An X-ray beam passes through tissue with μ = 0.10\ cm^-1. Find the half-value thickness. Take ln 2 = 0.693.

OCR OCR 2019-style practice: A parallel X-ray beam passes through 4.0\ cm of soft tissue with a linear attenuation coefficient of 0.20\ cm^-1. Calculate the fraction of the beam intensity transmitted, and the half-value thickness. Take ln 2 = 0.693.

Attenuation: I = I_0 e^-μ x, so the transmitted fraction is (I)/(I_0) = e^-μ x = e^-(0.20)(4.0) = e^-0.80. e^-0.80 = 0.45, so about 45 per cent is transmitted. Half-value thickness: x_1/2 = (ln 2)/(μ) = (0.693)/(0.20) = 3.5\ cm. Markers reward I = I_0 e^-μ x, the transmitted fraction about 0.45, and the half-value thickness about 3.5\ cm.

EnglandPhysicsSyllabus dot point

How do X-rays, gamma cameras, PET and ultrasound let us see inside the body?

Medical imaging: the production and attenuation of X-rays, the gamma camera and PET scanning, and ultrasound with acoustic impedance and the Doppler effect.

A focused answer to the OCR H556 medical imaging content, covering the production of X-rays and their exponential attenuation, the gamma camera and PET scanning with positron annihilation, and ultrasound imaging with acoustic impedance, the intensity reflection coefficient and the Doppler effect for blood flow.

Generated by Claude Opus 4.812 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

X-rays are produced when fast electrons hit a metal target, and their intensity is attenuated exponentially as they pass through tissue, $I = I_0 e^{-\mu x}$ , where $\mu$ is the linear attenuation coefficient (the half-value thickness is $x_{1/2} = \frac{\ln 2}{\mu}$ ); denser tissue such as bone attenuates more, giving contrast. A gamma camera detects gamma photons from an injected tracer to image function. PET uses a positron-emitting tracer: each positron annihilates with an electron to give two back-to-back $511\ \text{keV}$ photons detected in coincidence. Ultrasound is high-frequency sound; the fraction reflected at a boundary depends on the acoustic impedances $Z = \rho c$ through $\frac{I_r}{I_i} = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$ , and the Doppler shift of reflected ultrasound measures blood flow.

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What this dot point is asking

OCR wants you to describe the production and exponential attenuation of X-rays, explain the gamma camera and PET scanning, and describe ultrasound imaging using acoustic impedance, the intensity reflection coefficient and the Doppler effect.

The answer

X-ray production and attenuation

The gamma camera and PET

Ultrasound and the Doppler effect

Formula

Ultrasound is sound at frequencies above human hearing (above $20\ \text{kHz}$ , typically megahertz for imaging). At a boundary between two media, part of the beam is reflected; the fraction depends on the acoustic impedances $Z = \rho c$ (density times speed of sound) through the intensity reflection coefficient $\frac{I_r}{I_i} = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2$ . A large mismatch (as between air and tissue) reflects nearly all the beam, which is why a coupling gel is used to match impedances. The Doppler effect shifts the frequency of ultrasound reflected from moving blood, $\Delta f = \frac{2vf_0\cos\theta}{c}$ , allowing blood-flow speed to be measured.

Examples in context

X-rays and CT scans image bones, lungs and dense structures and guide surgery, while PET and gamma-camera scans reveal function, such as blood flow in the heart or the metabolic activity of a tumour. Ultrasound is the safe, non-ionising choice for imaging soft tissue and monitoring pregnancy, and Doppler ultrasound measures blood flow to detect blocked arteries. Each method trades resolution, penetration and safety, which is why doctors choose between them according to the tissue and the clinical question.

Try this

Q1. State the exponential attenuation law for X-rays and define each term. [2 marks]

Cue. $I = I_0 e^{-\mu x}$ : transmitted intensity, initial intensity, attenuation coefficient and thickness.

Q2. An X-ray beam passes through tissue with $\mu = 0.10\ \text{cm}^{-1}$ . Find the half-value thickness. Take $\ln 2 = 0.693$ . [2 marks]

Cue. $x_{1/2} = \frac{\ln 2}{\mu} = \frac{0.693}{0.10} = 6.9\ \text{cm}$ .

Q3. Explain why a coupling gel is used in ultrasound scanning. [2 marks]

Cue. Without it, the large impedance mismatch between air and skin would reflect almost all the ultrasound; the gel matches the impedances so the beam enters the body.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20194 marksA parallel X-ray beam passes through

4.0\ \text{cm}

of soft tissue with a linear attenuation coefficient of

0.20\ \text{cm}^{-1}

. Calculate the fraction of the beam intensity transmitted, and the half-value thickness. Take

\ln 2 = 0.693

Show worked answer →

Attenuation: $I = I_0 e^{-\mu x}$ , so the transmitted fraction is $\frac{I}{I_0} = e^{-\mu x} = e^{-(0.20)(4.0)} = e^{-0.80}$ .

$e^{-0.80} = 0.45$ , so about 45 per cent is transmitted.

Half-value thickness: $x_{1/2} = \frac{\ln 2}{\mu} = \frac{0.693}{0.20} = 3.5\ \text{cm}$ .

Markers reward $I = I_0 e^{-\mu x}$ , the transmitted fraction about 0.45, and the half-value thickness about $3.5\ \text{cm}$ .

OCR 20214 marksUltrasound passes from soft tissue (acoustic impedance

1.6 \times 10^{6}\ \text{kg m}^{-2}\ \text{s}^{-1}

) to bone (acoustic impedance

6.4 \times 10^{6}\ \text{kg m}^{-2}\ \text{s}^{-1}

). Calculate the fraction of the intensity reflected at the boundary, and explain why a coupling gel is used between the transducer and the skin.

Show worked answer →

Intensity reflection coefficient: $\frac{I_r}{I_i} = \left(\frac{Z_2 - Z_1}{Z_2 + Z_1}\right)^2 = \left(\frac{6.4 - 1.6}{6.4 + 1.6}\right)^2 = \left(\frac{4.8}{8.0}\right)^2$ .

$= (0.60)^2 = 0.36$ , so 36 per cent of the intensity is reflected at the tissue-bone boundary.

A coupling gel is used because the large impedance mismatch between air and skin would otherwise reflect almost all the ultrasound; the gel matches the impedances so the beam enters the body.

Markers reward the reflection-coefficient equation, the value 0.36, and an explanation that the gel removes the air gap and its huge impedance mismatch.

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Sources & how we know this

OCR AS and A Level Physics A (H156, H556) specification — OCR (2015)