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3.3.2Stoichiometry

Ultrasound — Cambridge IGCSE Chemistry 0620 Revision Notes & Topical Questions

5 min read 5 parts Lesson 6 of 28 Unit quiz

📚 Cambridge IGCSE Chemistry 0620 Hub / Topical Revision
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Curriculum Equation Key Wave Speed & Doppler Effect FormulasFast Exam Solution
v = f × λ | f_obs = f × (v ± v_o) / (v ∓ v_s) | T = 1 / f
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Part 1 of 5Ultrasound Basics

3.3.2 Ultrasound
Definition
  • Ultrasound is sound with frequencies above the range of human hearing, above 20 000 Hz.
Transducer
  • A transducer both emits ultrasound pulses and receives their echoes, converting the echoes into electrical signals.
Partial reflection
  • At each boundary between different materials some ultrasound is reflected while the rest is transmitted onward.
Speed in media
  • Ultrasound travels at different speeds through different materials, so the speed must be known for accurate distances.

Part 2 of 5Building an Ultrasound Image

1
The transducer emits a short pulse of ultrasound into the body or the material being tested.
2
Echoes from boundaries between tissues return to the transducer at different times.
3
The scanner uses the speed and the time of each echo to calculate the distance to each boundary.
4
Measurements taken across an area are combined to build a two-dimensional image.

Part 3 of 5Applications

Applications
Foetal imaging
  • Ultrasound is non-invasive and considered harmless, so it is widely used to image a developing foetus.
Organ imaging
  • It produces images of soft organs, but bone blocks the beam, so structures behind bone cannot be imaged.
Kidney stones
  • Focused ultrasound can break up kidney stones into smaller pieces that pass out of the body.
Industrial testing
  • Ultrasound finds cracks inside metal blocks by detecting echoes that return earlier than the far face.

Part 4 of 5Worked Example

Question

  • Ultrasound travels at 6000 m/s in a steel bolt.
  • An echo from an internal crack is detected 0.000040 s after the pulse is emitted.
  • Calculate the distance from the head of the bolt to the crack, in centimetres.

  1. Total distance = 6000 × 0.000040 = 0.24 m
  2. The pulse travels to the crack and back, so distance to crack = 0.24 ÷ 2 = 0.12 m
  3. Convert to centimetres0.12 × 100 = 12 cm

Answer The crack is 12 cm from the head of the bolt.

Part 5 of 5Exam Checklist & Tips

3.3.2 – Key Points for the Exam

  • Ultrasound Sound with a frequency above 20 000 Hz.
  • Transducer A device that converts electrical signals into ultrasound and back again.
  • Echo A reflected pulse of ultrasound returned from a boundary.
  • Boundary A surface between two materials at which part of the ultrasound is reflected.
  • Non-invasive A technique that does not need to break the skin or enter the body.

Common Mistakes

  • Forgetting to halve the distance, since the pulse travels to the boundary and back.
  • Confusing ultrasound with infrasound, which has frequencies below 20 Hz.
  • Saying ultrasound can image organs behind bone, when bone blocks the beam.
  • Using the same speed of sound for all materials when calculating distances.

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