The intensity of X–rays of wavelength 0.5Å reduces to one fourth on passing through 3.5 mm. Thickness of a metal foil. The coefficient of absorption of metal will be–

Options

A. `0.2 mm^(-1)`
B. `0.4 mm^(-1)`
C. `0.6 mm^(-1)`
D. `0.8mm^(-1)`

Answer

To find the coefficient of absorption (μ) of the metal foil, we can use the formula related to the intensity of X-rays passing through a material:

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

Where:
- \( I \) is the final intensity after passing through the material,
- \( I_0 \) is the initial intensity,
- \( \mu \) is the coefficient of absorption,
- \( x \) is the thickness of the material.

Given that the intensity of X-rays reduces to one fourth, we can express this as:

\[ I = \frac{I_0}{4} \]

Now, substituting this into the equation:

\[ \frac{I_0}{4} = I_0 e^{-\mu x} \]

Dividing both sides by \( I_0 \):

\[ \frac{1}{4} = e^{-\mu x} \]

Taking the natural logarithm of both sides:

\[ \ln\left(\frac{1}{4}\right) = -\mu x \]

We can rewrite \( \frac{1}{4} \) as \( 4^{-1} \):

\[ \ln(4^{-1}) = -\mu x \]

Using the property of logarithms:

\[ -\ln(4) = -\mu x \]

Thus:

\[ \mu x = \ln(4) \]

Now, substituting \( x = 3.5 \, \text{mm} \):

\[ \mu \cdot 3.5 = \ln(4) \]

Now, we know that \( \ln(4) = 2 \ln(2) \). The value of \( \ln(2) \) is approximately \( 0.693 \):

\[ \ln(4) = 2 \cdot 0.693 = 1.386 \]

So we have:

\[ \mu \cdot 3.5 = 1.386 \]

Now, solving for \( \mu \):

\[ \mu = \frac{1.386}{3.5} \]

Calculating this gives:

\[ \mu \approx 0.396 \, \text{mm}^{-1} \]

Thus, the coefficient of absorption of the metal is approximately \( 0.396 \, \text{mm}^{-1} \).

### Final Answer:
The coefficient of absorption of the metal is \( 0.396 \, \text{mm}^{-1} \).

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