AS & A Level Physics (9702)
Medical physics Banker
28 questions · about 3% of the marks we indexed · examined in 9 of 9 sessions.
Practice questions
Original questions written by HomiLabs in the style of the exam, with answers and mark schemes. They are not
Cambridge past paper questions and have not been reviewed by Cambridge. Try each one before opening the answer.
Q1X-rays, attenuation & CT · Core · Paper 4 style · 5 marks- (a) Describe how X-rays are produced in an X-ray tube. [3]
- (b) The accelerating potential difference is 80 kV. Calculate the minimum wavelength of the X-rays produced. [2]
Show answer
- (a) Electrons from a heated cathode are accelerated through a large p.d. to the anode; they strike a metal target and decelerate rapidly, emitting X-ray photons (braking radiation); characteristic lines come from inner-shell electron transitions in the target atoms.
B1: electrons emitted by thermionic emission and accelerated through a high p.d. B1: electrons hit a (high-melting-point, e.g. tungsten) metal target. B1: rapid deceleration produces X-ray photons (continuous spectrum) / electron transitions in target atoms give characteristic lines.
- (b) 1.55 × 10⁻¹¹ m
C1: eV = hc/λ_min (all KE of one electron → one photon). A1: λ_min = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / (1.60 × 10⁻¹⁹ × 80 × 10³) = 1.55 × 10⁻¹¹ m.
A parallel beam of X-rays passes through 4.0 cm of muscle. The linear attenuation coefficient of muscle for these X-rays is 0.20 cm⁻¹.
- (a) State what is meant by the linear attenuation coefficient µ. [1]
- (b) Calculate the fraction of the incident intensity that emerges from the muscle. [2]
- (c) State one reason why a parallel beam is assumed in this calculation. [1]
Show answer
- (a) The fraction (proportion) of the intensity of the beam absorbed/removed per unit thickness of the material (µ in I = I₀e^(−µx)).
B1: fractional decrease in intensity per unit distance travelled (or the constant µ in I = I₀e^(−µx)).
- (b) 0.449
C1: I/I₀ = e^(−0.20 × 4.0). A1: 0.449.
- (c) So that any decrease in intensity is due only to attenuation, not to the beam spreading out.
B1: no reduction in intensity due to divergence of the beam.
A metal filter used in an X-ray machine has a linear attenuation coefficient of 0.46 mm⁻¹ for X-rays of a particular energy.
- (a) Show that the half-value thickness x½ is related to µ by x½ = ln 2 / µ. [2]
- (b) Calculate the half-value thickness of the filter material. [1]
- (c) Calculate the thickness that reduces the intensity to 1/32 of its incident value. [1]
Show answer
- (a) ½ = e^(−µx½), so µx½ = ln 2
M1: I = I₀/2 when x = x½: ½ = e^(−µx½). A1: ln 2 = µx½.
- (b) 1.51 mm
A1: x½ = 0.693/0.46 = 1.51 mm.
- (c) 7.53 mm
A1: 1/32 = (½)⁵, so 5 half-value thicknesses = 7.53 mm.
- (a) State what is meant by the sharpness of an X-ray image, and suggest one way it is improved. [2]
- (b) State what is meant by the contrast of an X-ray image. [1]
- (c) Explain why a thin aluminium filter is placed in the X-ray beam before it reaches the patient. [2]
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- (a) Sharpness is how clearly the edges/boundaries of structures are defined; improved by a smaller area of the target hit by electrons (smaller focal spot), or a lead grid / aperture to produce a narrower, more parallel beam.
B1: edges well defined / how clearly boundaries are seen. B1: small anode area (focal spot) / collimate beam / lead grid in front of detector.
- (b) The difference in degree of blackening (brightness) between different areas/tissues of the image.
B1: difference in blackening between structures.
- (c) It absorbs the low-energy (long-wavelength) X-rays, which would be absorbed in the skin and tissue near the surface without reaching the detector, so they would increase the patient's dose without improving the image.
B1: removes low-energy X-ray photons. B1: these would only be absorbed by the patient (skin) — dose without contributing to the image.
A dental X-ray tube operates with an accelerating potential difference of 30 kV.
- (a) Calculate the maximum energy, in joules, of an X-ray photon produced. [1]
- (b) The tube current is 25 mA and 1.0% of the electron beam's energy is converted to X-rays. Calculate the rate at which thermal energy is produced in the anode. [2]
- (c) Suggest why the anode in many X-ray tubes is made to rotate. [1]
Show answer
- (a) 4.80 × 10⁻¹⁵ J
A1: E = eV = 1.60 × 10⁻¹⁹ × 30 × 10³ = 4.80 × 10⁻¹⁵ J.
- (b) 743 W (742.5 W)
C1: power of beam = IV = 25 × 10⁻³ × 30 × 10³ = 750 W. A1: thermal power = 0.99 × 750 = 742.5 W ≈ 743 W (accept 742 W).
- (c) So that the heating is spread over a larger area of the target, preventing it from overheating/melting.
B1: spreads the heat over a larger area / reduces local temperature.
A parallel X-ray beam passes through a patient's arm along two paths of the same total length, 6.0 cm. Path 1 is through soft tissue only. Path 2 is through 2.5 cm of bone and 3.5 cm of soft tissue. Linear attenuation coefficients: soft tissue 0.24 cm⁻¹, bone 0.62 cm⁻¹.
- (a) Calculate, for each path, the fraction of the incident intensity that emerges. [3]
- (b) Calculate the ratio of these intensities and state whether the bone will be clearly visible on the image. [2]
Show answer
- (a) tissue only: 0.237; bone path: 0.0916
C1: I/I₀ = e^(−µx), adding the exponents for successive layers. A1: tissue only: e^(−1.44) = 0.237. A1: bone path: e^(−(1.55 + 0.84)) = 0.0916.
- (b) 2.59; yes, there is good contrast
A1: 0.237/0.0916 = 2.59. B1: large ratio → clear difference in blackening, good contrast.
A block of tissue-equivalent plastic 3.5 cm thick transmits 18% of the intensity of a parallel X-ray beam.
- (a) Calculate the linear attenuation coefficient of the material. [2]
- (b) Calculate the half-value thickness of the material. [1]
- (c) State and explain how µ would change if X-rays of higher photon energy were used. [2]
Show answer
- (a) 0.490 cm⁻¹
C1: 0.18 = e^(−µ × 3.5). A1: µ = ln(1/0.18)/3.5 = 0.490 cm⁻¹.
- (b) 1.41 cm
A1: x½ = ln 2/µ = 1.41 cm.
- (c) µ would decrease: higher-energy photons are more penetrating, so a smaller fraction is absorbed per unit thickness.
B1: decreases. B1: more penetrating / less absorption per unit thickness.
- (a) Explain how a CT scan produces a three-dimensional image of a section of the body. [4]
- (b) Suggest one advantage and one disadvantage of a CT scan compared with a single X-ray image. [2]
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- (a) Images of many thin slices are taken; for each slice, an X-ray tube rotates around the patient and detectors on the opposite side record the intensity from many directions; a computer calculates the attenuation for each voxel and builds an image of each slice; images of successive slices are combined to give a 3D image that can be viewed from any angle.
B1: image of a thin slice (section) taken. B1: X-ray tube (and detectors) rotate around the patient, many directions in the plane of the slice. B1: computer deduces value for each voxel/pixel of the slice. B1: many slices combined into 3D image that can be rotated.
- (b) Advantage: 3D image / soft tissues and structures at different depths can be distinguished (no overlapping shadows). Disadvantage: much larger radiation dose (and higher cost, longer time).
B1: advantage, e.g. 3D, can be viewed from any angle, better contrast for soft tissue, no overlap. B1: disadvantage, e.g. higher dose.
A simple model of a CT slice has four voxels arranged in a square: P (top left), Q (top right), R (bottom left) and S (bottom right). Each detector reading equals the sum of the voxel values along the beam.Beam along top row (P, Q): 10Beam along bottom row (R, S): 14Beam down left column (P, R): 8Beam along the diagonal (P, S): 12
- (a) Explain why at least four readings are needed to find the four voxel values. [1]
- (b) Determine the values of P, Q, R and S. [4]
Show answer
- (a) There are four unknowns, so four independent equations (readings) are needed.
B1: four unknowns need four independent equations.
- (b) P = 3, Q = 7, R = 5, S = 9
C1: from P + R = 8 and P + S = 12: R = 8 − P, S = 12 − P. C1: R + S = 14 → 20 − 2P = 14. A1: P = 3. A1: Q = 10 − 3 = 7, R = 5, S = 9.
A lead apron used by a radiographer has a linear attenuation coefficient of 0.62 mm⁻¹ for the scattered X-rays in the room.
- (a) Calculate the thickness of lead needed to reduce the intensity of the beam to 5.0% of its incident value. [2]
- (b) Calculate how many half-value thicknesses this represents. [1]
- (c) Explain why the intensity of the beam never becomes exactly zero, however thick the lead. [1]
Show answer
- (a) 4.83 mm
C1: 0.050 = e^(−0.62x). A1: x = ln 20/0.62 = 4.83 mm.
- (b) 4.32
A1: n = ln 20/ln 2 = 4.32.
- (c) Attenuation is exponential: each extra thickness removes the same fraction, so the intensity approaches but never reaches zero.
B1: exponential decrease — constant fraction removed per unit thickness.
Linear attenuation coefficients for two X-ray photon energies are:40 keV: bone 1.30 cm⁻¹, soft tissue 0.27 cm⁻¹100 keV: bone 0.40 cm⁻¹, soft tissue 0.17 cm⁻¹
- (a) For each photon energy, calculate the ratio (intensity through 3.0 cm of soft tissue)/(intensity through 3.0 cm of bone). [3]
- (b) Calculate the fraction transmitted through 3.0 cm of bone at each energy. [1]
- (c) Using your answers, discuss the choice of photon energy for imaging a bone fracture. [2]
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- (a) 40 keV: 22.0; 100 keV: 1.99
C1: ratio = e^(−µ_t x)/e^(−µ_b x) = e^((µ_b − µ_t)x). A1: 40 keV: e^(1.03 × 3.0) = 22.0. A1: 100 keV: e^(0.23 × 3.0) = 1.99.
- (b) 40 keV: 0.0202; 100 keV: 0.301
A1: e^(−3.9) = 0.0202; e^(−1.2) = 0.301.
- (c) Lower energy gives much better contrast between bone and tissue, but far less is transmitted, so a larger dose is needed for enough intensity at the detector; a compromise energy is chosen.
B1: low energy → better contrast. B1: but more absorbed by patient / higher dose / weak signal through thick parts — compromise.
A parallel X-ray beam passes through a patient's leg. In order, it crosses 0.80 cm of skin and fat (µ = 0.18 cm⁻¹), 2.0 cm of bone (µ = 0.55 cm⁻¹) and 3.0 cm of muscle (µ = 0.23 cm⁻¹).
- (a) Calculate the percentage of the incident intensity that leaves the leg. [3]
- (b) Calculate the thickness of muscle alone that would produce the same attenuation. [2]
- (c) The beam is not parallel but spreads out from the tube. State the effect on the measured intensity at the detector and explain why this does not change the contrast between two neighbouring paths. [2]
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- (a) 14.5%
C1: total exponent = 0.18 × 0.80 + 0.55 × 2.0 + 0.23 × 3.0 = 1.934. C1: I/I₀ = e^(−1.934). A1: 14.5%.
- (b) 8.41 cm
C1: 0.23x = 1.934. A1: x = 8.41 cm.
- (c) The measured intensity is lower (beam spreads over a larger area), but both neighbouring paths are reduced by the same factor, so their ratio (contrast) is unchanged.
B1: lower intensity due to divergence. B1: same factor for both paths, so ratio unchanged.
A hospital physicist measures the intensity I (arbitrary units) of a narrow parallel X-ray beam after it passes through thickness x of a test absorber. It is suggested that I = I₀e^(−µx).x / mm: 0, 2, 4, 6, 8, 10I: 1000, 793, 633, 500, 399, 317
- (a) Explain why a graph of ln I against x should be a straight line, and state how µ is found from it. [2]
- (b) Calculate ln I for x = 6 mm and x = 10 mm, giving your answers to 3 decimal places. [1]
- (c) Using the first and last points (which lie on the best-fit line), determine µ. [2]
- (d) Calculate the half-value thickness of the absorber. [1]
- (e) Each reading of I has an uncertainty of ±10 (arbitrary units). Calculate the absolute uncertainty in ln I for the reading at x = 10 mm. [2]
Show answer
- (a) ln I = ln I₀ − µx, which has the form y = c + mx; µ = −gradient.
B1: ln I = −µx + ln I₀ (linear in x). B1: µ = −gradient (intercept = ln I₀).
- (b) 6.215 and 5.759
A1: ln 500 = 6.215; ln 317 = 5.759 (3 d.p., matching the 3 s.f. data).
- (c) 0.115 mm⁻¹
C1: gradient = (5.759 − 6.908)/(10 − 0) = −0.115 mm⁻¹. A1: µ = 0.115 mm⁻¹.
- (d) 6.03 mm
A1: x½ = ln 2/µ = 6.03 mm.
- (e) 0.032
C1: Δ(ln I) = ΔI/I. A1: 10/317 = ±0.032.
For the X-rays used, the linear attenuation coefficient of soft tissue is 0.25 cm⁻¹ and that of a barium-sulfate meal is 4.2 cm⁻¹.
- (a) Explain why a plain X-ray image of the intestine shows poor contrast with the surrounding tissue. [2]
- (b) A patient swallows a barium meal. Calculate the fraction of incident intensity transmitted through 1.5 cm of barium meal and through 1.5 cm of soft tissue, and the ratio of the two. [3]
- (c) Suggest why barium is effective as a contrast medium. [1]
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- (a) Intestine walls and surrounding soft tissues have similar densities/atomic numbers and therefore similar attenuation coefficients, so they absorb X-rays to a similar extent.
B1: similar attenuation coefficients (similar density / Z). B1: so similar transmitted intensities, little difference in blackening.
- (b) barium: 0.00184; tissue: 0.687; ratio 374
C1: e^(−4.2 × 1.5) = 0.00184. C1: e^(−0.25 × 1.5) = 0.687. A1: ratio = 374.
- (c) It has a high atomic number (proton number), so it absorbs X-rays strongly.
B1: high Z → large attenuation coefficient.
- (a) Explain how a piezoelectric transducer produces a pulse of ultrasound. [3]
- (b) Explain how the same transducer detects the reflected pulse. [2]
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- (a) An alternating p.d. at ultrasonic frequency is applied across a piezoelectric crystal; the crystal changes shape (expands and contracts) with the p.d.; it vibrates (resonantly when the frequency matches its natural frequency), sending out ultrasound waves.
B1: alternating p.d. applied across the crystal. B1: crystal changes shape/thickness with the p.d. B1: it vibrates at the frequency of the p.d. (ultrasonic, >20 kHz), producing ultrasound (large amplitude at resonance).
- (b) The returning ultrasound makes the crystal vibrate (change shape), which produces an alternating p.d. across it that is detected and processed.
B1: reflected wave makes the crystal vibrate / deform. B1: this generates an alternating e.m.f. across the crystal.
Data for two tissues:muscle: density 1075 kg m⁻³, speed of ultrasound 1590 m s⁻¹fat: density 925 kg m⁻³, speed of ultrasound 1450 m s⁻¹
- (a) Define specific acoustic impedance. [1]
- (b) Calculate the specific acoustic impedance of each tissue. [2]
- (c) Calculate the intensity reflection coefficient at a fat–muscle boundary. [2]
Show answer
- (a) The product of the density of a medium and the speed of sound (ultrasound) in it: Z = ρc.
B1: Z = density × speed of sound in the medium.
- (b) muscle 1.71 × 10⁶ kg m⁻² s⁻¹; fat 1.34 × 10⁶ kg m⁻² s⁻¹
A1: muscle: 1075 × 1590 = 1.71 × 10⁶ kg m⁻² s⁻¹. A1: fat: 925 × 1450 = 1.34 × 10⁶ kg m⁻² s⁻¹.
- (c) 0.0146
C1: α = (Z₂ − Z₁)²/(Z₂ + Z₁)². A1: α = 0.0146.
An ultrasound pulse is sent into a patient's abdomen. The echo from a boundary between two organs is received 65 µs after the pulse is emitted. The average speed of ultrasound in soft tissue is 1540 m s⁻¹.
- (a) Calculate the depth of the boundary below the skin. [2]
- (b) Suggest why the ultrasound is sent in short pulses rather than as a continuous wave. [1]
Show answer
- (a) 5.01 cm
C1: pulse travels to the boundary and back: 2d = ct. A1: d = 1540 × 65 × 10⁻⁶ / 2 = 0.0501 m = 5.01 cm (accept 5.0 cm).
- (b) So that the time between emission and the arrival of each echo can be measured (the transducer can act as a receiver between pulses).
B1: to time the echo / allow the transducer to receive between pulses.
Specific acoustic impedances: air 430 kg m⁻² s⁻¹, skin 1.6 × 10⁶ kg m⁻² s⁻¹, coupling gel 1.5 × 10⁶ kg m⁻² s⁻¹.
- (a) Calculate the intensity reflection coefficient at an air–skin boundary. [2]
- (b) Calculate the intensity reflection coefficient when a gel is placed between the transducer and the skin. [2]
- (c) Use your answers to explain the purpose of the gel. [2]
Show answer
- (a) 0.9989
C1: α = ((1.6 × 10⁶ − 430)/(1.6 × 10⁶ + 430))². A1: α = 0.9989 (almost total reflection).
- (b) 0.00104
A1: ((1.6 − 1.5)/(1.6 + 1.5))² = 0.00104.
- (c) Without gel, almost all the ultrasound is reflected at the skin surface and very little enters the body; the gel has an acoustic impedance close to that of skin, so almost all the ultrasound is transmitted.
B1: air gap reflects nearly all (99.9%) of the ultrasound. B1: gel impedance-matches, so nearly all is transmitted into the body.
Ultrasound is attenuated exponentially in soft tissue with an attenuation coefficient of 48 m⁻¹. A boundary 4.0 cm below the skin has an intensity reflection coefficient of 0.0080. Ignore reflection at the skin.
- (a) State the total distance travelled through the tissue by the ultrasound that returns to the transducer from the boundary. [1]
- (b) Calculate the ratio (intensity of the echo received at the transducer)/(intensity transmitted into the tissue). [3]
- (c) Explain why the received echoes from deeper boundaries are amplified more before display. [1]
Show answer
- (a) 8.0 cm (0.080 m)
B1: there and back: 2 × 4.0 cm.
- (b) 1.72 × 10⁻⁴
C1: attenuation factor e^(−48 × 0.080) = 0.0215. C1: multiply by α = 0.0080. A1: 1.72 × 10⁻⁴.
- (c) They have travelled further through tissue and been attenuated more, so amplification compensates for the extra attenuation.
B1: greater attenuation over a longer path.
An A-scan of a patient's thigh shows the transmitted pulse at time zero, an echo from the fat–muscle boundary at 26 µs and an echo from the muscle–bone boundary at 92 µs. The transducer is in contact with the fat layer. Speeds of ultrasound: fat 1450 m s⁻¹, muscle 1590 m s⁻¹.
- (a) Calculate the thickness of the fat layer. [2]
- (b) Calculate the thickness of the muscle layer. [2]
- (c) Suggest why the echo from the muscle–bone boundary is much stronger than that from the fat–muscle boundary. [1]
Show answer
- (a) 1.89 cm
C1: d = ct/2 = 1450 × 26 × 10⁻⁶ / 2. A1: 0.01885 m = 1.89 cm (accept 1.88 cm).
- (b) 5.25 cm
C1: time in muscle = 92 − 26 = 66 µs (there and back). A1: d = 1590 × 66 × 10⁻⁶ / 2 = 5.25 cm.
- (c) The difference in acoustic impedance between muscle and bone is much greater, so a much larger fraction is reflected.
B1: larger difference in Z → larger reflection coefficient.
Ultrasound in soft tissue travels at 1540 m s⁻¹. The attenuation coefficient is 40 m⁻¹ at 3.5 MHz and 115 m⁻¹ at 10 MHz.
- (a) Calculate the wavelength in soft tissue at each frequency. [2]
- (b) Calculate, for each frequency, the fraction of intensity remaining after the ultrasound travels to an organ 6.0 cm deep and back, considering attenuation only. [2]
- (c) Use your answers to explain why 10 MHz is used for imaging an eye or a tendon close to the skin but 3.5 MHz is used for imaging an organ deep in the abdomen. [3]
Show answer
- (a) 3.5 MHz: 0.440 mm; 10 MHz: 0.154 mm
A1: λ = c/f = 1540/3.5 × 10⁶ = 0.440 mm. A1: 1540/10 × 10⁶ = 0.154 mm.
- (b) 3.5 MHz: 0.00823; 10 MHz: 1.02 × 10⁻⁶
C1: path 0.12 m, I/I₀ = e^(−µx). A1: e^(−4.8) = 0.00823; e^(−13.8) = 1.02 × 10⁻⁶.
- (c) Higher frequency → shorter wavelength → finer detail (better resolution) but much greater attenuation, so only usable for shallow structures; deep organs need lower frequency to give a detectable echo, accepting poorer resolution.
B1: shorter wavelength resolves smaller details. B1: high frequency attenuated much more (echo ~10⁻⁶, undetectable from deep). B1: so trade-off: high f for shallow, low f for deep.
Ultrasound enters a fat layer from a transducer, then passes into muscle and is partly reflected at bone. Specific acoustic impedances: fat 1.34 × 10⁶, muscle 1.71 × 10⁶, bone 7.80 × 10⁶ kg m⁻² s⁻¹.
- (a) Calculate the intensity reflection coefficients at the fat–muscle and muscle–bone boundaries. [2]
- (b) Ignoring absorption, calculate the fraction of the intensity entering the fat that returns to the transducer as the echo from the muscle–bone boundary. [3]
- (c) Explain why structures lying behind a bone are difficult to image with ultrasound. [1]
Show answer
- (a) fat–muscle 0.0147; muscle–bone 0.410
A1: ((1.71 − 1.34)/(1.71 + 1.34))² = 0.0147. A1: ((7.80 − 1.71)/(7.80 + 1.71))² = 0.410.
- (b) 0.398
C1: transmitted into muscle: 1 − 0.0147. C1: reflected at bone then transmitted back through the fat–muscle boundary: × 0.410 × (1 − 0.0147). A1: 0.398.
- (c) A large fraction of the ultrasound is reflected at the bone surface (and bone absorbs strongly), so little passes beyond it.
B1: large reflection at the bone boundary — little intensity transmitted.
A coupling gel has specific acoustic impedance 1.50 × 10⁶ kg m⁻² s⁻¹. At the boundary between the gel and a tissue of greater acoustic impedance, the intensity reflection coefficient is measured to be 0.0025.
- (a) Show that (Z_T − Z_G)/(Z_T + Z_G) = 0.050, where Z_T and Z_G are the specific acoustic impedances of the tissue and the gel. [1]
- (b) Calculate Z_T. [2]
- (c) The speed of ultrasound in the tissue is 1560 m s⁻¹. Calculate the density of the tissue. [1]
- (d) State the fraction of the incident intensity transmitted into the tissue. [1]
Show answer
- (a) √0.0025 = 0.050 (Z_T > Z_G so the positive root)
B1: α = ((Z_T − Z_G)/(Z_T + Z_G))², square root = 0.050.
- (b) 1.66 × 10⁶ kg m⁻² s⁻¹
C1: Z_T − Z_G = 0.050(Z_T + Z_G) → Z_T = Z_G × 1.05/0.95. A1: Z_T = 1.66 × 10⁶ kg m⁻² s⁻¹.
- (c) 1060 kg m⁻³
A1: ρ = Z/c = 1.66 × 10⁶/1560 = 1060 kg m⁻³.
- (d) 0.9975 (99.75%)
B1: 1 − α = 0.9975.
- (a) State what is meant by a tracer. [1]
- (b) Describe what happens when a positron emitted by the tracer meets an electron. [2]
- (c) Explain how detecting the gamma photons allows an image of tracer concentration to be produced. [2]
Show answer
- (a) A radioactive substance (radionuclide attached to a compound) introduced into the body that is absorbed by tissue and whose emissions show where it has gone.
B1: substance containing a radioactive nuclide that is absorbed by the tissue being studied.
- (b) They annihilate: the total mass of the positron and electron is converted into energy in the form of two gamma photons.
B1: annihilation of particle and antiparticle. B1: all mass converted to energy of two γ-ray photons.
- (c) The two photons travel in opposite directions and are detected at the same time by detectors in a ring around the patient; the event lies on the line joining the two detectors; the difference in arrival times locates the point, and many such events build up an image.
B1: photons detected simultaneously by detectors on opposite sides of the ring → event lies on the line between them. B1: arrival-time difference / many lines intersecting give the location; number of events shows concentration.
In a PET scanner, a positron and an electron, both effectively at rest, annihilate to produce two gamma photons. Mass of electron and of positron = 9.11 × 10⁻³¹ kg.
- (a) Calculate the energy of each gamma photon, in J and in MeV. [3]
- (b) Calculate the wavelength of each gamma photon. [2]
Show answer
- (a) 8.20 × 10⁻¹⁴ J = 0.512 MeV
C1: total energy 2m_ec², shared between two photons, so each has m_ec². A1: 9.11 × 10⁻³¹ × (3.00 × 10⁸)² = 8.20 × 10⁻¹⁴ J. A1: 8.20 × 10⁻¹⁴/1.60 × 10⁻¹³ = 0.512 MeV.
- (b) 2.43 × 10⁻¹² m
C1: λ = hc/E. A1: λ = 6.63 × 10⁻³⁴ × 3.00 × 10⁸ / 8.20 × 10⁻¹⁴ = 2.43 × 10⁻¹² m.
- (a) Explain, using a conservation law, why the two gamma photons travel in opposite directions. [2]
- (b) The tracer contains fluorine-18, of half-life 110 minutes. Calculate the fraction of the fluorine-18 nuclei remaining 3.0 hours after the tracer is made. [2]
- (c) Suggest one reason why a short half-life is chosen for a PET tracer. [1]
Show answer
- (a) The total momentum before annihilation is (approximately) zero; momentum is conserved, so the two photons must have equal and opposite momenta.
B1: initial momentum zero (particles at rest). B1: conservation of momentum → equal and opposite momenta → opposite directions (and equal energies).
- (b) 0.322
C1: N/N₀ = (½)^(t/t½) with t = 180 min. A1: (½)^(180/110) = 0.322.
- (c) The tracer's activity falls quickly after the scan, so the patient's total radiation dose is small.
B1: small dose to patient / tracer does not stay active in the body for long.
In a PET scanner, two detectors A and B lie directly opposite each other on the ring, 0.80 m apart. Two gamma photons from one annihilation event are detected; the photon at A arrives 0.60 ns before the photon at B. Speed of gamma photons = 3.00 × 10⁸ m s⁻¹.
- (a) Calculate the distance of the annihilation point from the centre of the line joining the detectors, and state which detector it is nearer to. [3]
- (b) Explain why detectors on opposite sides of the ring record an event only if the two photons arrive within a few nanoseconds of each other. [2]
Show answer
- (a) 9.00 cm from the centre, nearer detector A
C1: if the event is a distance x from the centre, one photon travels x further than the other: path difference = 2x. C1: 2x = cΔt = 3.00 × 10⁸ × 0.60 × 10⁻⁹. A1: x = 9.00 cm, nearer A (which receives first).
- (b) Photons from the same annihilation arrive almost simultaneously; requiring coincidence rejects unrelated photons (from different events or scattered photons), so only genuine pairs define a line.
B1: photons from one annihilation reach the detectors at nearly the same time (maximum difference ~ ring diameter/c). B1: rejects photons from separate events / scatter.
A patient is injected with a fluorine-18 glucose tracer of activity 370 MBq. The half-life of fluorine-18 is 110 minutes.
- (a) Calculate the activity of the tracer 75 minutes after injection, when the scan begins. [2]
- (b) Calculate the number of fluorine-18 nuclei in the dose at the time of injection. [2]
- (c) Explain why the patient is asked to rest quietly between injection and scan when imaging glucose uptake in the brain. [1]
Show answer
- (a) 231 MBq
C1: A = A₀(½)^(t/t½). A1: A = 370 × (½)^(75/110) = 231 MBq.
- (b) 3.52 × 10¹²
C1: λ = ln 2/(110 × 60 s) = 1.05 × 10⁻⁴ s⁻¹. A1: N = A/λ = 370 × 10⁶ / λ = 3.52 × 10¹².
- (c) Muscle activity would take up the glucose-based tracer, reducing the amount reaching the brain and adding unwanted signal from muscles.
B1: active muscles absorb the tracer / tracer goes where glucose is used.
In a PET scan the positron and electron are at rest when they annihilate. Mass of electron = 9.11 × 10⁻³¹ kg; speed of light = 3.00 × 10⁸ m s⁻¹.
- (a) Calculate the momentum of each gamma photon. [2]
- (b) Each decay produces one positron, which annihilates. Calculate the total power emitted as gamma radiation when the activity of the tracer in the body is 200 MBq. [3]
- (c) Suggest why it is an advantage that gamma photons of this energy are produced, compared with lower-energy photons. [2]
Show answer
- (a) 2.73 × 10⁻²² N s
C1: p = E/c = m_ec²/c = m_ec. A1: p = 9.11 × 10⁻³¹ × 3.00 × 10⁸ = 2.73 × 10⁻²² N s.
- (b) 3.28 × 10⁻⁵ W
C1: energy per annihilation = 2m_ec² = 1.64 × 10⁻¹³ J. C1: P = activity × energy per decay. A1: P = 200 × 10⁶ × 1.64 × 10⁻¹³ = 3.28 × 10⁻⁵ W.
- (c) High-energy photons are very penetrating, so most leave the body without being absorbed; this gives a strong signal at the detectors and less energy deposited in the patient.
B1: penetrating — most escape the body. B1: detectable signal / lower absorbed dose / less scattering.
A PET tracer uses fluorine-18, half-life 110 minutes.
- (a) The tracer is produced in a cyclotron and takes 2.5 hours to reach the hospital and be injected. Calculate the activity it must have when it leaves the cyclotron so that 370 MBq is available at injection. [3]
- (b) Suggest one problem with using a positron emitter with a half-life of only 2 minutes, and one problem with a half-life of 2 days. [2]
- (c) State what is meant by annihilation in terms of mass and energy. [1]
Show answer
- (a) 952 MBq
C1: A = A₀(½)^(t/t½), so A₀ = A × 2^(t/t½). C1: t = 150 min, t/t½ = 1.36. A1: A₀ = 370 × 2^(150/110) = 952 MBq.
- (b) 2 minutes: decays before it can be transported and absorbed by the tissue (needs a cyclotron on site); 2 days: remains active in the body long after the scan, giving the patient an unnecessary dose.
B1: very short: too little activity left by the time of the scan / needs on-site production. B1: long: prolonged exposure / higher dose to patient (and others).
- (c) A particle meets its antiparticle and their total mass (together with their kinetic energy) is converted into energy in the form of photons (two γ-ray photons).
B1: particle and antiparticle; all mass converted to photon energy.
