Almost all of naturally occurring uranium is `._92^238U` with a half-life of `4.468xx10^9` yr. Most of the rest of natural uranium is `._92^235`U with a half-life of `7.038xx10^8` yr.Today a sample contains 0.72% `._92^235`U
(a)What was this percentage 1.0 billion years ago ?
(b)What percentage of the sample would be `._92^235`U in 100 million years ?

Answer

To solve the problem, we will break it down into two parts as specified in the question.

### Part (a): What was the percentage of `._92^235U` 1.0 billion years ago?

1. **Identify the current amount of `._92^235U`:**
- Given that the current percentage of `._92^235U` is 0.72%, we can assume a sample size for easier calculations. Let's assume we have 100 grams of uranium.
- Therefore, the current amount of `._92^235U` is:
\[ n = 0.72 \text{ grams} \]

2. **Calculate the initial amount of `._92^235U` using the decay formula:**
- The decay formula is given by:
\[ n = n_0 e^{-\frac{0.693 t}{t_{1/2}}} \]
- Rearranging this to find \( n_0 \):
\[ n_0 = n e^{\frac{0.693 t}{t_{1/2}}} \]
- Here, \( t = 1 \times 10^9 \) years (1 billion years) and \( t_{1/2} = 7.038 \times 10^8 \) years (half-life of `._92^235U`).
- Substituting the values:
\[ n_0 = 0.72 e^{\frac{0.693 \times 1 \times 10^9}{7.038 \times 10^8}} \]

3. **Calculate the exponent:**
- First, calculate \( \frac{0.693 \times 1 \times 10^9}{7.038 \times 10^8} \):
\[ \frac{0.693 \times 10^9}{7.038 \times 10^8} \approx 0.0983 \]
- Therefore:
\[ n_0 \approx 0.72 e^{0.0983} \]

4. **Calculate \( e^{0.0983} \):**
- Using a calculator, we find:
\[ e^{0.0983} \approx 1.103 \]
- Thus:
\[ n_0 \approx 0.72 \times 1.103 \approx 0.793 \text{ grams} \]

5. **Calculate the total uranium amount (including `._92^238U`):**
- The amount of `._92^238U` is:
\[ 100 - 0.72 = 99.28 \text{ grams} \]

6. **Calculate the percentage of `._92^235U` 1 billion years ago:**
- The total amount of uranium is:
\[ n_0 + 99.28 = 0.793 + 99.28 \approx 100.073 \text{ grams} \]
- The percentage of `._92^235U` is:
\[ \text{Percentage} = \left(\frac{0.793}{100.073}\right) \times 100 \approx 0.792\% \]

### Part (b): What percentage of the sample would be `._92^235U` in 100 million years?

1. **Use the decay formula again:**
- We will use the same formula:
\[ n = n_0 e^{-\frac{0.693 t}{t_{1/2}}} \]
- Here, \( t = 1 \times 10^8 \) years (100 million years).
- Using the initial amount \( n_0 = 0.72 \) grams:
\[ n = 0.72 e^{-\frac{0.693 \times 1 \times 10^8}{7.038 \times 10^8}} \]

2. **Calculate the exponent:**
- First, calculate \( \frac{0.693 \times 1 \times 10^8}{7.038 \times 10^8} \):
\[ \frac{0.693 \times 10^8}{7.038 \times 10^8} \approx 0.0983 \times \frac{1}{10} \approx 0.00983 \]
- Therefore:
\[ n \approx 0.72 e^{-0.00983} \]

3. **Calculate \( e^{-0.00983} \):**
- Using a calculator, we find:
\[ e^{-0.00983} \approx 0.9902 \]
- Thus:
\[ n \approx 0.72 \times 0.9902 \approx 0.712 \text{ grams} \]

4. **Calculate the amount of `._92^238U` after 100 million years:**
- The amount of `._92^238U` will remain approximately the same since its half-life is much longer:
\[ n_{238} \approx 99.28 \text{ grams} \]

5. **Calculate the total amount of uranium:**
- Total amount:
\[ n + n_{238} \approx 0.712 + 99.28 \approx 99.992 \text{ grams} \]

6. **Calculate the percentage of `._92^235U`:**
- The percentage of `._92^235U` is:
\[ \text{Percentage} = \left(\frac{0.712}{99.992}\right) \times 100 \approx 0.712\% \]

### Final Answers:
(a) The percentage of `._92^235U` 1.0 billion years ago was approximately **0.792%**.
(b) The percentage of `._92^235U` in 100 million years would be approximately **0.712%**.