The mass absorption coefficient of aluminium for X-rays of a certain energy is \(0.027 \mathrm{~m}^{2} 7 \mathrm{~kg}\). Calculate the half-value thickness of aluminium for these \(X\) -rays. What thickness of aluminium would attenuate the X-rays by \(80 \%\) ? Density of aluminium is \(2700 \mathrm{~kg} / \mathrm{m}^{3}\).

The mass absorption coefficient plays a crucial role in understanding how materials absorb X-rays. It is denoted by the symbol \( \beta \) and has units of \( \frac{m^2}{kg} \). This parameter indicates how much energy a material can absorb per unit mass. To find it, you typically need experimental data. For instance, in our problem, the mass absorption coefficient for aluminum is \( 0.027 \) \( \frac{m^2}{kg} \).

• The higher the value, the more X-rays are absorbed by the material.
• This coefficient helps to calculate other important factors, such as the linear absorption coefficient and half-value thickness.

Understanding this coefficient is the first step in analyzing X-ray attenuation.

The linear absorption coefficient, often represented as \(\beta_{linear}\), describes how much the intensity of the X-ray beam reduces per unit thickness of the material. It is calculated using the formula:

\[ \beta_{linear} = \beta \times \rho \]

Here, \( \beta \) is the mass absorption coefficient and \( \rho \) is the material's density. For aluminum with \( \beta = 0.027 \frac{m^2}{kg}\) and \( \rho = 2700 \frac{kg}{m^3}\):

\[ \beta_{linear} = 0.027 \times 2700 = 72 \text{ m}^{-1} \] Understanding the linear absorption coefficient allows us to further calculate the material's half-value thickness and attenuation properties.

Half-value thickness, or \( T_{1/2} \), is a measure of the thickness required to reduce the X-ray intensity by 50%. The formula to calculate half-value thickness is:

\[ T_{1/2} = \frac{\text{ln}(2)}{\beta_{linear}} \]

For aluminum, with \(\beta_{linear}= 72 \text{ m}^{-1}\), we substitute the values:

\[ T_{1/2} = \frac{0.693}{72} ≈ 0.0096 \text{ m} \]

• This essentially means you need around 0.0096 meters of aluminum to halve the X-ray intensity.
• Knowing the half-value thickness aids in designing appropriate shielding and protection against X-rays.

Density (\rho\text{\rho}\rho) is a fundamental property of materials, defined as mass per unit volume. It is measured in \(kg/m^3\). The density of a material affects how it attenuates X-rays. In our problem, the density of aluminum is \(\rho=2700\) \frac{kg}{m^3} \.

• A higher density means more atoms per unit volume to absorb X-ray photons, increasing attenuation.
• It's essential to know the density for calculating linear absorption coefficients and other attenuation properties.

Understanding density helps gauge how different materials perform as X-ray shields.

The exponential attenuation formula describes how the intensity \(I \) of an X-ray beam decreases as it passes through a material. It is given by:

\[ I = I_0 \text{e}^{- \beta_{\text{linear}} x} \]

Where:

• \( I_0 \) is the initial intensity.
• \( \beta_{\text{linear}} \) is the linear absorption coefficient.
• \( x \) is the material thickness.

For our problem, to attenuate the X-rays by 80%, we set the transmitted fraction to 0.2 and solve for \ x \:

\[ \text{ln}(0.2) = - \beta_{\text{linear}} x \]

Substituting \(\text{ln}(0.2) ≈ -1.609 \) and \( \beta_{\text{linear}} = 72 \text{ m}^{-1}\):

\[ x = \frac{-1.609}{-72} ≈ 0.0223 \text{ m} \] This calculation shows you need approximately 0.0223 meters of aluminum for 80% attenuation, demonstrating how the exponential formula helps in practical applications like determining required material thickness for shielding.