Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.
69 original questions across 13 selected topic areas. No account or payment required to practise.
6 questions in this topic. Exact English revisions checked in an AI-assisted academic review; independent teacher approval is still pending.
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Original English concept practice addressing selected identified gaps. Not complete syllabus coverage or practical competence.
Selected subtopics: Beta decay and quarks · Ultrasound echoes · X-ray and PET physics · Stellar properties · Redshift and expansion
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Question 1A free neutron undergoes beta-minus decay. Using up-quark charge +2e/3 and down-quark charge -e/3, identify the neutron/proton quark compositions and the quark change. Write the decay products and explain why the electron energy need not have one fixed value.4 indicative marks
Explain using the evidence4 indicative marksNo calculator
A free neutron undergoes beta-minus decay. Using up-quark charge +2e/3 and down-quark charge -e/3, identify the neutron/proton quark compositions and the quark change. Write the decay products and explain why the electron energy need not have one fixed value.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked quark charges, correct beta-minus antineutrino and continuous-energy explanation against 11.1/11.2, not merely alpha bookkeeping. This is conceptual placement, not completion of the entire referenced subsection.
Sum udd charges: 2/3 - 1/3 - 1/3 = 0. Replacing one d by u gives 2/3 + 2/3 - 1/3 = 1.
The proton's extra positive charge is balanced by the emitted electron. The antineutrino also carries energy/momentum, so conservation does not require every emitted electron to have the same kinetic energy.
Self-marking points
Neutron udd has charge 0; proton uud has charge +e.
One down quark changes to an up quark.
n -> p + electron + electron antineutrino conserves total charge: 0 = +e-e+0.
Energy is shared with the antineutrino and recoiling proton, so beta electrons have a continuous energy range.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 2In a simplified ultrasound model, a pulse echo returns from a boundary after 40.0 microseconds. Sound speed in the tissue is 1500 m s^-1. Find boundary depth. The impedances are Z1 = 1.50 x 10^6 and Z2 = 3.00 x 10^6 kg m^-2 s^-1. Find the reflected intensity fraction using (Z2-Z1)^2/(Z2+Z1)^2 and explain why an air gap can impair imaging. No medical advice or experiment is requested.3 indicative marks
Calculate and explain3 indicative marksCalculator allowed
In a simplified ultrasound model, a pulse echo returns from a boundary after 40.0 microseconds. Sound speed in the tissue is 1500 m s^-1. Find boundary depth. The impedances are Z1 = 1.50 x 10^6 and Z2 = 3.00 x 10^6 kg m^-2 s^-1. Find the reflected intensity fraction using (Z2-Z1)^2/(Z2+Z1)^2 and explain why an air gap can impair imaging. No medical advice or experiment is requested.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Independently checked two-way travel and squared impedance ratio, keeping numerical example separate from qualitative air mismatch. This is conceptual placement, not completion of the entire referenced subsection.
9702: 24.1: Ultrasound transducers and attenuation
The measured echo time includes outward and return travel: 1500 x 40.0e-6/2 = 0.0300 m.
The common 10^6 impedance factors cancel in the ratio. Coupling reduces a strong surface mismatch; the given 11.1% value applies to the supplied two impedances, not to an unspecified air interface.
Self-marking points
Depth = ct/2 = 0.0300 m.
Reflected fraction = (1.50/4.50)^2 = 1/9, approximately 0.111.
A large impedance mismatch at an air interface reflects much of the incident intensity rather than transmitting it into tissue.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 3Electrons accelerated through 50.0 kV strike an idealised X-ray target. Use h=6.63 x 10^-34 J s, c=3.00 x 10^8 m s^-1 and e=1.60 x 10^-19 C. Calculate the minimum X-ray wavelength. If mu=20.0 m^-1 and material thickness is 0.0500 m, calculate I/I0 using exponential attenuation. Explain why CT needs multiple projection angles.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
Electrons accelerated through 50.0 kV strike an idealised X-ray target. Use h=6.63 x 10^-34 J s, c=3.00 x 10^8 m s^-1 and e=1.60 x 10^-19 C. Calculate the minimum X-ray wavelength. If mu=20.0 m^-1 and material thickness is 0.0500 m, calculate I/I0 using exponential attenuation. Explain why CT needs multiple projection angles.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked 50 kV energy conversion, wavelength limit 24.9 pm and e^-1 transmission; CT explanation follows explicit reconstruction content, without radiation-use instructions. This is conceptual placement, not completion of the entire referenced subsection.
The minimum wavelength corresponds to the maximum possible photon energy, not the energy of every X-ray. Divide 1.989e-25 J m by 8.00e-15 J.
The attenuation exponent is -1.00 and is dimensionless. A single projection superposes structures along a beam; multiple directions supply information needed to reconstruct their locations.
Self-marking points
Maximum single-photon energy eV = 8.00 x 10^-15 J.
lambda_min = hc/(eV) = 2.49 x 10^-11 m.
I/I0 = exp(-20.0 x 0.0500) = 0.368.
Different angular projections reconstruct a section, and successive sections can form a three-dimensional image.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 4In an idealised PET event an electron and positron annihilate with negligible initial kinetic energy, producing two opposite photons. With me=9.11 x 10^-31 kg and c=3.00 x 10^8 m s^-1, find total photon energy and energy per photon in joules. Explain the opposite directions and how coincident detections constrain location.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
In an idealised PET event an electron and positron annihilate with negligible initial kinetic energy, producing two opposite photons. With me=9.11 x 10^-31 kg and c=3.00 x 10^8 m s^-1, find total photon energy and energy per photon in joules. Explain the opposite directions and how coincident detections constrain location.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked factor of two between total and each photon and energy/momentum conservation; no unsupported claim of a unique point from one pair. This is conceptual placement, not completion of the entire referenced subsection.
Each particle contributes rest energy 9.11e-31 x (3.00e8)^2 = 8.199e-14 J. Two rest masses give twice this value, shared equally in this centre-of-momentum approximation.
One detection pair alone does not identify a unique point along the line. Arrival-time and many-event information constrain the reconstruction; this question does not prescribe tracer preparation or administration.
Self-marking points
Total energy = 2me c^2 = 1.64 x 10^-13 J.
Equal photon energies are each me c^2 = 8.20 x 10^-14 J.
Opposite equal photon momenta conserve the approximately zero initial total momentum.
A detector pair identifies a line consistent with the event; many events and processing reconstruct tracer distribution.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 5A star has luminosity 4.00 x 10^26 W. Its spectrum peaks at 5.00 x 10^-7 m; use lambda_max T = 2.90 x 10^-3 m K and sigma=5.67 x 10^-8 W m^-2 K^-4. Estimate surface temperature and radius using L=4pi sigma r^2 T^4. If its observed radiant flux is 1.00 x 10^-8 W m^-2, estimate distance. Treat radiation as isotropic and ignore absorption.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A star has luminosity 4.00 x 10^26 W. Its spectrum peaks at 5.00 x 10^-7 m; use lambda_max T = 2.90 x 10^-3 m K and sigma=5.67 x 10^-8 W m^-2 K^-4. Estimate surface temperature and radius using L=4pi sigma r^2 T^4. If its observed radiant flux is 1.00 x 10^-8 W m^-2, estimate distance. Treat radiation as isotropic and ignore absorption.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Checked Wien temperature, Stefan-Boltzmann fourth power and distinct stellar-radius/observer-distance spheres; no real-star calibration asserted. This is conceptual placement, not completion of the entire referenced subsection.
Divide the displacement constant by the peak wavelength. With T=5800 K, T^4 = 1.1316496e15 K^4; substitution gives a radius about 7.043e8 m.
Radiant flux is luminosity spread over a sphere at the observer, not the stellar surface. Its inverse-square relation gives d about 5.642e16 m under the stated isotropic/no-absorption model.
Self-marking points
T = 5800 K.
r = sqrt(L/(4pi sigma T^4)) = 7.04 x 10^8 m.
d = sqrt(L/(4pi F)) = 5.64 x 10^16 m.
Known luminosity allows a flux-based standard-candle distance; absorption would bias an uncorrected estimate.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
Question 6A galaxy's line with laboratory wavelength 500 nm is observed at 505 nm. In the small-redshift approximation use z=delta lambda/lambda=v/c, c=3.00 x 10^8 m s^-1 and H0=2.20 x 10^-18 s^-1. Estimate recession speed, Hubble-law distance and the timescale 1/H0 in seconds. Explain one limitation of calling that timescale the exact age of the Universe.4 indicative marks
Calculate and explain4 indicative marksCalculator allowed
A galaxy's line with laboratory wavelength 500 nm is observed at 505 nm. In the small-redshift approximation use z=delta lambda/lambda=v/c, c=3.00 x 10^8 m s^-1 and H0=2.20 x 10^-18 s^-1. Estimate recession speed, Hubble-law distance and the timescale 1/H0 in seconds. Explain one limitation of calling that timescale the exact age of the Universe.
Cambridge 2027: Direct conceptual match
Original targeted concept exercise: Verified small-z sign/denominator, SI H0 distance and reciprocal timescale; distinction from exact age avoids an unsupported cosmology claim. This is conceptual placement, not completion of the entire referenced subsection.
Use the rest wavelength in the denominator. A 1% shift yields 1% of c in the supplied low-z approximation; no relativistic high-redshift formula is implied.
Divide by H0 with units s^-1 to obtain metres, and invert H0 separately for seconds. Hubble-law expansion supports a finite expansion timescale but an exact cosmological age requires the expansion history.
Self-marking points
z=(505-500)/500 = 0.0100; v = 3.00 x 10^6 m s^-1.
d = v/H0 = 1.36 x 10^24 m.
1/H0 = 4.55 x 10^17 s.
The reciprocal timescale assumes a simple expansion history; the expansion rate need not have stayed constant.
Examiner tip
These are original indicative marking points, not an official Cambridge answer key. State assumptions and connect each conclusion to the supplied evidence.
