Medical Physics · 29.1
X-Ray Imaging
Slam a beam of fast electrons into a tungsten block and about one percent of their energy re-emerges as X-ray photons — the other ninety-nine percent is heat. Aim those photons through a patient at a detector, and the shadow the skeleton casts becomes the oldest medical image of all.
Build the model
Connect the measurement to the mechanism.
An X-ray tube boils electrons off a hot cathode by thermionic emission, accelerates them through a p.d. of tens of kilovolts in vacuum, and crashes them into a tungsten anode. The violent deceleration radiates a continuous braking spectrum whose sharp short-wavelength cutoff comes from an electron converting its entire kinetic energy into one photon: eV = hc/λmin. Electrons that knock out inner tungsten electrons add characteristic spike lines at fixed wavelengths on top of the continuum.
Only about 1% of the beam power becomes X-rays, so the anode spins to spread the heat. Inside the body the beam attenuates exponentially, I = I₀e^(−μx), and the attenuation coefficient μ rises steeply with atomic number and falls with photon energy: calcium-rich bone absorbs far more than soft tissue, so the transmitted beam prints a shadowgraph. Soft structures that barely differ in μ are made visible by filling them with a high-Z contrast medium such as a barium meal.
- Simple definition
- An X-ray tube converts the kinetic energy of electrons accelerated through a p.d. V into X-ray photons — maximum photon energy eV, so λmin = hc/eV. The body attenuates the beam as I = I₀e^(−μx), and differences in μ between tissues create the image contrast.
- Example
- At 100 kV no photon can carry more than 100 keV, so λmin = hc/eV ≈ 12.4 pm. Raising the voltage shifts the whole spectrum to shorter, more penetrating wavelengths; raising the current only makes more of the same photons.
The most extreme electron gives everything to a single photon. Nothing in the tube can beat that, so the spectrum stops dead at λmin — a cutoff you can move with the voltage knob.
λmin = hc/eV — set by the tube p.d. alone
Each layer of tissue removes the same fraction of whatever intensity reaches it — so intensity falls exponentially, never to zero, exactly like radioactive decay with depth playing the role of time.
μ is the linear attenuation coefficient
A 100 kV tube gives each electron 100 keV. Tungsten is chosen for its high Z (efficient X-ray production) and high melting point (it must survive the other 99%).
≈ 99% ends as heat in the tungsten target
Bone is rich in calcium (Z = 20) while soft tissue averages Z ≈ 7 — that Z³ gap is the whole radiograph. Barium (Z = 56) borrows the same trick to paint the gut visible.
roughly as Z³ at diagnostic energies
Inside the tube
A heated filament releases electrons (thermionic emission); a p.d. of 30–150 kV accelerates them across an evacuated gap; a tungsten anode stops them. The tube current sets how many electrons arrive per second and hence the beam intensity; the tube voltage sets each photon's energy budget. Because ~99% of the power becomes heat, the anode is a spinning disc, presenting fresh metal to the beam thousands of times a second.
Two spectra in one beam
The continuous spectrum comes from braking radiation: electrons decelerating in the target radiate every wavelength down to the λmin cutoff. The line spectrum comes from tungsten itself: a beam electron ejects an inner-shell electron, an outer electron falls into the hole, and a photon of fixed, element-specific energy leaves. The lines only appear once the tube voltage exceeds the inner-shell binding energy — about 70 kV for tungsten's K lines.
From shadow to image
Sharpness demands a small focal spot and a patient close to the detector, so each point casts a crisp shadow instead of a blurred penumbra. Contrast demands tissues of different μ in the path — given for bone, engineered with barium for the gut. An aluminium filter hardens the beam by absorbing the long-wavelength photons that would only dose the skin without ever reaching the detector, and digital flat-panel detectors have replaced film, trading grains of silver for pixels and slashing the dose needed.
Change one variable at a time
Make the relationship visible.
Raise the p.d. and the cutoff λmin = hc/eV slides left — plus, past ≈ 70 kV, tungsten’s characteristic spikes switch on. The filter eats the long-wavelength tail that would only dose the skin. Current scales the whole curve but never moves the cutoff.
Cutoff λmin = hc/eV15.5 pm · max photon 80 keV
Characteristic linesexcited — p.d. beats the K-shell binding
Beam power P = VI16 kW
X-rays ≈ 1% · heat ≈ 99%160 W out · 15.8 kW of heat
Catch the common trap
Explain before calculating.
Doubling an X-ray tube's accelerating p.d. does what to the minimum wavelength λmin?
Choose an answer to test the model.
Worked examples
State the rule, substitute, then check units.
EasyFind the minimum X-ray wavelength produced by a tube operating at 80 kV.
- The most energetic photon carries the whole electron energy: eV = hc/λmin.
- λmin = hc/eV = (6.63 × 10⁻³⁴ × 3.00 × 10⁸) ÷ (1.60 × 10⁻¹⁹ × 8.0 × 10⁴).
- λmin = 1.55 × 10⁻¹¹ m ≈ 16 pm — deep in the X-ray band.
Answerλmin ≈ 1.6 × 10⁻¹¹ m (16 pm)
MediumA beam passes through 5.0 cm of muscle with μ = 0.20 cm⁻¹. What fraction of the intensity emerges?
- I/I₀ = e^(−μx) = e^(−0.20 × 5.0) = e^(−1.0).
- e^(−1.0) = 0.37.
- About 37% of the intensity survives — attenuation removes a fixed fraction per centimetre, so the answer would be 0.37² ≈ 14% for 10 cm.
Answer≈ 37% transmitted
HardOne ray crosses 4.0 cm of soft tissue (μ = 0.22 cm⁻¹). A neighbouring ray crosses 3.0 cm of tissue plus 1.0 cm of bone (μ = 0.60 cm⁻¹). Find the intensity ratio between the two rays — the contrast in the image.
- Tissue-only ray: I₁/I₀ = e^(−0.22 × 4.0) = e^(−0.88) = 0.415.
- Tissue + bone ray: exponents add along the path: I₂/I₀ = e^(−0.22 × 3.0 − 0.60 × 1.0) = e^(−1.26) = 0.284.
- Ratio I₂/I₁ = 0.284 ÷ 0.415 = 0.68 — the bone path arrives about 32% dimmer, and that difference is the shadow the radiologist reads.
AnswerI₂/I₁ ≈ 0.68 — one centimetre of bone dims the ray by about a third
ChallengingA tube runs at 90 kV and 250 mA. Find the electrical power, the approximate X-ray output, the number of electrons striking the target per second — and explain the rotating anode.
- P = VI = 9.0 × 10⁴ × 0.250 = 22.5 kW of electron beam power.
- At ~1% efficiency, only ≈ 225 W leaves as X-rays; ≈ 22.3 kW becomes heat in a focal spot barely a millimetre across.
- Electrons per second = I/e = 0.250 ÷ 1.60 × 10⁻¹⁹ = 1.6 × 10¹⁸ s⁻¹.
- No stationary target survives 22 kW on a millimetre: the anode is a disc spinning at thousands of rpm, so each part of the rim is heated only briefly and radiates its heat between passes.
Answer22.5 kW in, ≈ 225 W of X-rays out, 1.6 × 10¹⁸ electrons per second — hence the spinning anode