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How do SI prefixes and order-of-magnitude reasoning let us handle quantities spanning more than forty powers of ten?

Use SI prefixes from pico to tera, convert between prefixed units consistently, and make order-of-magnitude estimates to check whether a numerical answer is physically reasonable

A focused answer to the H2 Physics Measurement learning outcome on prefixes and estimation. The common SI prefixes, how to convert safely between them, and how order-of-magnitude estimates catch unreasonable answers.

Generated by Claude Opus 4.87 min answer

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

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
  4. Try this

What this dot point is asking

SEAB wants you to use SI prefixes fluently, to convert between prefixed units without arithmetic slips, and to make order-of-magnitude estimates that sanity-check a calculated answer. Estimation appears in Paper 1 multiple choice and as a discipline that protects you from absurd answers in every structured question.

The answer

The common SI prefixes

Each prefix is a power-of-ten multiplier attached to a unit.

Prefix Symbol Factor
tera T 101210^{12}
giga G 10910^{9}
mega M 10610^{6}
kilo k 10310^{3}
centi c 10−210^{-2}
milli m 10−310^{-3}
micro μ\mu 10−610^{-6}
nano n 10−910^{-9}
pico p 10−1210^{-12}

Reading a prefixed quantity means replacing the prefix by its factor: 5 nm=5×10−9 m5\ \text{nm} = 5 \times 10^{-9}\ \text{m}.

Converting between prefixed units

Convert to base units first, do the arithmetic, then re-prefix if needed. This avoids the classic error of mishandling squared or cubed prefixes.

For powers, the prefix factor is also raised to that power:

1 cm3=(10−2 m)3=10−6 m31\ \text{cm}^3 = (10^{-2}\ \text{m})^3 = 10^{-6}\ \text{m}^3

1 km2=(103 m)2=106 m21\ \text{km}^2 = (10^{3}\ \text{m})^2 = 10^{6}\ \text{m}^2

This is the single most common conversion slip in volume and area problems.

Order-of-magnitude estimates

An order of magnitude is the nearest power of ten to a quantity. To estimate:

  1. Round each input to one significant figure (or the nearest power of ten).
  2. Combine the powers of ten using index laws.
  3. Quote the answer as a single power of ten.

Estimation tells you whether a precise answer is plausible. If a current calculation yields 109 A10^{9}\ \text{A} for a torch bulb, the order of magnitude alone tells you something is wrong.

Standard form discipline

Always express very large or very small results in standard form a×10na \times 10^{n} with 1≤a<101 \le a < 10. This keeps significant figures explicit and makes order-of-magnitude comparison immediate.

Examples in context

Example 1. Atomic spacing. The spacing of atoms in a crystal is about 0.3 nm=3×10−10 m0.3\ \text{nm} = 3 \times 10^{-10}\ \text{m}. A 1 mm1\ \text{mm} thick foil therefore stacks about 10−33×10−10≈3×106\dfrac{10^{-3}}{3 \times 10^{-10}} \approx 3 \times 10^{6} atoms across its thickness, an order of magnitude of 10610^{6} to 10710^{7} atoms.

Example 2. Power station output. A power station rated at 1.2 GW1.2\ \text{GW} delivers 1.2×109 W1.2 \times 10^{9}\ \text{W}. Over one day (8.64×104 s8.64 \times 10^{4}\ \text{s}) it produces 1.2×109×8.64×104≈1.0×1014 J1.2 \times 10^{9} \times 8.64 \times 10^{4} \approx 1.0 \times 10^{14}\ \text{J}, an order of magnitude of 1014 J10^{14}\ \text{J}. Checking the order of magnitude guards against a stray prefix error.

Try this

Q1. Convert 0.045 GHz0.045\ \text{GHz} to hertz in standard form. [1 mark]

  • Cue. 0.045×109=4.5×107 Hz0.045 \times 10^{9} = 4.5 \times 10^{7}\ \text{Hz}.

Q2. A cube has sides of 2.0 cm2.0\ \text{cm}. Find its volume in cubic metres. [2 marks]

  • Cue. V=(2.0×10−2)3=8.0×10−6 m3V = (2.0 \times 10^{-2})^3 = 8.0 \times 10^{-6}\ \text{m}^3.

Q3. Estimate the order of magnitude of the number of heartbeats in an average human lifetime, stating your assumptions. [3 marks]

  • Cue. About 7070 beats per minute over 8080 years: 70×60×24×365×80≈3×10970 \times 60 \times 24 \times 365 \times 80 \approx 3 \times 10^{9}, order of magnitude 10910^{9}.

Exam-style practice questions

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

Original3 marksA capacitor is labelled 470 μF470\ \mu\text{F} and is charged to 12 V12\ \text{V}. (a) Convert the capacitance to farads in standard form. (b) Estimate the order of magnitude of the energy stored, using E=12CV2E = \tfrac{1}{2}CV^2.
Show worked answer →

(a) The prefix μ\mu means 10−610^{-6}, so 470 μF=470×10−6 F=4.70×10−4 F470\ \mu\text{F} = 470 \times 10^{-6}\ \text{F} = 4.70 \times 10^{-4}\ \text{F}.

(b) E=12(4.70×10−4)(12)2=12(4.70×10−4)(144)≈3.4×10−2 JE = \tfrac{1}{2}(4.70 \times 10^{-4})(12)^2 = \tfrac{1}{2}(4.70 \times 10^{-4})(144) \approx 3.4 \times 10^{-2}\ \text{J}.

To one order of magnitude this is about 10−2 J10^{-2}\ \text{J}, a few hundredths of a joule.

Markers reward the correct prefix conversion to standard form, correct substitution into the energy formula, and a final order-of-magnitude statement (10−2 J10^{-2}\ \text{J}).

Original2 marksEstimate the order of magnitude of the number of breaths a typical person takes in a year. State the assumptions you make.
Show worked answer →

Assume about 1515 breaths per minute.

Per year: 15×60×24×365≈15×5.26×105≈7.9×10615 \times 60 \times 24 \times 365 \approx 15 \times 5.26 \times 10^5 \approx 7.9 \times 10^6 breaths.

To one order of magnitude this is about 10710^7 breaths per year.

Markers reward stating a sensible breathing rate, the chain of conversions (per minute to per year), and a final order-of-magnitude figure. Any answer of 10610^6 to 10710^7 with clear assumptions is creditable.

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