SI Prefixes and Unit Conversions Revision Notes | A Level AQA Physics

1. Why Prefixes and Conversions Matter in Physics

The diameter of a proton is roughly 1.7 × 10⁻¹⁵ m (about 1.7 fm). The distance from Earth to the nearest star is about 40 000 000 000 000 000 m. Writing either of those numbers in full in a calculation is a recipe for errors. SI prefixes and standard form exist to solve exactly that problem.

At A Level, you will encounter quantities spanning more than 30 orders of magnitude — from femtometre-scale nuclear radii to terawatt-scale power outputs. Fluency with prefixes is not a preliminary skill; it is a core tool that appears in every topic from mechanics to particle physics.

2. The SI Prefix Table

Each prefix is a shorthand for a power of ten. The table below covers every prefix required by the AQA specification. Notice that the prefixes above the base unit increase by factors of 1000 (kilo, mega, giga, tera), while those below decrease by factors of 1000 (milli, micro, nano, pico, femto) — with centi (10⁻²) as the one exception you need to know.

A reliable memory anchor: the prefixes above the base unit go up in steps of 3 (kilo = 3, mega = 6, giga = 9, tera = 12), and those below go down in steps of 3 (milli = −3, micro = −6, nano = −9, pico = −12, femto = −15). Centi (−2) is the only one that breaks the pattern, and it appears almost exclusively with metres (cm) and occasionally with litres.

3. Converting Between Prefixed Units and SI Base Units

The core conversion rule is straightforward: to convert a prefixed quantity into its SI base unit, multiply by the prefix's power of ten. To go the other way — from base unit to a prefixed unit — divide by that power of ten. The challenge is keeping track of which direction you are travelling.

A useful mental model: think of the prefix as a label on a box. "450 nm" means the box is labelled "nano-metres", and inside the box there are 450 of them. To find the total in plain metres, you multiply: $450 \times 10^{-9} = 4.50 \times 10^{-7}$ m. The label tells you the conversion factor.

Worked example: Converting Prefixed Units to SI Base Units

A capacitor has a capacitance of 47 μF. Express this in farads (F) in standard form.

1. μ = micro = $10^{-6}$ — Read the prefix symbol from the table. Micro always means $10^{-6}$. [B1]
2. $47 \text{ μF} = 47 \times 10^{-6} \text{ F}$ — Multiply the numerical value by the prefix multiplier. [M1]
3. $47 \times 10^{-6} = 4.7 \times 10^{1} \times 10^{-6} = 4.7 \times 10^{-5} \text{ F}$ — Standard form requires the leading number to be between 1 and 10. Add the exponents: $1 + (-6) = -5$. [A1]
4. $4.7 \times 10^{-5}$ is a very small number — correct for a capacitance in farads. — Capacitances in everyday circuits are always much less than 1 F, so a negative power of ten is expected here.

Answer: $47 \text{ μF} = 4.7 \times 10^{-5} \text{ F}$

Now consider the reverse: converting from a base unit into a prefixed unit. Suppose a calculation gives you a wavelength of $6.5 \times 10^{-7}$ m and you want to express it in nanometres. Divide by $10^{-9}$ (which is the same as multiplying by $10^{9}$): $6.5 \times 10^{-7} \times 10^{9} = 6.5 \times 10^{2} = 650$ nm. Visible light sits between roughly 400 nm and 700 nm, so 650 nm is a physically sensible answer for red light.

4. Multi-Step Prefix Conversions and Standard Form

Examination questions often require you to convert between two prefixed units rather than between a prefix and the base unit. The safest strategy is always to convert both quantities to SI base units first, carry out the calculation, and then convert the result to whatever unit the question requests. Trying to shortcut between two prefixes in one step is where sign errors creep in.

Standard form is your consistency tool throughout. When every number in a calculation is written as $a \times 10^{n}$ (where $1 \leq a < 10$), you can handle the powers of ten separately from the numerical coefficients, which dramatically reduces arithmetic errors.

Worked example: Converting Between Two Prefixed Units

A signal has a frequency of 2.45 GHz. Express this frequency in MHz.

1. $2.45 \text{ GHz} = 2.45 \times 10^{9} \text{ Hz}$ — Giga = $10^{9}$, so multiply by $10^{9}$. [M1]
2. MHz: mega = $10^{6}$, so 1 MHz = $10^{6}$ Hz — To express in MHz, divide the value in Hz by $10^{6}$.
3. $\frac{2.45 \times 10^{9}}{10^{6}} = 2.45 \times 10^{9-6} = 2.45 \times 10^{3} \text{ MHz}$ — Subtract the exponents when dividing powers of ten: $9 - 6 = 3$. [A1]
4. $2.45 \text{ GHz} = 2.45 \times 10^{3} \text{ MHz} = 2450 \text{ MHz}$ — Both forms are acceptable; standard form is preferred in examination answers unless the question specifies otherwise.

Answer: $2.45 \text{ GHz} = 2450 \text{ MHz}$ (or $2.45 \times 10^{3}$ MHz)

5. Converting Between Non-SI Units: Joules and Electronvolts

Not all energy values in physics are conveniently expressed in joules. When you are working at the atomic or subatomic scale — ionisation energies, photon energies, particle kinetic energies — the joule produces numbers like $3.2 \times 10^{-19}$ J, which are cumbersome to compare. The electronvolt (eV) was introduced specifically for this context.

The conversion factor $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$ is given in the AQA data booklet, so you do not need to memorise the numerical value — but you must know which direction to apply it. To convert eV to J, multiply by $1.6 \times 10^{-19}$. To convert J to eV, divide by $1.6 \times 10^{-19}$.

Worked example: Converting Between Joules and Electronvolts

The ionisation energy of a hydrogen atom is $2.18 \times 10^{-18}$ J. Express this in electronvolts.

1. $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$ — This is given in the AQA data booklet. To go from J to eV, divide by $1.6 \times 10^{-19}$. [B1]
2. $E = \frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}} \text{ eV}$ — Dividing joules by (joules per electronvolt) gives electronvolts. [M1]
3. $\frac{2.18}{1.6} = 1.3625$ — Carry out the numerical division: $2.18 \div 1.6 = 1.3625$.
4. $\frac{10^{-18}}{10^{-19}} = 10^{-18-(-19)} = 10^{-18+19} = 10^{1}$ — Subtracting a negative exponent: $-18 - (-19) = -18 + 19 = 1$.
5. $E = 1.3625 \times 10^{1} \approx 13.6 \text{ eV}$ — The ionisation energy of hydrogen is the well-known value of 13.6 eV — a useful cross-check that the calculation is correct. [A1]

Answer: $E \approx 13.6 \text{ eV}$

You will also encounter keV ($10^{3}$ eV), MeV ($10^{6}$ eV), and GeV ($10^{9}$ eV) in particle physics contexts. These combine the eV conversion with the prefix multipliers you already know — so the skills from Section 3 and Section 5 work together directly.

6. Converting Between Non-SI Units: Joules and Kilowatt-Hours

At the other end of the energy scale, the joule becomes inconveniently small. A domestic household uses roughly 3 600 000 000 J of electrical energy per year. Energy suppliers use the kilowatt-hour (kW h) instead, because it produces manageable numbers for billing and engineering contexts.

The derivation of the conversion factor is worth understanding, not just memorising. Power in watts multiplied by time in seconds gives energy in joules: $E = P \times t$. One kilowatt is $1 \times 10^{3}$ W, and one hour is $60 \times 60 = 3600$ s. Multiplying these: $1 \times 10^{3} \times 3600 = 3.6 \times 10^{6}$ J. That derivation is a one-mark answer in its own right if an examiner asks you to show where the conversion factor comes from.

Worked example: Converting Between kW h and Joules

An electric shower uses 2.5 kW h of energy during a week. Calculate this energy in joules.

1. $1 \text{ kW h} = 3.6 \times 10^{6} \text{ J}$ — Derived from $1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^{6}$ J. [B1]
2. $E = 2.5 \times 3.6 \times 10^{6} \text{ J}$ — To convert kW h to J, multiply by $3.6 \times 10^{6}$. [M1]
3. $2.5 \times 3.6 = 9.0$ — Numerical multiplication: $2.5 \times 3.6 = 2.5 \times 3 + 2.5 \times 0.6 = 7.5 + 1.5 = 9.0$.
4. $E = 9.0 \times 10^{6} \text{ J}$ — The answer is already in standard form. Always include units in the final answer. [A1]

Answer: $E = 9.0 \times 10^{6}$ J

For the reverse conversion — from joules to kW h — divide by $3.6 \times 10^{6}$. For example, a battery storing $7.2 \times 10^{7}$ J holds $\frac{7.2 \times 10^{7}}{3.6 \times 10^{6}} = \frac{7.2}{3.6} \times 10^{7-6} = 2.0 \times 10^{1} = 20$ kW h. Recognising that 20 kW h is a physically plausible battery capacity for an electric vehicle is the kind of order-of-magnitude check that distinguishes confident exam answers from uncertain ones.

7. Exam Technique and Common Errors

Unit conversion errors are among the most common sources of lost marks at A Level — not because students do not know the prefixes, but because they apply them in the wrong direction under time pressure. The strategies below address the specific traps that appear repeatedly in AQA mark schemes.

A final point on significant figures: your converted answer should carry the same number of significant figures as the least precise value in the original data. If a question gives you 2.5 kW h (2 significant figures), your answer in joules should also be given to 2 significant figures: $9.0 \times 10^{6}$ J, not $9.00 \times 10^{6}$ J.

Mastery of SI prefixes, standard form, and the J ↔ eV and J ↔ kW h conversions underpins accurate quantitative work across every topic in AQA A Level Physics — from calculating photon energies in quantum physics to comparing power outputs in nuclear and electrical contexts.