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S6 Physics | Medical Physics Lesson-Prep Guide (for tutors) | Ultrasound, X-ray attenuation, radioisotopes
Students can say that ultrasound images organs but cannot link echo time to wave speed; half-value layers and half-lives feel like two unrelated formulas. This lesson folds both into one exponential decay.
Objectives — Students can: (1) use v = f times lambda for wavelength and 2d over v for echo time; (2) use Z = rho c and the reflection coefficient to explain why soft tissue and bone reflect strongly; (3) use I equals I0 times 0.5 to the n, or e to the minus mu x, and combine half-lives into an effective half-life.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Wave equation | In v = f times lambda, is f in Hz or MHz | Hz: 5 MHz is 5 times 10 to the 6 Hz |
| Distance and time | How many trips does an echo make | Two: out and back |
| Indices | What is 0.5 cubed | 0.125 |
02 Lesson flow (60 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-8 min | Wavelength | v is 1540 m/s, f is 5 MHz | 0.308 mm |
| 8-20 min | Pulse echo | 60 mm depth, two trips | 77.9 microseconds |
| 20-30 min | Impedance and reflection | Z = rho c, bone 7.75, tissue 1.62 MRayl | 42.9% reflected |
| 30-42 min | X-ray attenuation | HVL 3 cm, thickness 9 cm | 12.5% transmitted |
| 42-54 min | Half-life and activity | Tc-99m half-life 6 h, after 24 h | 50 MBq |
| 54-60 min | Exit check | Three questions collected | Independent |
03 Teaching script (read it out)
Scene 1: Ultrasound is a ruler, not a camera
Ultrasound travels at about 1540 m/s in soft tissue, so at 5 MHz the wavelength is 1540 divided by 5 times 10 to the 6, which is 0.308 mm. Shorter wavelength resolves finer detail, so high frequency suits shallow work and low frequency suits deep work.
The pulse travels out and back, so the path is 2d. At 60 mm depth the time is 2 times 0.06 divided by 1540, which is 77.9 microseconds; the machine halves that time and multiplies by speed. That is how a depth is drawn from a clock reading.
Attenuation must be counted too: about 5 dB per cm at 5 MHz, so a 12 cm round trip loses 60 dB, a factor of a million in intensity. Deep imaging therefore uses lower frequency.
The pulse travels out and back, so the path is 2d. At 60 mm depth the time is 2 times 0.06 divided by 1540, which is 77.9 microseconds; the machine halves that time and multiplies by speed. That is how a depth is drawn from a clock reading.
Attenuation must be counted too: about 5 dB per cm at 5 MHz, so a 12 cm round trip loses 60 dB, a factor of a million in intensity. Deep imaging therefore uses lower frequency.
Have students convert 77.9 microseconds to milliseconds and explain the factor of two.
Scene 2: Impedance difference sets the echo
Acoustic impedance is Z = rho c. Soft tissue gives 1050 times 1540, which is 1.617 MRayl; bone gives 1900 times 4080, which is 7.752 MRayl.
The intensity reflection coefficient is the square of the impedance difference over the sum: ((7.752 minus 1.617) over (7.752 plus 1.617)) squared, which is 0.429, so 42.9% is reflected and only 57.1% enters.
Clinical consequence: bone reflects almost everything, which is why an acoustic shadow forms behind it, and gel is needed because air and skin differ even more. Without gel the beam never enters.
Between soft tissues the impedances are close, so only a few percent reflects, and that is what makes organs visible.
The intensity reflection coefficient is the square of the impedance difference over the sum: ((7.752 minus 1.617) over (7.752 plus 1.617)) squared, which is 0.429, so 42.9% is reflected and only 57.1% enters.
Clinical consequence: bone reflects almost everything, which is why an acoustic shadow forms behind it, and gel is needed because air and skin differ even more. Without gel the beam never enters.
Between soft tissues the impedances are close, so only a few percent reflects, and that is what makes organs visible.
Have students explain gel as impedance matching, not lubrication.
Scene 3: X-rays and isotopes: one decay law
For X-rays, I equals I0 times e to the minus mu x, and the half-value layer is ln2 over mu. With HVL of 3 cm, mu is 0.231 per cm. Nine centimetres is three half-value layers, so the transmitted fraction is 0.5 cubed, which is 12.5%, matching e to the minus 2.079.
Radioisotopes use the same law on time. Tc-99m has a 6 hour half-life, so 24 hours is four half-lives and activity falls from 800 to 50 MBq. F-18 at 110 minutes falls to 92.5 MBq after 220 minutes.
Inside a body add biological clearance: one over effective half-life equals one over physical plus one over biological, so 6 hours and 12 hours give 4 hours. Clearance and decay are parallel exits, so the effective half-life is the shorter.
Radiotherapy uses fractionation: 2 Gy per fraction over 30 fractions is 60 Gy, and splitting the dose lets normal tissue repair.
Radioisotopes use the same law on time. Tc-99m has a 6 hour half-life, so 24 hours is four half-lives and activity falls from 800 to 50 MBq. F-18 at 110 minutes falls to 92.5 MBq after 220 minutes.
Inside a body add biological clearance: one over effective half-life equals one over physical plus one over biological, so 6 hours and 12 hours give 4 hours. Clearance and decay are parallel exits, so the effective half-life is the shorter.
Radiotherapy uses fractionation: 2 Gy per fraction over 30 fractions is 60 Gy, and splitting the dose lets normal tissue repair.
Warn that half-lives never add, and that unit prefixes must be unified first.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Calculation | Wavelength and echo time | lambda = v over f; t = 2d over v | 0.308 mm, 77.9 microseconds |
| Calculation | Reflection coefficient | R is the square of the impedance ratio | 42.9% |
| Calculation | X-ray transmission | I over I0 equals 0.5 to the power of thickness over HVL | 12.5% |
| Calculation | Activity and effective half-life | A equals A0 times 0.5 to the power of t over half-life | 50 MBq, 4 hours |
Sanity check (teach this): The echo time must be under a millisecond: 77.9 microseconds is plausible, while 10 ms means a missed factor of two.
A reflection coefficient must lie between 0 and 1: 0.429 is fine, and a negative value means the ratio is upside down.
A transmitted fraction of 12.5% agrees with three half-value layers; treating mu times x as a percentage is wrong.
Activity must fall, never rise: 50 MBq is below 800 MBq.
The effective half-life, 4 hours, must be shorter than the physical half-life, 6 hours.
A reflection coefficient must lie between 0 and 1: 0.429 is fine, and a negative value means the ratio is upside down.
A transmitted fraction of 12.5% agrees with three half-value layers; treating mu times x as a percentage is wrong.
Activity must fall, never rise: 50 MBq is below 800 MBq.
The effective half-life, 4 hours, must be shorter than the physical half-life, 6 hours.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Echo time at 60 mm depth | 77.9 microseconds | Use 2d over v |
| Percentage reflected at a bone interface | 42.9% | Find Z first |
| Transmission through 9 cm at HVL 3 cm | 12.5% | Three half-value layers |
| Activity of Tc-99m after 24 hours | 50 MBq | Four half-lives, so one sixteenth |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Using a one-way path | Answers 39.0 microseconds | An echo travels out and back, so 2d over v. |
| Forgetting to square the amplitude ratio | Answers 65.5% | Intensity is amplitude squared. |
| Treating mu x as a percentage | Answers 208% | Mu x is an exponent, not a percentage. |
| Adding half-lives | Answers 18 hours | Half-lives combine as reciprocals, never by addition. |
07 Homework (paper, section and question numbers)
- Secondary Physics chapter index — Medical physics unit, all sections
- Radioactivity chapter — Compare decay with attenuation
- Lesson-prep hub — Compare elective chapters
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| Wavelength of 5 MHz ultrasound in tissue | 0.308 mm | Wrong means redo v = f times lambda |
| Transmission through 9 cm at HVL 3 cm | 12.5% | Wrong means return to act two |
| The effective half-life formula | One over T equals one over physical plus one over biological | Two wrong means repeat act three |
09 Tutor FAQ
Why is ultrasound not used for lungs?
Air and tissue differ so much in impedance that almost everything reflects, so the beam cannot enter; lungs are imaged with X-ray and CT.
Why does mu change with photon energy?
Higher energy means less photoelectric absorption, so mu falls and a thicker layer transmits more.
Can the effective half-life exceed the physical one?
No. Biological clearance only adds an exit route, so the effective half-life is shorter than either component.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.
