The intensity of gamma radiation from a given source is `I`
On passing through `36 mm` of lead , it is reduced to `(1)/(8)` . The thickness of lead which will reduce the intensity to `(1)/(2)` will be

Options

A. 6mm
B. 9mm
C. 18mm
D. 12mm

Explanation

To solve the problem, we need to determine the thickness of lead that will reduce the intensity of gamma radiation from a given source to half its original value. We are given that the intensity is reduced to \( \frac{1}{8} \) after passing through \( 36 \, \text{mm} \) of lead.

### Step-by-Step Solution:

1. **Understanding the Intensity Reduction**:
The intensity of gamma radiation decreases exponentially as it passes through a material. The relationship can be expressed as:
\[ I = I_0 e^{-\mu d} \]
where:
- \( I \) is the final intensity,
- \( I_0 \) is the initial intensity,
- \( \mu \) is the linear attenuation coefficient,
- \( d \) is the thickness of the material.

2. **Setting Up the Equation for Given Conditions**:
From the problem, when \( d = 36 \, \text{mm} \), the intensity reduces to \( \frac{1}{8} I_0 \). Thus, we can write:
\[ \frac{1}{8} I_0 = I_0 e^{-\mu \cdot 36} \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{8} = e^{-\mu \cdot 36} \]

3. **Taking the Natural Logarithm**:
Taking the natural logarithm of both sides:
\[ \ln\left(\frac{1}{8}\right) = -\mu \cdot 36 \]
We know that \( \ln\left(\frac{1}{8}\right) = \ln(8^{-1}) = -\ln(8) \), so:
\[ -\ln(8) = -\mu \cdot 36 \]
This simplifies to:
\[ \mu \cdot 36 = \ln(8) \]

4. **Finding the Thickness for Half Intensity**:
Now, we need to find the thickness \( d \) that will reduce the intensity to half, i.e., \( I = \frac{1}{2} I_0 \):
\[ \frac{1}{2} I_0 = I_0 e^{-\mu d} \]
Dividing by \( I_0 \):
\[ \frac{1}{2} = e^{-\mu d} \]
Taking the natural logarithm:
\[ \ln\left(\frac{1}{2}\right) = -\mu d \]
This simplifies to:
\[ -\ln(2) = -\mu d \quad \Rightarrow \quad \mu d = \ln(2) \]

5. **Relating the Two Equations**:
From the first equation, we have:
\[ \mu = \frac{\ln(8)}{36} \]
Substituting this into the second equation:
\[ \frac{\ln(8)}{36} d = \ln(2) \]

6. **Solving for \( d \)**:
Rearranging gives:
\[ d = \frac{36 \ln(2)}{\ln(8)} \]
Since \( \ln(8) = 3 \ln(2) \):
\[ d = \frac{36 \ln(2)}{3 \ln(2)} = \frac{36}{3} = 12 \, \text{mm} \]

### Final Answer:
The thickness of lead which will reduce the intensity to \( \frac{1}{2} \) is \( 12 \, \text{mm} \).