A radioactive sample of `^(238)U` decay to Pb through a process for which the half is `4.5 xx 10^(9)` year. Find the ratio of number of nuclei of Pb to `^(238)U`after a time of `1.5 xx 10^(9)` year Given `(2)^(1//3)= 1.26`

Options

A. 0.12
B. 0.26
C. 1.2
D. 0.37

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the given information
We have a radioactive sample of Uranium-238 (U-238) that decays into Lead (Pb) with a half-life of \( T_{1/2} = 4.5 \times 10^9 \) years. We need to find the ratio of the number of nuclei of Pb to U-238 after a time of \( t = 1.5 \times 10^9 \) years.

### Step 2: Use the decay formula
The number of radioactive nuclei remaining after time \( t \) can be calculated using the formula:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{t}{T_{1/2}}} \]
Where:
- \( N(t) \) is the number of nuclei remaining after time \( t \),
- \( N_0 \) is the initial number of nuclei,
- \( T_{1/2} \) is the half-life.

### Step 3: Calculate the exponent
Substituting the values into the formula:
\[ N(t) = N_0 \left( \frac{1}{2} \right)^{\frac{1.5 \times 10^9}{4.5 \times 10^9}} = N_0 \left( \frac{1}{2} \right)^{\frac{1}{3}} \]

### Step 4: Simplify the expression
Using the property of exponents:
\[ N(t) = N_0 \cdot 2^{-\frac{1}{3}} = \frac{N_0}{2^{\frac{1}{3}}} \]

### Step 5: Substitute the value of \( 2^{\frac{1}{3}} \)
Given that \( 2^{\frac{1}{3}} = 1.26 \):
\[ N(t) = \frac{N_0}{1.26} \]

### Step 6: Calculate the number of Pb nuclei
The number of Pb nuclei formed will be the initial number of U-238 nuclei minus the number of U-238 nuclei remaining:
\[ N_{Pb} = N_0 - N(t) = N_0 - \frac{N_0}{1.26} = N_0 \left(1 - \frac{1}{1.26}\right) \]
Calculating the fraction:
\[ 1 - \frac{1}{1.26} = \frac{1.26 - 1}{1.26} = \frac{0.26}{1.26} \]
Thus,
\[ N_{Pb} = N_0 \cdot \frac{0.26}{1.26} \]

### Step 7: Find the ratio of Pb to U-238
Now we can find the ratio of the number of Pb nuclei to the number of U-238 nuclei remaining:
\[ \text{Ratio} = \frac{N_{Pb}}{N(t)} = \frac{N_0 \cdot \frac{0.26}{1.26}}{\frac{N_0}{1.26}} = \frac{0.26}{1} = 0.26 \]

### Final Answer
The ratio of the number of nuclei of Pb to U-238 after \( 1.5 \times 10^9 \) years is \( 0.26 \).

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