Production and use of X-rays
| English | Chinese | Pinyin |
|---|---|---|
| tungsten | 钨 | wū |
| target | 靶 | bǎ |
| anode | 阳极 | yáng jí |
| cathode | 阴极 | yīn jí |
| thermionic emission | 热电子发射 | rè diàn zi fā shè |
| kinetic energy | 动能 | dòng néng |
| photons | 光子 | guāng zi |
| Bremsstrahlung | 轫致辐射 | rèn zhì fú shè |
| hardness | 硬度 | yìng dù |
| characteristic | 特征 | tè zhēng |
| sharpness | 清晰度 | qīng xī dù |
| contrast | 对比度 | duì bǐ dù |
| attenuation | 衰减 | shuāi jiǎn |
| contrast medium | 造影剂 | zào yǐng jì |
| half-value thickness | 半值厚度 | bàn zhí hòu dù |
| computed tomography | 计算机断层扫描 | jì suàn jī duàn céng sǎo miáo |
A machine that wastes ninety-nine per cent of its power
- An X-ray tube run at $75\ \text{kV}$ and $30\ \text{mA}$ takes in $2250\ \text{W}$. About $99\%$ of that becomes heat in a lump of tungsten the size of a coin.
- Left alone, the target's temperature would climb at over $1000\ \text{K}$ per second and melt in a few seconds.
- That is why exposures are brief and the anode spins. The remaining one per cent is the picture.
- This lesson is how X-rays are produced, what sets their penetrating power, and how they make an image.
Production
- The four-mark answer runs in four moves, and each is a mark.
- One. Electrons are emitted from a heated filament (the cathode 阴极) by thermionic emission 热电子发射.
- Two. They are accelerated through a high p.d., tens of kV, across an evacuated tube towards a metal target 靶, the anode 阳极, usually tungsten 钨.
- Three. They decelerate rapidly when they strike the target.
- Four. The kinetic energy 动能 lost is emitted as X-ray photons 光子, this is Bremsstrahlung 轫致辐射 or braking radiation, with most of it becoming heat.
Heat one end, brake at the other
X-rays are produced when fast electrons:
The sudden deceleration on the target gives out X-ray photons (Bremsstrahlung).
Most of the electrons' kinetic energy in an X-ray tube becomes heat, not X-rays.
Only a small fraction becomes X-rays — which is why the target must be cooled.
Two controls, two different jobs
- Intensity, the energy per unit area per second, is set by the number of electrons hitting the target per second, so it is controlled by the filament current. A hotter filament emits more electrons.
- Hardness 硬度, the penetrating power, is set by the photon energies, so it is controlled by the accelerating p.d. A larger p.d. gives higher-energy, shorter-wavelength, more penetrating X-rays.
- Keep them apart. A question asking how to make the beam more penetrating is not answered by turning up the current.
- A metal filter removes the softest X-rays, which would be absorbed in the patient's skin without ever reaching the detector.
Match each control to what it changes.
A question asking how to make the beam more penetrating is not answered by turning up the current.
The minimum wavelength
- The most energy one photon can carry is the whole kinetic energy of one electron, lost in a single event:
- This is the short-wavelength cut-off, and it depends only on the accelerating p.d., never on the target metal or the current.
- Doubling the p.d. halves the minimum wavelength.
The minimum X-ray wavelength from a tube at accelerating voltage $V$ is:
From $hf_{\max} = eV$ and $c = f\lambda$: $\lambda_{\min} = \dfrac{hc}{eV}$.
A higher accelerating voltage gives a ____ minimum wavelength.
$\lambda_{\min} = \dfrac{hc}{eV}$, so a larger $V$ gives a smaller $\lambda_{\min}$ (more energetic X-rays).
Worked example: the cut-off at 75 kV
- Electrons are accelerated through $75\ \text{kV}$. Find the maximum photon energy and the minimum wavelength.
- Maximum energy $= eV = 75\ \text{keV} = 75000 \times 1.60\times10^{-19} = 1.2\times10^{-14}\ \text{J}$.
- $\lambda_{\text{min}} = \dfrac{hc}{eV} = \dfrac{(6.63\times10^{-34})(3.00\times10^{8})}{1.2\times10^{-14}} = 1.7\times10^{-11}\ \text{m}$.
- The maximum photon momentum is $p = E/c = 4.0\times10^{-23}\ \text{N s}$.
- Note that $eV$ in joules needs the p.d. in volts, not kilovolts. That factor of a thousand is the usual casualty.
Electrons are accelerated through 75 kV. What is the minimum X-ray wavelength, in units of 1e-11 m?
lambda_min = hc/(eV) = 1.7e-11 m. Use volts, not kilovolts. Doubling the p.d. halves the minimum wavelength, and the target metal makes no difference to it.
Reading the spectrum
- The spectrum is continuous because each electron may lose any fraction of its energy, in one deceleration or several, so photons of every energy up to the maximum appear.
- It has a sharp cut-off at $\lambda_{\text{min}}$ because no photon can carry more than one electron's energy, $eV$.
- The sharp peaks are characteristic 特征 of the target metal: an incoming electron knocks out an inner electron, an outer one falls into the vacancy, and the photon carries the difference between those two levels.
- So the curve's shape has three separate causes, and a question about it wants all three named.
A cliff, a hump, and two spikes
X-ray production route
Follow electrons from cathode to X-ray photons.
Why does an X-ray spectrum have the shape it does? Select all that apply.
The peaks belong to the target's energy levels, so only changing the target moves them. The p.d. moves the cut-off.
Sharpness and contrast are different things
- Sharpness 清晰度 is how well defined the edges are. It improves with a small source spot, a still patient, and the detector close to the patient.
- Contrast 对比度 is the difference in degree of blackening between neighbouring regions, produced by a difference in attenuation 衰减.
- The two are independent: an image can be sharp with poor contrast, or blurred with strong contrast.
- Where two soft tissues attenuate similarly, a contrast medium 造影剂 such as barium is swallowed or injected to outline one of them.
Same beam, very different shadows
An X-ray image with excellent sharpness must also have good contrast.
They are independent. Sharpness is edge definition, set by the source size and geometry; contrast is a difference in attenuation between tissues.
Worked example: why bone shows white
- X-rays cross $3.0\ \text{cm}$ of soft tissue ($\mu = 0.20\ \text{cm}^{-1}$) in one region and $3.0\ \text{cm}$ of bone ($\mu = 0.60\ \text{cm}^{-1}$) in another. Compare the transmitted intensities.
- Soft tissue: $I/I_0 = e^{-0.20\times3.0} = e^{-0.60} = 0.55$.
- Bone: $I/I_0 = e^{-0.60\times3.0} = e^{-1.80} = 0.17$.
- The tissue region receives more than three times the exposure, so bone appears white and tissue dark.
- The exposure ratio is $e^{(\mu_{\text{bone}} - \mu_{\text{tissue}})x}$, which grows with thickness: contrast improves for thicker structures, though the total attenuation grows too.
The attenuation law
- Higher-energy X-rays penetrate further, meaning a smaller $\mu$; bone has a much larger $\mu$ than soft tissue.
- For a thickness, take logs: $x = \dfrac{1}{\mu}\ln\dfrac{I_0}{I}$. The half-value thickness 半值厚度 is $x_{1/2} = \ln 2/\mu$, exactly like a half-life.
- Through two layers the exponentials multiply, so add the exponents: $I = I_0 e^{-(\mu_1 x_1 + \mu_2 x_2)}$. Never add the thicknesses or the coefficients, and note the order of the layers makes no difference.
- Read the question: "what fraction is absorbed" is $1 - e^{-\mu x}$, not $e^{-\mu x}$.
Bone looks white on an X-ray image because it:
Dense, high-$Z$ bone absorbs more ($I = I_0 e^{-\mu x}$ with large $\mu$), casting a stronger shadow.
A beam crosses 2.0 cm of P (mu = 0.35 /cm) then 1.5 cm of Q (mu = 0.90 /cm). What fraction of the intensity is transmitted?
The exponentials multiply, so add the exponents: e^-(0.70 + 1.35) = e^-2.05 = 0.13. Never add the thicknesses or the coefficients, and the order of the layers makes no difference.
Computed tomography
- A computed tomography 计算机断层扫描 (CT) scan builds a three-dimensional image, and the syllabus wants the construction described in order.
- One. Many X-ray images of the same section are taken from different angles around the patient.
- Two. A computer combines them into one two-dimensional image of that section.
- Three. The process is repeated for many sections along an axis.
- Four. The two-dimensional sections are combined into a three-dimensional image, which can be rotated or sliced on screen. The dose is far higher than a single radiograph, which is the trade-off.
A CT scan takes X-ray images from many angles to build a 3-D image.
Rotating the tube and detectors and reconstructing the slices separates overlapping tissues into a 3-D picture.
Put the construction of a CT image in order.
Angles first, then sections. The dose is far higher than a single radiograph, which is the trade-off for the 3-D view.
Marks that slip away
- Filament current sets intensity, accelerating p.d. sets penetrating power. Do not swap them.
- $\lambda_{\text{min}}$ depends only on the p.d., not on the target or the current.
- Sharpness and contrast are independent. A question about blurred edges is not answered by talking about attenuation.
- Two layers: add the exponents $\mu_1 x_1 + \mu_2 x_2$.
- Match the units of $\mu$ and $x$, and check whether the question wants what gets through or what is absorbed.
You've got it
- X-rays are produced when electrons from a heated cathode, accelerated through a high p.d., decelerate rapidly in a metal target, mostly as heat
- $\lambda_{\text{min}} = hc/(eV)$ is the cut-off set by the accelerating p.d. alone, with characteristic peaks from the target metal on a continuous curve
- sharpness is edge definition, contrast is the difference in attenuation, and they are independent
- $I = I_0 e^{-\mu x}$, with half-value thickness $\ln 2/\mu$ and exponents that add through successive layers; CT combines many angles into a section and many sections into a 3-D image