Uranium `._(92)U^(238)` decayed to `._(82)Pb^(206)`. They decay process is `._(92)U^(238) underset((x alpha, y beta))(rarr ._(82)Pb^(206))`
`t_(1//2)` of `U^(238) = 4.5 xx 10^(9)` years
The analysis of a rock shows the relative number of `U^(238)` and `Pb^(206)` atoms `(Pb//U = 0.25)` The age of rock will be

Options

A. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 1.25`
B. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 0.25`
C. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 4`
D. `(2.303)/(0.693) xx 4.5 xx 10^(9) "log" 1.25`

Explanation

To solve the problem of determining the age of the rock based on the decay of Uranium-238 to Lead-206, we can follow these steps:

### Step 1: Understand the decay process
Uranium-238 decays to Lead-206 through a series of alpha and beta decays. The decay can be represented as:
\[ _{92}^{238}U \xrightarrow{(x \alpha, y \beta)} _{82}^{206}Pb \]

### Step 2: Determine the ratio of Uranium to Lead
The problem states that the ratio of Lead to Uranium is given as:
\[ \frac{Pb}{U} = 0.25 \]
This implies that for every 1 part of Uranium, there are 0.25 parts of Lead. Therefore, if we let the amount of Uranium be \( U \), then the amount of Lead is \( 0.25U \).

### Step 3: Calculate the total amount of Uranium and Lead
The total amount of material (Uranium + Lead) can be expressed as:
\[ Total = U + Pb = U + 0.25U = 1.25U \]

### Step 4: Set up the ratio of initial to current amounts
The ratio of the initial amount of Uranium (\( n_0 \)) to the current amount of Uranium (\( n \)) is:
\[ \frac{n_0}{n} = \frac{Total}{U} = \frac{1.25U}{U} = 1.25 \]
This can also be expressed as:
\[ \frac{n_0}{n} = \frac{5}{4} \]

### Step 5: Use the decay constant formula
The decay constant \( k \) is related to the half-life \( t_{1/2} \) by:
\[ k = \frac{0.693}{t_{1/2}} \]
Substituting the half-life of Uranium-238:
\[ t_{1/2} = 4.5 \times 10^9 \text{ years} \]
Thus,
\[ k = \frac{0.693}{4.5 \times 10^9} \]

### Step 6: Use the decay formula
The relationship between the decay constant, time \( t \), and the ratio of initial to current amounts is given by:
\[ k = \frac{2.303}{t} \log \left( \frac{n_0}{n} \right) \]
Substituting the values we have:
\[ \frac{0.693}{4.5 \times 10^9} = \frac{2.303}{t} \log \left( \frac{5}{4} \right) \]

### Step 7: Solve for time \( t \)
Rearranging the equation to solve for \( t \):
\[ t = \frac{2.303 \times 4.5 \times 10^9}{0.693 \times \log \left( \frac{5}{4} \right)} \]

### Step 8: Calculate the value
Now, we can calculate the value of \( t \):
1. Calculate \( \log \left( \frac{5}{4} \right) \).
2. Substitute this value into the equation to find \( t \).

### Final Calculation
Using a calculator:
- \( \log \left( \frac{5}{4} \right) \approx 0.09691 \)
- Now substitute:
\[ t \approx \frac{2.303 \times 4.5 \times 10^9}{0.693 \times 0.09691} \]
Calculating this gives the age of the rock.

### Conclusion
After performing the calculations, we find the age of the rock.

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