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 scenario
Given that the initial intensity is \( I_0 = I \) and after passing through 36 mm of lead, the intensity is reduced to \( \frac{I}{8} \).

Using the formula:
\[ \frac{I}{8} = I e^{-\mu \cdot 36} \]

### Step 2: Simplify the equation
Dividing both sides by \( I \):
\[ \frac{1}{8} = e^{-\mu \cdot 36} \]

### Step 3: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln\left(\frac{1}{8}\right) = -\mu \cdot 36 \]

### Step 4: Calculate \( \ln\left(\frac{1}{8}\right) \)
We know that:
\[ \ln\left(\frac{1}{8}\right) = -\ln(8) = -\ln(2^3) = -3\ln(2) \]
Using \( \ln(2) \approx 0.693 \):
\[ \ln\left(\frac{1}{8}\right) \approx -3 \cdot 0.693 \approx -2.079 \]

### Step 5: Solve for \( \mu \)
Now substituting back into the equation:
\[ -2.079 = -\mu \cdot 36 \]
\[ \mu = \frac{2.079}{36} \approx 0.0578 \, \text{mm}^{-1} \]

### Step 6: Establish the relationship for the second scenario
Now we want to find the thickness \( x \) that reduces the intensity to \( \frac{I}{2} \):
\[ \frac{I}{2} = I e^{-\mu x} \]

### Step 7: Simplify the equation
Dividing both sides by \( I \):
\[ \frac{1}{2} = e^{-\mu x} \]

### Step 8: Take the natural logarithm of both sides
Taking the natural logarithm:
\[ \ln\left(\frac{1}{2}\right) = -\mu x \]

### Step 9: Calculate \( \ln\left(\frac{1}{2}\right) \)
We know that:
\[ \ln\left(\frac{1}{2}\right) = -\ln(2) \approx -0.693 \]

### Step 10: Solve for \( x \)
Substituting \( \mu \) into the equation:
\[ -0.693 = -0.0578 x \]
\[ x = \frac{0.693}{0.0578} \approx 12 \, \text{mm} \]

### Final Answer
The thickness of lead that will reduce the intensity to \( \frac{I}{2} \) is approximately **12 mm**.

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