24.2 Production and use of X-rays

Syllabus
9702–2028–2029
Topic
24.2
Level
A2

Learning objectives

Electron bombardment of a metal target produces X-rays with a minimum wavelength

Electrons emitted by a heated cathode accelerate through a large p.d. V toward a metal target. Rapid deceleration and atomic interactions at the target produce X-ray photons.

maximumelectronkineticenergy=eVmaximumphotonenergy=hfmax=hc/λmin=eVmaximum electron kinetic energy=eV maximum photon energy=hf_max=hc/λ_min=eV

For V=58 kV, λmin=hc/eV=(6.63×10⁻³⁴)(3.00×10⁸)/[(1.60×10⁻¹⁹)(58×10³)]=2.14×10⁻¹¹ m=21.4 pm.

Most electron energy becomes thermal energy, so the target needs a high melting point and heat removal; tungsten is commonly suitable.

The minimum wavelength is the rare limit where one photon receives all eV. Increasing V decreases λmin; tube current mainly changes the number/intensity of X-rays, not this limit.

X-ray contrast comes from differences in transmitted intensity

Send X-rays through the body and detect the transmitted beam. Different structures absorb/attenuate different fractions, so the detector records different transmitted intensities and forms an internal image.

Contrast is the difference in detector response or degree of blackening between image regions. Greater difference in transmitted intensity gives greater contrast.

Bone has a much larger attenuation coefficient than many soft tissues, so fewer X-rays reach the detector behind bone. Bone and soft tissue therefore give good contrast; tissues with similar μ give poor contrast.

On traditional film, more transmitted X-rays produce greater blackening; digital displays may map detector signal to brightness differently, but contrast still comes from signal differences.

Do not define contrast as brightness alone. It is a difference between regions produced by different attenuation/transmission through their material and thickness.

Calculate X-ray attenuation through one or several material layers

I=I0e(μx)μ=ln(I/I0)/x;x=ln(I/I0)/μI=I₀e^(−μx) μ=−ln(I/I₀)/x; x=−ln(I/I₀)/μ

I/I₀ is the transmitted fraction. If 80% is absorbed, 20%=0.20 remains and belongs on the left of the exponential equation.

For muscle μ=0.22 cm⁻¹ and 80% absorption: 0.20=e^(−0.22x), so x=−ln(0.20)/0.22=7.3 cm.

Forsuccessivelayers:I/I0=e(μ1x1)e(μ2x2)=e[Σ(μixi)]For successive layers: I/I₀=e^(−μ₁x₁)e^(−μ₂x₂)…=e^[−Σ(μᵢxᵢ)]

Equal thickness x through materials with μ=3.0 and 0.22 cm⁻¹ gives I/I₀=e^(−3.22x). If I/I₀=0.13, x=0.63 cm.

Add μx exponents, not transmitted fractions. Keep μ and x in reciprocal units and distinguish fraction transmitted from fraction absorbed=1−I/I₀.

CT builds a 3D volume from many-angle X-ray projections of many sections

Choose one thin body section. An X-ray source and detectors acquire many transmission projections through that same section from different angles.

A computer combines the angular attenuation data to reconstruct a two-dimensional map/image of that section.

Repeat the angular scan for successive sections along an axis, then combine/stack the reconstructed 2D section images to produce a three-dimensional representation.

manyanglesofonesectiononereconstructed2Dslicemanyadjacent2Dslicesone3Dvolumemany angles of one section → one reconstructed 2D slice many adjacent 2D slices → one 3D volume

Compared with one projection radiograph, CT separates overlapping structures and gives 3D localization, but multiple X-ray exposures generally increase ionising-radiation dose.

Different angles create one slice; different positions create different slices. A single rotating projection is not itself the final 3D image.