The intensity of `gamma` - radiation from a given source is I. On passing through 36 mm of lead, it is reduced to `I//8`. Assuming that the intensity in trasmitted radiation varies expontially with the thickness of the material, then the thickness of lead, which will reduce the intensity to `I//2` will be

Options

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

Answer

To solve the problem, we will use the exponential decay formula for the intensity of gamma radiation as it passes through a material. The formula is given by:

\[ I = I_0 e^{-\mu x} \]

Where:
- \( I \) is the transmitted intensity,
- \( I_0 \) is the initial intensity,
- \( \mu \) is the linear attenuation coefficient,
- \( x \) is the thickness of the material.

### Step 1: Establish the relationship for the first case
From the problem, we know that the initial intensity \( I_0 = I \) and after passing through 36 mm of lead, the intensity is reduced to \( \frac{I}{8} \). We can set up the equation:

\[ \frac{I}{8} = I e^{-\mu \cdot 36} \]

### Step 2: Simplify the equation
We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):

\[ \frac{1}{8} = e^{-\mu \cdot 36} \]

### Step 3: Take the natural logarithm of both sides
Taking the natural logarithm gives us:

\[ \ln\left(\frac{1}{8}\right) = -\mu \cdot 36 \]

### Step 4: Express \( \ln\left(\frac{1}{8}\right) \)
We can express \( \ln\left(\frac{1}{8}\right) \) as:

\[ \ln\left(\frac{1}{8}\right) = \ln(8^{-1}) = -\ln(8) = -\ln(2^3) = -3\ln(2) \]

### Step 5: Substitute back into the equation
Now substituting this back into the equation gives us:

\[ -3\ln(2) = -\mu \cdot 36 \]

### Step 6: Solve for \( \mu \)
Rearranging gives:

\[ \mu = \frac{3\ln(2)}{36} = \frac{\ln(2)}{12} \]

### Step 7: Establish the relationship for the second case
Now we want to find the thickness \( x \) that reduces the intensity to \( \frac{I}{2} \):

\[ \frac{I}{2} = I e^{-\mu x} \]

### Step 8: Simplify the equation
Cancelling \( I \) gives us:

\[ \frac{1}{2} = e^{-\mu x} \]

### Step 9: Take the natural logarithm of both sides
Taking the natural logarithm gives:

\[ \ln\left(\frac{1}{2}\right) = -\mu x \]

### Step 10: Express \( \ln\left(\frac{1}{2}\right) \)
We can express \( \ln\left(\frac{1}{2}\right) \) as:

\[ \ln\left(\frac{1}{2}\right) = -\ln(2) \]

### Step 11: Substitute \( \mu \) into the equation
Substituting \( \mu = \frac{\ln(2)}{12} \) gives:

\[ -\ln(2) = -\frac{\ln(2)}{12} x \]

### Step 12: Solve for \( x \)
Rearranging gives:

\[ x = 12 \]

Thus, the thickness of lead that will reduce the intensity to \( \frac{I}{2} \) is **12 mm**.

### Summary of Steps:
1. Set up the exponential decay formula for the first case.
2. Simplify and solve for the linear attenuation coefficient \( \mu \).
3. Set up the formula for the second case where intensity is reduced to \( \frac{I}{2} \).
4. Solve for the thickness \( x \).