Question

The half-life of radon-222 is 3.8 days. How much of a 100 gram sample is left after 15.2 days?
5. Carbon-14 has a half-life of 5,730 years. If a sample contained 70 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?
7. The half-life of I-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?

Answer 1

To solve these questions, we'll use the formula for radioactive decay, which is:

\[
N = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}}
\]

Where:
- \( N \) = remaining quantity of substance
- \( N_0 \) = initial quantity of substance
- \( t \) = time elapsed
- \( T_{1/2} \) = half-life of the substance

### 1. Radon-222 (half-life = 3.8 days)
Given:
- \( N_0 = 100 \) grams
- \( t = 15.2 \) days
- \( T_{1/2} = 3.8 \) days

Calculating the number of half-lives that have elapsed:
\[
\text{Number of half-lives} = \frac{t}{T_{1/2}} = \frac{15.2 \text{ days}}{3.8 \text{ days}} = 4
\]

Calculating the remaining quantity:
\[
N = 100 \left( \frac{1}{2} \right)^4 = 100 \times \frac{1}{16} = 6.25 \text{ grams}
\]

### 2. Carbon-14 (half-life = 5,730 years)
Given:
- \( N_0 = 70 \) mg
- \( t = 17,190 \) years
- \( T_{1/2} = 5,730 \) years

Calculating the number of half-lives that have elapsed:
\[
\text{Number of half-lives} = \frac{t}{T_{1/2}} = \frac{17,190 \text{ years}}{5,730 \text{ years}} = 3
\]

Calculating the remaining quantity:
\[
N = 70 \left( \frac{1}{2} \right)^3 = 70 \times \frac{1}{8} = 8.75 \text{ mg}
\]

### 3. Cobalt-60 (half-life = 5.26 years)
Given:
- \( N = 50 \) grams
- \( t = 15.78 \) years
- \( T_{1/2} = 5.26 \) years

Calculating the number of half-lives that have elapsed:
\[
\text{Number of half-lives} = \frac{t}{T_{1/2}} = \frac{15.78 \text{ years}}{5.26 \text{ years}} \approx 3
\]

Using the remaining quantity formula, we have:
\[
50 = N_0 \left( \frac{1}{2} \right)^3
\]

Multiply by \( 8 \):
\[
N_0 = 50 \times 8 = 400 \text{ grams}
\]

### 4. I-137 (half-life = 8.07 days)
Given:
- \( N = 25 \) grams
- \( t = 40.35 \) days
- \( T_{1/2} = 8.07 \) days

Calculating the number of half-lives that have elapsed:
\[
\text{Number of half-lives} = \frac{t}{T_{1/2}} = \frac{40.35 \text{ days}}{8.07 \text{ days}} \approx 5
\]

Using the remaining quantity formula, we have:
\[
25 = N_0 \left( \frac{1}{2} \right)^5
\]

Multiply by \( 32 \):
\[
N_0 = 25 \times 32 = 800 \text{ grams}
\]

### 5. Au-198 (decayed from 100 grams to 6.25 grams in 10.8 days)
Given:
- \( N_0 = 100 \) grams
- \( N = 6.25 \) grams
- \( t = 10.8 \) days

First, calculate the number of half-lives that have elapsed:
\[
\frac{N_0}{N} = \frac{100}{6.25} = 16
\]
Or, since \( \frac{1}{2}^n = \frac{1}{16} \):
\[
\Rightarrow 2^n = 16 \rightarrow n = 4
\]

Thus, 4 half-lives have elapsed. So:
\[
T_{1/2} = \frac{t}{4} = \frac{10.8 \text{ days}}{4} = 2.7 \text{ days}
\]

### Summary of Results:
1. Remaining radon-222: **6.25 grams**
2. Remaining carbon-14: **8.75 mg**
3. Original cobalt-60: **400 grams**
4. Original I-137: **800 grams**
5. Half-life of Au-198: **2.7 days**