The present day abundances (in moles) of the isotopes `U^(238)andU^(235)` are in the atio of 128 : 1. They have half lives of `4.5xx10^(9)` years and `7xx10^(8)` years respectively. If age of earth in `(49X)/(76)xx10^(7)` years, then calculate X (Assume equal moles of each isotope existed at the time of formation of the earth)

Answer

To solve the problem, we will use the concept of radioactive decay and the relationship between the initial and present abundances of isotopes.

### Step-by-Step Solution:

1. **Understanding the Problem:**
We have two isotopes of Uranium: \( U^{238} \) and \( U^{235} \). Their present-day abundances are in the ratio of 128:1. We are given their half-lives and need to find the value of \( X \) in the age of the Earth, given as \( \frac{49X}{76} \times 10^7 \) years.

2. **Define Variables:**
Let:
- \( N_0 \) = initial amount of \( U^{238} \) = initial amount of \( U^{235} \) (since they are equal at the time of formation)
- \( N_{238} \) = present amount of \( U^{238} \)
- \( N_{235} \) = present amount of \( U^{235} \)

3. **Decay Formula:**
The amount of a radioactive substance remaining after time \( t \) is given by:
\[ N = N_0 e^{-\lambda t} \]
where \( \lambda = \frac{\ln(2)}{T_{1/2}} \) and \( T_{1/2} \) is the half-life.

4. **Calculate Decay Constants:**
For \( U^{238} \):
\[ \lambda_{238} = \frac{\ln(2)}{4.5 \times 10^9} \text{ years}^{-1} \]
For \( U^{235} \):
\[ \lambda_{235} = \frac{\ln(2)}{7 \times 10^8} \text{ years}^{-1} \]

5. **Present Amounts:**
Using the decay formula:
\[ N_{238} = N_0 e^{-\lambda_{238} t} \]
\[ N_{235} = N_0 e^{-\lambda_{235} t} \]

6. **Ratio of Present Amounts:**
Given the ratio \( \frac{N_{238}}{N_{235}} = 128 \):
\[ \frac{N_0 e^{-\lambda_{238} t}}{N_0 e^{-\lambda_{235} t}} = 128 \]
Simplifying gives:
\[ e^{-(\lambda_{238} - \lambda_{235}) t} = 128 \]

7. **Taking Natural Logarithm:**
Taking the natural logarithm of both sides:
\[ -(\lambda_{238} - \lambda_{235}) t = \ln(128) \]

8. **Substituting Values:**
Substitute the values of \( \lambda_{238} \) and \( \lambda_{235} \):
\[ -\left(\frac{\ln(2)}{4.5 \times 10^9} - \frac{\ln(2)}{7 \times 10^8}\right) t = \ln(128) \]

9. **Calculating \( t \):**
Rearranging gives:
\[ t = \frac{\ln(128)}{\left(\frac{\ln(2)}{4.5 \times 10^9} - \frac{\ln(2)}{7 \times 10^8}\right)} \]

10. **Finding Age of Earth:**
We know that the age of the Earth is given as \( \frac{49X}{76} \times 10^7 \). Set this equal to \( t \) and solve for \( X \).

11. **Final Calculation:**
After calculating \( t \) and comparing it to \( \frac{49X}{76} \times 10^7 \), we can solve for \( X \).