3.1.1 - Use of SI Units and Their Prefixes

3.1.1 - Use of SI Units and Their Prefixes

A physical answer needs a value and a unit. Learn how base units combine, how prefixes change a number's scale, and how to convert energy units without changing the physical quantity.

The SI base units

The SI system gives scientists a shared set of units so that measurements can be compared directly. A quantity is what is measured; its unit is the agreed amount used for comparison. For example, a length of 2.0 m2.0\,\mathrm m contains the value 2.0 and the unit metre.

Base quantitySI base unitUnit symbol
Masskilogramkg
Lengthmetrem
Timeseconds
Electric currentampereA
Thermodynamic temperaturekelvinK
Amount of substancemolemol

These are the six base quantities required here. SI also contains luminous intensity and the candela, but that base unit is excluded from this specification. You need the quantity–unit pairs, not the formal definitions used to realise the units.

Unit symbols are case-sensitive: A\mathrm A means ampere, while K\mathrm K means kelvin. The symbol for a quantity can differ from its unit: current II is measured in amperes. Kelvin is written without a degree sign.

The kilogram is unusual because its name already contains a prefix. Further mass prefixes attach to gram, so 1 mg=10−3 g=10−6 kg1\,\mathrm{mg}=10^{-3}\,\mathrm g=10^{-6}\,\mathrm{kg}. Using 10−3 kg10^{-3}\,\mathrm{kg} for one milligram would make the mass a thousand times too large.

Building derived units from equations

A derived unit is built by multiplying or dividing base units in the same way that an equation combines its quantities. You can work out the units even before you have numerical values.

Velocity is displacement divided by time, so its unit is m s−1\mathrm{m\,s^{-1}}. Acceleration is change in velocity divided by time, giving m s−2\mathrm{m\,s^{-2}}. A negative exponent means division: s−2=1/s2\mathrm{s^{-2}}=1/\mathrm{s^2}.

Force and energy

For a constant mass, the resultant force is F=maF=ma. Substituting units gives

[F]=kg×m s−2=kg m s−2=N.[F]=\mathrm{kg}\times\mathrm{m\,s^{-2}} =\mathrm{kg\,m\,s^{-2}}=\mathrm N.

Square brackets here mean “the unit of”. The newton is a special name for this combination, not an additional base unit.

When a constant force acts along a displacement, the work done is W=FsW=Fs. Consequently

[W]=N×m=kg m2 s−2=J.[W]=\mathrm N\times\mathrm m =\mathrm{kg\,m^2\,s^{-2}}=\mathrm J.

Potential difference

Potential difference is energy transferred per unit charge: V=E/QV=E/Q. Charge is current multiplied by time, Q=ItQ=It, so [Q]=A s[Q]=\mathrm{A\,s}. Hence

[V]=kg m2 s−2A s=kg m2 s−3 A−1.[V]=\frac{\mathrm{kg\,m^2\,s^{-2}}}{\mathrm{A\,s}} =\mathrm{kg\,m^2\,s^{-3}\,A^{-1}}.

Divide by s\mathrm s by subtracting one from its exponent, so −2−1=−3-2-1=-3. This method uses the defining equation and ordinary unit algebra; formal dimensional analysis is not required.

Prefixes, standard form and powered units

A prefix multiplies a unit by a power of ten. Replacing the prefix by that multiplier is safer than guessing which way to move a decimal point.

PrefixSymbolMultiplier
teraT101210^{12}
gigaG10910^9
megaM10610^6
kilok10310^3
centic10−210^{-2}
millim10−310^{-3}
microμ\mu10−610^{-6}
nanon10−910^{-9}
picop10−1210^{-12}
femtof10−1510^{-15}

These are the required prefixes, not the complete SI list. Upper- and lower-case symbols matter: MHz\mathrm{MHz} and mHz\mathrm{mHz} differ by a factor of 10910^9.

Standard form writes a non-zero magnitude as a×10na\times10^n, with 1≤a<101\leq a<10 and integer nn. For example,

73 μs=73×10−6 s=7.3×10−5 s.73\,\mu\mathrm s=73\times10^{-6}\,\mathrm s =7.3\times10^{-5}\,\mathrm s.

To change between prefixes, pass through the unprefixed unit. Since 1 nF=10−9 F1\,\mathrm{nF}=10^{-9}\,\mathrm F and 1 pF=10−12 F1\,\mathrm{pF}=10^{-12}\,\mathrm F, one nanofarad is a thousand picofarads. The smaller unit needs the larger number.

Squaring or cubing a conversion factor

A centimetre is 10−2 m10^{-2}\,\mathrm m. A square centimetre is a square with both sides that length:

1 cm2=(10−2 m)2=10−4 m2,1\,\mathrm{cm^2}=(10^{-2}\,\mathrm m)^2=10^{-4}\,\mathrm{m^2}, 1 cm3=10−6 m3.1\,\mathrm{cm^3}=10^{-6}\,\mathrm{m^3}.

Thus 2.4 cm2=2.4×10−4 m22.4\,\mathrm{cm^2}=2.4\times10^{-4}\,\mathrm{m^2}: the prefix factor is squared, not the stated area 2.4. For a compound unit, convert each factor. For example,

1 g cm−3=10−3 kg10−6 m3=103 kg m−3.1\,\mathrm{g\,cm^{-3}} =\frac{10^{-3}\,\mathrm{kg}}{10^{-6}\,\mathrm{m^3}} =10^3\,\mathrm{kg\,m^{-3}}.

Converting energy units consistently

A conversion changes the numerical value and unit together, while leaving the energy itself unchanged. Multiplying by a ratio equal to one shows why the operation is valid and lets unwanted units cancel.

Electronvolts and joules

An electronvolt is the energy gained by an electron when it accelerates through a potential difference of one volt. Using the rounded conversion 1 eV=1.60×10−19 J1\,\mathrm{eV}=1.60\times10^{-19}\,\mathrm J,

76 MeV=76×106 eV×1.60×10−19 J1 eV=1.216×10−11 J.76\,\mathrm{MeV} =76\times10^6\,\mathrm{eV} \times\frac{1.60\times10^{-19}\,\mathrm J}{1\,\mathrm{eV}} =1.216\times10^{-11}\,\mathrm J.

The supplied energy has two significant figures, so a suitable final answer is 1.2×10−11 J1.2\times10^{-11}\,\mathrm J. Keep extra digits until the final step. To reverse the conversion, divide the energy in joules by the number of joules in one electronvolt. A prefix such as mega must still be handled separately.

Kilowatt-hours and joules

Power is energy transferred per second: 1 W=1 J s−11\,\mathrm W=1\,\mathrm{J\,s^{-1}}. A kilowatt-hour is therefore an energy, not a power:

1 kWh=(1000 J s−1)(3600 s)=3.6×106 J.1\,\mathrm{kWh} =(1000\,\mathrm{J\,s^{-1}})(3600\,\mathrm s) =3.6\times10^6\,\mathrm J.

For a 2.5 kW2.5\,\mathrm{kW} heater used for 45 minutes, E=PtE=Pt can be used in two consistent ways:

  • Watts and seconds: E=2500×2700=6.75×106 JE=2500\times2700=6.75\times10^6\,\mathrm J.
  • Kilowatts and hours: E=2.5×0.75=1.875 kWhE=2.5\times0.75=1.875\,\mathrm{kWh}.

Converting the second result using 3.6×106 J kWh−13.6\times10^6\,\mathrm{J\,kWh^{-1}} gives the first. To two significant figures these are 6.8×106 J6.8\times10^6\,\mathrm J and 1.9 kWh1.9\,\mathrm{kWh}. They are the same energy; the unrounded values show the equivalence more clearly.

Use coherent SI units with SI constants unless the equation and chosen units support another consistent route. You can correctly use kilowatts and hours to find kilowatt-hours; mixing kilowatts with seconds and labelling the answer joules introduces a factor-of-1000 error.