understand that annihilation occurs when a particle interacts with its antiparticle and that mass–energy and momentum are conserved in the process

Production and Use of X‑rays – A‑Level Physics 9702

Learning Objective

Explain how X‑rays are produced by electron bombardment of a metal target, calculate the minimum wavelength of the emitted radiation, describe how X‑rays are attenuated in matter, and outline the main medical and industrial applications (including PET, which uses positron‑electron annihilation). Demonstrate that annihilation occurs when a particle meets its antiparticle and that both mass–energy and momentum are conserved.

1. The X‑ray tube

• Cathode – heated filament that emits electrons by thermionic emission.
• Anode (target) – a high‑Z metal (usually tungsten) where the electrons are abruptly decelerated.
• Vacuum envelope – prevents electrons from being scattered by air molecules.
• High‑voltage supply – creates an accelerating potential \$V\$ (typically 20–150 kV) that gives each electron kinetic energy

  \[

  K_e = eV

  \]

  where \$e = 1.60\times10^{-19}\,\text{C}\$.

2. Production of X‑rays

2.1 Bremsstrahlung (braking radiation)

• When a high‑energy electron passes close to a nucleus it is deflected by the strong electric field. The change in velocity causes the electron to lose kinetic energy as photons.
• The emitted photon energies form a continuous spectrum ranging from \$0\$ up to a maximum equal to the electron’s kinetic energy.

Maximum photon energy and minimum wavelength

According to the syllabus the maximum photon energy is

\[

E_{\max}=eV

\]

and the corresponding minimum wavelength is obtained from \$E=hc/\lambda\$:

\[

\lambda_{\min}= \frac{hc}{eV}

\]

where \$h=6.626\times10^{-34}\,\text{J·s}\$ and \$c=3.00\times10^{8}\,\text{m s}^{-1}\$.

Worked example

For a 30 kV tube:

\[

\lambda_{\min}= \frac{6.626\times10^{-34}\times3.00\times10^{8}}

{1.60\times10^{-19}\times3.0\times10^{4}}

\approx 4.1\times10^{-11}\,\text{m}=0.041\;\text{nm}

\]

2.2 Characteristic X‑ray emission

• If an incident electron ejects an inner‑shell electron, an outer‑shell electron drops down to fill the vacancy.
• The photon energy equals the difference between the binding energies of the two shells involved; this produces sharp spectral lines.

3. Attenuation of X‑rays in matter

• The intensity \$I\$ of a narrow‑beam of X‑rays passing through a thickness \$x\$ of material follows the exponential law

  \[

  I = I_{0}\,e^{-\mu x}

  \]

  where \$\mu\$ is the linear attenuation coefficient (units m⁻¹ or mm⁻¹).

• Often the mass attenuation coefficient** \$\displaystyle\frac{\mu}{\rho}\$ (units m² kg⁻¹ or cm² g⁻¹) is tabulated. The relationship is

  \[

  \mu = \left(\frac{\mu}{\rho}\right)\rho .

  \]

• The half‑value layer (HVL) is the thickness that reduces the intensity to \$I_{0}/2\$:

  \[

  \text{HVL} = \frac{\ln 2}{\mu}\; .

  \]

Example

For 100 keV X‑rays in aluminium, \$\mu = 0.20\;\text{mm}^{-1}\$.

\[

\text{HVL}= \frac{0.693}{0.20}=3.5\;\text{mm}

\]

4. Imaging with X‑rays

• Radiography – a single projection; contrast depends on differences in \$\mu\$ of the tissues and on exposure settings (tube voltage, current, time).
• Computed Tomography (CT) – many narrow‑beam projections are taken around the patient; computer reconstruction yields cross‑sectional images.
• Key image‑quality factors: spatial resolution, contrast resolution, noise, and artefacts.

5. Positron–electron annihilation (second X‑ray source)

• A positron (\$e^{+}\$) is the antiparticle of the electron (\$e^{-}\$). When they meet they annihilate, converting their rest‑mass energy into photons.
• Conservation of energy and momentum requires at least two photons; a single photon cannot satisfy momentum conservation in the centre‑of‑mass frame.

Centre‑of‑mass frame (both particles at rest)

\[

E{\gamma 1}=E{\gamma 2}=m_{e}c^{2}=511\;\text{keV},\qquad

\mathbf{p}{\gamma 1}=-\mathbf{p}{\gamma 2}.

\]

General case – moving positron, stationary electron

\[

2m{e}c^{2}+K{e^{+}} = \sum{i=1}^{2}E{\gamma i},

\qquad

\mathbf{p}{e^{+}} = \sum{i=1}^{2}\mathbf{p}_{\gamma i}.

\]

One photon may have \$E>511\,\$keV while the other has \$E<511\,\$keV, but the total energy always equals the sum of the rest‑mass energies plus the kinetic energy of the positron. Momentum vectors of the two photons are opposite in the centre‑of‑mass frame and are not collinear when the positron has kinetic energy.

6. Applications of X‑rays

• Medical imaging – radiography, CT, fluoroscopy.
• Positron Emission Tomography (PET) – detects the two 511 keV annihilation photons to map metabolic activity.
• Crystallography – determination of crystal structures via diffraction.
• Industrial inspection – weld testing, non‑destructive testing, security scanning.
• X‑ray fluorescence (XRF) – elemental analysis based on characteristic X‑ray emission.

7. Summary

• X‑rays are produced when high‑energy electrons are abruptly decelerated in a high‑Z target (bremsstrahlung) or when they cause inner‑shell transitions (characteristic lines). The maximum photon energy is \$eV\$, giving a minimum wavelength \$\lambda_{\min}=hc/eV\$.
• As X‑rays pass through matter their intensity follows \$I=I_{0}e^{-\mu x}\$; the linear attenuation coefficient \$\mu\$ (m⁻¹) is related to the mass attenuation coefficient by \$\mu=(\mu/\rho)\rho\$. The half‑value layer \$\text{HVL}=\ln2/\mu\$ quantifies shielding effectiveness.
• Imaging techniques exploit differences in attenuation; CT reconstructs cross‑sections from many projections.
• When a particle meets its antiparticle (e.g. \$e^{+}+e^{-}\$) annihilation occurs. Conservation of mass–energy and momentum forces the conversion of the combined rest‑mass (plus any kinetic energy) into at least two photons, each of 511 keV in the centre‑of‑mass frame.
• The 511 keV annihilation photons are the basis of PET scanning, providing a powerful tool for functional medical imaging.
• These principles underpin a wide range of scientific, medical and industrial applications.