Uranium- `238` decays to thorium-`234` with half-life `5xx10^(9) yr`. The resulting nucleus is in the excited state and hence further emits `gamma`-rays to come to the ground state. It emits `20 gamma`-rays per second. The emission rate will drop to `5 gamma`-rays per second in

Options

A. (a)`.25xx10^(9)yr`
B. (b)`10^(10)yr`
C. (c)`10^(-8)yr`
D. (d)`1.25xx10^(-9)s`

Answer

To solve the problem, we need to determine the time it takes for the emission rate of gamma rays from the excited state of Thorium-234 to drop from 20 gamma rays per second to 5 gamma rays per second.

### Step-by-step Solution:

1. **Understanding the decay process**:
- Uranium-238 decays to Thorium-234, which is in an excited state and emits gamma rays to reach its ground state.
- The initial emission rate of gamma rays is 20 gamma rays per second.

2. **Establishing the relationship**:
- The emission rate of gamma rays is proportional to the number of excited nuclei present.
- We denote the initial emission rate as \( N_0 = 20 \) gamma rays/second and the final emission rate as \( N = 5 \) gamma rays/second.

3. **Using the decay formula**:
- The decay of the excited nuclei can be expressed as:
\[ \frac{N}{N_0} = \frac{5}{20} = \frac{1}{4} \]
- This means that after some time \( t \), the number of excited nuclei has decreased to one-fourth of its initial value.

4. **Applying the natural logarithm**:
- We can use the relationship of decay:
\[ \ln\left(\frac{N}{N_0}\right) = -\lambda t \]
- Substituting the values we found:
\[ \ln\left(\frac{1}{4}\right) = -\lambda t \]

5. **Finding the decay constant \( \lambda \)**:
- The half-life \( T_{1/2} \) of Uranium-238 is given as \( 5 \times 10^9 \) years.
- The decay constant \( \lambda \) is related to the half-life by:
\[ \lambda = \frac{\ln(2)}{T_{1/2}} \]
- Substituting the half-life:
\[ \lambda = \frac{\ln(2)}{5 \times 10^9} \]

6. **Substituting \( \lambda \) into the decay equation**:
- Now we can substitute \( \lambda \) back into the equation:
\[ \ln\left(\frac{1}{4}\right) = -\left(\frac{\ln(2)}{5 \times 10^9}\right)t \]
- Since \( \ln\left(\frac{1}{4}\right) = -2 \ln(2) \):
\[ -2 \ln(2) = -\left(\frac{\ln(2)}{5 \times 10^9}\right)t \]

7. **Solving for \( t \)**:
- Canceling \( \ln(2) \) from both sides:
\[ 2 = \frac{t}{5 \times 10^9} \]
- Rearranging gives:
\[ t = 2 \times 5 \times 10^9 = 10 \times 10^9 = 10^{10} \text{ years} \]

### Final Answer:
The emission rate will drop to 5 gamma rays per second in \( 10^{10} \) years.