Absorption of X-rays (A-level only)

Welcome to the physics of medical imaging! When you look at an X-ray scan, you are looking at shadows. To understand how these shadows are formed, we need to look at how X-rays are absorbed as they travel through the body.

In this topic, you will learn:

What "exponential attenuation" means.
How to calculate the reduction in X-ray intensity using the linear and mass attenuation coefficients.
How to calculate the half-value thickness of a material.
Why different tissues (like bone and muscle) absorb X-rays differently, allowing us to see inside the body.

Exponential Attenuation

When a beam of X-rays passes through matter, some X-ray photons pass straight through, but many are absorbed or scattered by the atoms in the material. This reduction in the energy of the beam is called attenuation.

Definition

Intensity

Intensity ( $I$ ) is the power per unit area of the X-ray beam, measured in watts per square metre ( $m−2\text{W m}^{-2}$ ).

X-ray attenuation is a random process, much like radioactive decay. In any given tiny slice of material, a constant fraction of the remaining X-ray photons will be absorbed. Because the rate of absorption is proportional to the current intensity, the intensity of the X-ray beam decays exponentially as it travels deeper into the material.

X-ray attenuation diagram and exponential decay graph

The Attenuation Equation

The exponential decay of X-ray intensity is described by the following formula:

I_0 e^{-\mu x}

Where:

$I$ is the transmitted intensity of the beam.
$I_0$ is the initial intensity of the beam before it enters the material.
$μ\mu$ is the linear attenuation coefficient of the material.
$x$ is the thickness of the material the beam has passed through.

Definition

Linear Attenuation Coefficient ( $μ$ )

The linear attenuation coefficient represents the fraction of X-rays absorbed per unit thickness of a specific material. Its SI unit is $m−1\text{m}^{-1}$ (though you will often see it in $mm−1\text{mm}^{-1}$ or $cm−1\text{cm}^{-1}$ ).

Common Mistake

Mismatched Units

The product $μx\mu x$ in the exponent must be dimensionless (it cannot have any units). If the thickness $x$ is given in centimetres ( $cm\text{cm}$ ), your linear attenuation coefficient $μ\mu$ must be in $cm−1\text{cm}^{-1}$ . Always check they match before pressing buttons on your calculator!

Example

Calculating transmitted intensity

An X-ray beam with an initial intensity of $\text{ W m}^{-2}$ passes through $\text{ mm}$ of soft tissue. The linear attenuation coefficient of the tissue is $\text{ mm}^{-1}$ . Calculate the intensity of the transmitted beam.

Identify the known values: $m−2I_0 = 450 \text{ W m}^{-2}$ $mm−1\mu = 0.030 \text{ mm}^{-1}$ $\text{ mm}$
Check the units of $μ\mu$ and $x$ match. Both are in millimetres, so we can use them directly.
Substitute the values into the attenuation equation:

I_0 e^{-\mu x}

\times e^{-(0.030 \times 15)}

Calculate the exponent first: $\times 15) = -0.45$
Find the final intensity:

\times e^{-0.45} \approx 287 \text{ W m}^{-2}

Half-Value Thickness

Because X-rays decay exponentially, there is no single thickness that stops all the X-rays. Instead, we often measure how much material is needed to halve the intensity.

Definition

Half-Value Thickness

The half-value thickness ( $x_{1/2}$ ) is the thickness of a given material required to reduce the intensity of an X-ray beam to exactly half of its initial value.

This is mathematically identical to the concept of half-life in radioactivity. If you add one half-value thickness, the intensity halves. If you add a second, it halves again (down to a quarter), and so on.

You can calculate the half-value thickness using the linear attenuation coefficient. At the half-value thickness, $I = 0.5 I_0$ . Plugging this into our attenuation equation:

I_0 = I_0 e^{-\mu x_{1/2}}

e^{-\mu x_{1/2}}

\ln(0.5) = -\mu x_{1/2}

-\ln(2) = -\mu x_{1/2}

Which gives us the useful relationship:

x_{1/2} = \frac{\ln(2)}{\mu}

Tip

Quick Powers of Two

If a question asks for the transmitted intensity after $3$ half-value thicknesses, you don't always need the full exponential equation. Just halve the intensity $3$ times: $I_0 \times \left(\frac{1}{2}\right)^3 = \frac{1}{8} I_0$ .

Mass Attenuation Coefficient

The linear attenuation coefficient ( $μ\mu$ ) is useful, but it has a flaw: it depends heavily on the density of the material. For example, liquid water and water vapour are made of the exact same molecules, but liquid water will absorb far more X-rays per centimetre simply because the molecules are packed tighter together.

To get a measurement that describes how well the substance itself absorbs X-rays, regardless of its physical state or density, we use the mass attenuation coefficient ( $μm\mu_m$ ).

\mu_m = \frac{\mu}{\rho}

Where:

$μm\mu_m$ is the mass attenuation coefficient in $kg−1\text{m}^2\text{ kg}^{-1}$ .
$μ\mu$ is the linear attenuation coefficient in $m−1\text{m}^{-1}$ .
$ρ\rho$ is the density of the material in $m−3\text{kg m}^{-3}$ .

Key Idea

Why use the mass attenuation coefficient?

The mass attenuation coefficient depends only on the atomic number ( $Z$ ) of the atoms in the material and the energy of the X-ray photons. It removes density from the equation, giving a true measure of the material's atomic stopping power.

Example

Using the mass attenuation coefficient

The mass attenuation coefficient of a specific grade of bone for a $\text{ keV}$ X-ray beam is $\text{ m}^2\text{ kg}^{-1}$ . The density of this bone is $\text{ kg m}^{-3}$ . Calculate the percentage of X-rays transmitted through a $\text{ cm}$ thickness of this bone.

Identify the known values: $kg−1\mu_m = 0.024 \text{ m}^2\text{ kg}^{-1}$ $m−3\rho = 1850 \text{ kg m}^{-3}$ $\text{ cm} = 0.040 \text{ m}$
Calculate the linear attenuation coefficient ( $μ\mu$ ) by rearranging $μm=μρ\mu_m = \frac{\mu}{\rho}$ :

\mu = \mu_m \times \rho

\mu = 0.024 \times 1850 = 44.4 \text{ m}^{-1}

Set up the ratio for the percentage transmitted using the exponential equation:

\frac{I}{I_0} = e^{-\mu x}

Substitute the values of $μ\mu$ and $x$ (ensure both are in metres):

\frac{I}{I_0} = e^{-(44.4 \times 0.040)}

\frac{I}{I_0} = e^{-1.776} \approx 0.169

Convert the decimal to a percentage: $0.169 \times 100 = 16.9\%$ transmission.

Differential Tissue Absorption

X-ray images are essentially shadowgraphs. The detector (like a digital sensor or photographic film) behind the patient turns black where X-rays hit it, and remains white where X-rays are blocked.

For an image to be useful, there must be contrast — a visible difference between different types of tissue. This relies on differential tissue absorption.

How much a tissue absorbs X-rays depends heavily on two things:

Density ( $ρ\rho$ ): Denser tissues pack more atoms into the same space.
Atomic Number ( $Z$ ): The linear attenuation coefficient is roughly proportional to the cube of the atomic number ( $Z^3$ ) of the elements making up the tissue.

Bone vs Soft Tissue: Soft tissue (like muscle, fat, and organs) is mostly composed of elements with low atomic numbers, such as hydrogen ( $Z = 1$ ), carbon ( $Z = 6$ ), and oxygen ( $Z = 8$ ).

Bone, on the other hand, contains a large amount of calcium ( $Z = 20$ ) and phosphorus ( $Z = 15$ ). Because attenuation scales dramatically with atomic number, bone has a vastly higher attenuation coefficient than soft tissue. Bone is also significantly denser.

As a result, bone absorbs almost all the X-rays that hit it, casting a sharp "shadow" that leaves the X-ray film bright white. Soft tissue absorbs very few X-rays, letting them pass through to turn the film dark grey or black. This differential absorption is what gives X-ray images their diagnostic power!

Exam technique

In the exam

Watch your units: Examiners love to give $μm\mu_m$ in $g−1\text{cm}^2\text{ g}^{-1}$ but density in $m−3\text{kg m}^{-3}$ , or thickness $x$ in $mm\text{mm}$ while $μ\mu$ is in $cm−1\text{cm}^{-1}$ . Always convert everything to standard SI units ( $m\text{m}$ , $kg\text{kg}$ ) or ensure they match perfectly before multiplying.
Rearranging with natural logs: If you are asked to find the thickness $x$ , you will need to isolate the exponent. Divide by $I_0$ first, then take the natural logarithm ( $ln⁡\ln$ ) of both sides. $ln⁡(II0)=−μx\ln\left(\frac{I}{I_0}\right) = -\mu x$
Intensity vs Number of Photons: The equation $I_0 e^{-\mu x}$ works for intensity, but you can also use it for the number of X-ray photons: $N_0 e^{-\mu x}$ .
Differential absorption explanations: If asked why bone is clearly visible against muscle, mention both the higher density and the higher atomic number (calcium) of bone, and state that this leads to a higher attenuation coefficient.

Self review

Check yourself

Can you write down the equation for exponential attenuation and define every term?
How is the mass attenuation coefficient calculated from the linear attenuation coefficient?
If a material has a half-value thickness of $\text{ cm}$ , what fraction of the original X-ray intensity remains after $\text{ cm}$ ?
Why do bones appear white on an X-ray image while lungs appear black?

In this topic, you will learn:

What "exponential attenuation" means.
How to calculate the reduction in X-ray intensity using the linear and mass attenuation coefficients.
How to calculate the half-value thickness of a material.
Why different tissues (like bone and muscle) absorb X-rays differently, allowing us to see inside the body.

Exponential Attenuation

Definition

Intensity

Intensity ( $I$ ) is the power per unit area of the X-ray beam, measured in watts per square metre ( $m−2\text{W m}^{-2}$ ).

X-ray attenuation diagram and exponential decay graph

The Attenuation Equation

The exponential decay of X-ray intensity is described by the following formula:

I_0 e^{-\mu x}

Where:

$I$ is the transmitted intensity of the beam.
$I_0$ is the initial intensity of the beam before it enters the material.
$μ\mu$ is the linear attenuation coefficient of the material.
$x$ is the thickness of the material the beam has passed through.

Definition

Linear Attenuation Coefficient ( $μ$ )

Common Mistake

Mismatched Units

Example

Calculating transmitted intensity

Identify the known values: $m−2I_0 = 450 \text{ W m}^{-2}$ $mm−1\mu = 0.030 \text{ mm}^{-1}$ $\text{ mm}$
Check the units of $μ\mu$ and $x$ match. Both are in millimetres, so we can use them directly.
Substitute the values into the attenuation equation:

I_0 e^{-\mu x}

\times e^{-(0.030 \times 15)}

Calculate the exponent first: $\times 15) = -0.45$
Find the final intensity:

\times e^{-0.45} \approx 287 \text{ W m}^{-2}

Half-Value Thickness

Because X-rays decay exponentially, there is no single thickness that stops all the X-rays. Instead, we often measure how much material is needed to halve the intensity.

Definition

Half-Value Thickness

The half-value thickness ( $x_{1/2}$ ) is the thickness of a given material required to reduce the intensity of an X-ray beam to exactly half of its initial value.

You can calculate the half-value thickness using the linear attenuation coefficient. At the half-value thickness, $I = 0.5 I_0$ . Plugging this into our attenuation equation:

I_0 = I_0 e^{-\mu x_{1/2}}

e^{-\mu x_{1/2}}

\ln(0.5) = -\mu x_{1/2}

-\ln(2) = -\mu x_{1/2}

Which gives us the useful relationship:

x_{1/2} = \frac{\ln(2)}{\mu}

Tip

Quick Powers of Two

Mass Attenuation Coefficient

To get a measurement that describes how well the substance itself absorbs X-rays, regardless of its physical state or density, we use the mass attenuation coefficient ( $μm\mu_m$ ).

\mu_m = \frac{\mu}{\rho}

Where:

$μm\mu_m$ is the mass attenuation coefficient in $kg−1\text{m}^2\text{ kg}^{-1}$ .
$μ\mu$ is the linear attenuation coefficient in $m−1\text{m}^{-1}$ .
$ρ\rho$ is the density of the material in $m−3\text{kg m}^{-3}$ .

Key Idea

Why use the mass attenuation coefficient?

Example

Using the mass attenuation coefficient

Identify the known values: $kg−1\mu_m = 0.024 \text{ m}^2\text{ kg}^{-1}$ $m−3\rho = 1850 \text{ kg m}^{-3}$ $\text{ cm} = 0.040 \text{ m}$
Calculate the linear attenuation coefficient ( $μ\mu$ ) by rearranging $μm=μρ\mu_m = \frac{\mu}{\rho}$ :

\mu = \mu_m \times \rho

\mu = 0.024 \times 1850 = 44.4 \text{ m}^{-1}

Set up the ratio for the percentage transmitted using the exponential equation:

\frac{I}{I_0} = e^{-\mu x}

Substitute the values of $μ\mu$ and $x$ (ensure both are in metres):

\frac{I}{I_0} = e^{-(44.4 \times 0.040)}

\frac{I}{I_0} = e^{-1.776} \approx 0.169

Convert the decimal to a percentage: $0.169 \times 100 = 16.9\%$ transmission.

Differential Tissue Absorption

X-ray images are essentially shadowgraphs. The detector (like a digital sensor or photographic film) behind the patient turns black where X-rays hit it, and remains white where X-rays are blocked.

For an image to be useful, there must be contrast — a visible difference between different types of tissue. This relies on differential tissue absorption.

How much a tissue absorbs X-rays depends heavily on two things:

Density ( $ρ\rho$ ): Denser tissues pack more atoms into the same space.
Atomic Number ( $Z$ ): The linear attenuation coefficient is roughly proportional to the cube of the atomic number ( $Z^3$ ) of the elements making up the tissue.

Bone vs Soft Tissue: Soft tissue (like muscle, fat, and organs) is mostly composed of elements with low atomic numbers, such as hydrogen ( $Z = 1$ ), carbon ( $Z = 6$ ), and oxygen ( $Z = 8$ ).

Exam technique

In the exam

Watch your units: Examiners love to give $μm\mu_m$ in $g−1\text{cm}^2\text{ g}^{-1}$ but density in $m−3\text{kg m}^{-3}$ , or thickness $x$ in $mm\text{mm}$ while $μ\mu$ is in $cm−1\text{cm}^{-1}$ . Always convert everything to standard SI units ( $m\text{m}$ , $kg\text{kg}$ ) or ensure they match perfectly before multiplying.
Rearranging with natural logs: If you are asked to find the thickness $x$ , you will need to isolate the exponent. Divide by $I_0$ first, then take the natural logarithm ( $ln⁡\ln$ ) of both sides. $ln⁡(II0)=−μx\ln\left(\frac{I}{I_0}\right) = -\mu x$
Intensity vs Number of Photons: The equation $I_0 e^{-\mu x}$ works for intensity, but you can also use it for the number of X-ray photons: $N_0 e^{-\mu x}$ .
Differential absorption explanations: If asked why bone is clearly visible against muscle, mention both the higher density and the higher atomic number (calcium) of bone, and state that this leads to a higher attenuation coefficient.

Self review

Check yourself

Can you write down the equation for exponential attenuation and define every term?
How is the mass attenuation coefficient calculated from the linear attenuation coefficient?
If a material has a half-value thickness of $\text{ cm}$ , what fraction of the original X-ray intensity remains after $\text{ cm}$ ?
Why do bones appear white on an X-ray image while lungs appear black?

Exponential Attenuation

The Attenuation Equation

Half-Value Thickness

Mass Attenuation Coefficient

Differential Tissue Absorption

Review Absorption of X-rays (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Absorption of X-rays (A-level only)

Exponential Attenuation

The Attenuation Equation

Half-Value Thickness

Mass Attenuation Coefficient

Differential Tissue Absorption

Review Absorption of X-rays (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Absorption of X-rays (A-level only)

Exponential Attenuation

The Attenuation Equation

Half-Value Thickness

Mass Attenuation Coefficient

Differential Tissue Absorption

Review Absorption of X-rays (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Absorption of X-rays (A-level only) Revision Guide

Absorption of X-rays (A-level only)

Exponential Attenuation

The Attenuation Equation

Half-Value Thickness

Mass Attenuation Coefficient

Differential Tissue Absorption

Review Absorption of X-rays (A-level only) by teaching Genie

Recap your knowledge with an interactive lesson

Remember key concepts with flashcards

Absorption of X-rays (A-level only) Revision Guide