15.2}{3.8}}
]
[
A = 100 \times \left(\frac{1}{2}\right)^{4}
]
[
A = 100 \times \frac{1}{16} = 6.25 \, \text{g}
]
So, after 15.2 days, 6.25 grams of Radon-222 remain.
3. Half-life of Carbon-14 (5730 years)
Given:
- Initial amount, ( A_0 = 90 \, \text{mg} ).
- Time, ( t = 17,190 \, \text{years} ).
- Half-life, ( T = 5730 \, \text{years} ).
Using the decay formula:
[
A = 90 \times \left(\frac{1}{2}\right)^{\frac{17,190}{5730}}
]
[
A = 90 \times \left(\frac{1}{2}\right)^{3}
]
[
A = 90 \times \frac{1}{8} = 11.25 \, \text{mg}
]
So, after 17,190 years, 11.25 mg of Carbon-14 remain.
4. Half-life of Cobalt-60 (5.26 years)
Given:
- Remaining amount, ( A = 50 \, \text{g} ).
- Time, ( t = 15.78 \, \text{years} ).
- Half-life, ( T = 5.26 \, \text{years} ).
- We need to find the initial amount ( A_0 ).
Using the decay formula:
[
50 = A_0 \times \left(\frac{1}{2}\right)^{\frac{15.78}{5.26}}
]
[
50 = A_0 \times \left(\frac{1}{2}\right)^{3}
]
[
50 = A_0 \times \frac{1}{8}
]
[
A_0 = 50 \times 8 = 400 \, \text{g}
]
So, the initial amount of Cobalt-60 was 400 grams.
5. Half-life of Iodine-137 (8.07 days)
Given:
- Remaining amount, ( A = 25 \, \text{g} ).
- Time, ( t = 40.35 \, \text{days} ).
- We need to find the initial amount ( A_0 ).
Using the decay formula:
[
25 = A_0 \times \left(\frac{1}{2}\right)^{\frac{40.35}{8.07}}
]
[
25 = A_0 \times \left(\frac{1}{2}\right)^{5}
]
[
25 = A_0 \times \frac{1}{32}
]
[
A_0 = 25 \times 32 = 800 \, \text{g}
]
So, the initial amount of Iodine-137 was 800 grams.
6. Half-life of Gold-198 (Au-198)
Given:
- Initial amount, ( A_0 = 100 \, \text{g} ).
- Remaining amount, ( A = 6.25 \, \text{g} ).
- Time, ( t = 10.8 \, \text{days} ).
- We need to find the half-life ( T ).
Using the decay formula:
[
6.25 = 100 \times \left(\frac{1}{2}\right)^{\frac{10.8}{T}}
]
[
\frac{6.25}{100} = \left(\frac{1}{2}\right)^{\frac{10.8}{T}}
]
[
0.0625 = \left(\frac{1}{2}\right)^{\frac{10.8}{T}}
]
Since ( 0.0625 = \left(\frac{1}{2}\right)^{4} ), we have:
[
\frac{10.8}{T} = 4
]
[
T = \frac{10.8}{4} = 2.7 \, \text{days}
]
So, the half-life of Au-198 is 2.7 days.
Summary:
- Plutonium-239: 73,000 years to decay to 1 kg.
- Radon-222: 6.25 grams remain after 15.2 days.
- Carbon-14: 11.25 mg remain after 17,190 years.
- Cobalt-60: 400 grams were in the original sample.
- Iodine-137: 800 grams were in the original sample.
- Gold-198: The half-life is 2.7 days.