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1.3 Units

1.3 Units (ESAE)

Imagine you had to make curtains and needed to buy fabric. The shop assistant would need to know how much fabric you needed. Telling her you need fabric \(\text{2}\) wide and \(\text{6}\) long would be insufficient — you have to specify the unit (i.e. \(\text{2}\) metres wide and \(\text{6}\) metres long). Without the unit the information is incomplete and the shop assistant would have to guess. If you were making curtains for a doll's house the dimensions might be \(\text{2}\) centimetres wide and \(\text{6}\) centimetres long!

It is not just lengths that have units, all physical quantities have units (e.g. time, temperature, distance, etc.).

Physical Quantity

A physical quantity is anything that you can measure. For example, length, temperature, distance and time are physical quantities.

There are many different systems of units. The main systems of units are:

  • SI units

  • c.g.s units

  • Imperial units

  • Natural units

SI units (ESAF)

We will be using the SI units in this course. SI units are the internationally agreed upon units.

SI Units

The name SI units comes from the French Système International d'Unités, which means international system of units.

There are seven base SI units. These are listed in Table 1.1. All physical quantities have units which can be built from these seven base units. So, it is possible to create a different set of units by defining a different set of base units.

These seven units are called base units because none of them can be expressed as combinations of the other six. These base units are like the \(\text{26}\) letters of the alphabet for English. Many different words can be formed by using these letters.

Base quantity












electric current






amount of substance



luminous intensity



Table 1.1: SI base units

The other systems of units (ESAG)

The SI Units are not the only units available, but they are most widely used. In Science there are three other sets of units that can also be used. These are mentioned here for interest only.

  • c.g.s. units

    In the c.g.s. system, the metre is replaced by the centimetre and the kilogram is replaced by the gram. This is a simple change but it means that all units derived from these two are changed. For example, the units of force and work are different. These units are used most often in astrophysics and atomic physics.

  • Imperial units

    Imperial units arose when kings and queens decided the measures that were to be used in the land. All the imperial base units, except for the measure of time, are different to those of SI units. This is the unit system you are most likely to encounter if SI units are not used. Examples of imperial units are pounds, miles, gallons and yards. These units are used by the Americans and British. As you can imagine, having different units in use from place to place makes scientific communication very difficult. This was the motivation for adopting a set of internationally agreed upon units.

  • Natural units

    This is the most sophisticated choice of units. Here the most fundamental discovered quantities (such as the speed of light) are set equal to \(\text{1}\). The argument for this choice is that all other quantities should be built from these fundamental units. This system of units is used in high energy physics and quantum mechanics.

Combinations of SI base units (ESAH)

To make working with units easier, some combinations of the base units are given special names, but it is always correct to reduce everything to the base units. Table 1.2 lists some examples of combinations of SI base units that are assigned special names. Do not be concerned if the formulae look unfamiliar at this stage - we will deal with each in detail in the chapters ahead (as well as many others)!

It is very important that you are able to recognise the units correctly. For example, the newton (\(\text{N}\)) is another name for the kilogram metre per second squared (\(\text{kg·m·s$^{-2}$}\)), while the kilogram metre squared per second squared (\(\text{kg·m$^{2}$·s$^{-2}$}\)) is called the joule (\(\text{J}\)).



Unit expressed in base units

Name of combination




\(\text{N}\) (Newton)




\(\text{Hz}\) (Hertz)




\(\text{J}\) (Joule)

Table 1.2: Some examples of combinations of SI base units assigned special names

Prefixes of base units (ESAI)

Now that you know how to write numbers in scientific notation, another important aspect of units is the prefixes that are used with the units. In the case of units, the prefixes have a special use. The kilogram (\(\text{kg}\)) is a simple example. \(\text{1}\) \(\text{kg}\) is equal to \(\text{1 000}\) \(\text{g}\) or \(\text{1} \times \text{10}^{\text{3}}\) \(\text{g}\). Grouping the \(\text{10}^{\text{3}}\) and the \(\text{g}\) together we can replace the \(\text{10}^{\text{3}}\) with the prefix k (kilo). Therefore the k takes the place of the \(\text{10}^{\text{3}}\). The kilogram is unique in that it is the only SI base unit containing a prefix.

When writing combinations of base SI units, place a dot (·) between the units to indicate that different base units are used. For example, the symbol for metres per second is correctly written as \(\text{m·s$^{-1}$}\), and not as \(\text{ms$^{-1}$}\) or \(\text{m/s}\). Although the last two options will be accepted in tests and exams, we will only use the first one in this book.

In science, all the prefixes used with units are some power of \(\text{10}\). Table 1.3 lists some of these prefixes. You will not use most of these prefixes, but those prefixes listed in bold should be learnt. The case of the prefix symbol is very important. Where a letter features twice in the table, it is written in uppercase for exponents bigger than one and in lowercase for exponents less than one. For example M means mega (\(\text{10}^{\text{6}}\)) and m means milli (\(\text{10}^{-\text{3}}\)).



































































Table 1.3: Unit prefixes

There is no space and no dot between the prefix and the symbol for the unit.

Here are some examples of the use of prefixes:

  • \(\text{40 000}\) \(\text{m}\) can be written as \(\text{40}\) \(\text{km}\) (kilometre)

  • \(\text{0,001}\) \(\text{g}\) is the same as \(\text{10}^{-\text{3}}\) \(\text{g}\) and can be written as \(\text{1}\) \(\text{mg}\) (milligram)

  • \(\text{2,5} \times \text{10}^{\text{6}}\) \(\text{N}\) can be written as \(\text{2,5}\) \(\text{MN}\) (meganewton)

  • \(\text{250 000}\) \(\text{A}\) can be written as \(\text{250}\) \(\text{kA}\) (kiloampere) or \(\text{0,250}\) \(\text{MA}\) (megaampere)

  • \(\text{0,000000075}\) \(\text{s}\) can be written as \(\text{75}\) \(\text{ns}\) (nanoseconds)

  • \(\text{3} \times \text{10}^{-\text{7}}\) \(\text{mol}\) can be rewritten as \(\text{0,3} \times \text{10}^{-\text{6}}\) \(\text{mol}\), which is the same as \(\text{0,3} \times \text{10}^{-\text{6}}\) \(\text{μmol}\) (micromol)

Scientific notation

Textbook Exercise 1.1

Carry out the following calculations:

  1. \(\text{1,63} \times \text{10}^{\text{5}} + \text{4,32} \times \text{10}^{\text{6}} - \text{8,53} \times \text{10}^{\text{5}}\)

  2. \(\text{7,43} \times \text{10}^{\text{3}} \div \text{6,54} \times \text{10}^{\text{7}} \times \text{3,33} \times \text{10}^{\text{5}}\)

  3. \(\text{6,21434534} \times \text{10}^{-\text{5}} \times \text{3,2555} \times \text{10}^{-\text{3}} + \text{6,3} \times \text{10}^{-\text{4}}\)

Solution not available at present.

Write the following in scientific notation using Table 1.3 as a reference.

  1. \(\text{0,511}\) \(\text{MV}\)

  2. \(\text{10}\) \(\text{cℓ}\)

  3. \(\text{0,5}\) \(\text{μm}\)

  4. \(\text{250}\) \(\text{nm}\)

  5. \(\text{0,00035}\) \(\text{hg}\)

  1. \(\text{0,511} \times \text{10}^{\text{6}} =\) \(\text{5,11} \times \text{10}^{\text{5}}\) \(\text{V}\)
  2. \(\text{10} \times \text{10}^{-\text{2}} =\) \(\text{0,1}\) \(\text{L}\)
  3. \(\text{0,5} \times \text{10}^{-\text{6}} =\) \(\text{5} \times \text{10}^{-\text{7}}\) \(\text{m}\)
  4. \(\text{250} \times \text{10}^{-\text{9}} =\) \(\text{2,50} \times \text{10}^{-\text{7}}\) \(\text{m}\)
  5. \(\text{0,00035} \times \text{10}^{\text{2}} =\) \(\text{3,5} \times \text{10}^{-\text{2}}\) \(\text{g}\)

Write the following using the prefixes in Table 1.3.

  1. \(\text{1,602} \times \text{10}^{-\text{19}}\) \(\text{C}\)

  2. \(\text{1,992} \times \text{10}^{\text{6}}\) \(\text{J}\)

  3. \(\text{5,98} \times \text{10}^{\text{4}}\) \(\text{N}\)

  4. \(\text{25} \times \text{10}^{-\text{4}}\) \(\text{A}\)

  5. \(\text{0,0075} \times \text{10}^{\text{6}}\) \(\text{m}\)

Solution not available at present.

The importance of units (ESAJ)

Without units much of our work as scientists would be meaningless. We need to express our thoughts clearly and units give meaning to the numbers we measure and calculate. Depending on which units we use, the numbers are different. For example if you have \(\text{12}\) water, it means nothing. You could have \(\text{12}\) \(\text{mL}\) of water, \(\text{12}\) litres of water, or even \(\text{12}\) bottles of water. Units are an essential part of the language we use. Units must be specified when expressing physical quantities. Imagine that you are baking a cake, but the units, like grams and millilitres, for the flour, milk, sugar and baking powder are not specified!

Importance of units

Work in groups of \(\text{5}\) to discuss other possible situations where using the incorrect set of units can be to your disadvantage or even dangerous. Look for examples at home, at school, at a hospital, when travelling and in a shop.

The importance of units

Read the following extract from CNN News 30 September 1999 and answer the questions below.

NASA: Human error caused loss of Mars orbiter November 10, 1999

Failure to convert English measures to metric values caused the loss of the Mars Climate Orbiter, a spacecraft that smashed into the planet instead of reaching a safe orbit, a NASA investigation concluded Wednesday.

The Mars Climate Orbiter, a key craft in the space agency's exploration of the red planet, vanished on 23 September after a 10 month journey. It is believed that the craft came dangerously close to the atmosphere of Mars, where it presumably burned and broke into pieces.

An investigation board concluded that NASA engineers failed to convert English measures of rocket thrusts to newton, a metric system measuring rocket force. One English pound of force equals \(\text{4,45}\) newtons. A small difference between the two values caused the spacecraft to approach Mars at too low an altitude and the craft is thought to have smashed into the planet's atmosphere and was destroyed.

The spacecraft was to be a key part of the exploration of the planet. From its station about the red planet, the Mars Climate Orbiter was to relay signals from the Mars Polar Lander, which is scheduled to touch down on Mars next month.

“The root cause of the loss of the spacecraft was a failed translation of English units into metric units and a segment of ground-based, navigation-related mission software,” said Arthur Stephenson, chairman of the investigation board.


  1. Why did the Mars Climate Orbiter crash? Answer in your own words.

  2. How could this have been avoided?

  3. Why was the Mars Orbiter sent to Mars?

  4. Do you think space exploration is important? Explain your answer.

How to change units (ESAK)

It is very important that you are aware that different systems of units exist. Furthermore, you must be able to convert between units. Being able to change between units (for example, converting from millimetres to metres) is a useful skill in Science.

The following conversion diagrams will help you change from one unit to another.


Figure 1.1: The distance conversion table

If you want to change millimetre to metre, you divide by \(\text{1 000}\) (follow the arrow from \(\text{mm}\) to \(\text{m}\)); or if you want to change kilometre to millimetre, you multiply by \(\text{1 000}\) × \(\text{1 000}\).

The same method can be used to change millilitre to litre or kilolitre. Use Figure 1.2 to change volumes:


Figure 1.2: The volume conversion table

Worked example 3: Conversion 1

Express \(\text{3 800}\) \(\text{mm}\) in metres.

Use the conversion table

Use Figure 1.1. Millimetre is on the left and metre in the middle.

Decide which direction you are moving

You need to go from \(\text{mm}\) to \(\text{m}\), so you are moving from left to right.

Write the answer

\[\text{3 800}\text{ mm} \div \text{1 000} = \text{3,8}\text{ m}\]

Worked example 4: Conversion 2

Convert \(\text{4,56}\) \(\text{kg}\) to \(\text{g}\).

Find the two units on the conversion diagram.

Use Figure 1.1. Kilogram is the same as kilometre and gram is the same as metre.

Decide whether you are moving to the left or to the right.

You need to go from \(\text{kg}\) to \(\text{g}\), so it is from right to left.

Read from the diagram what you must do and find the answer.

\[\text{4,56}\text{ kg} \times \text{1 000} = \text{4 560}\text{ g}\]

Two other useful conversions

Very often in science you need to convert speed and temperature. The following two rules will help you do this:

  1. Converting speed

    When converting \(\text{km·h$^{-1}$}\) to \(\text{m·s$^{-1}$}\) you multiply by \(\text{1 000}\) and divide by \(\text{3 600}\) \(\left(\frac{\text{1 000}\text{ m}}{\text{3 600}\text{ s}}\right)\). For example \(\text{72}\text{ km·h$^{-1}$} \div \text{3,6} = \text{20}\text{ m·s$^{-1}$}\).

    When converting \(\text{m·s$^{-1}$}\) to \(\text{km·h$^{-1}$}\), you multiply by \(\text{3 600}\) and divide by \(\text{1 000}\) \(\left(\frac{\text{3 600}\text{ s}}{\text{1 000}\text{ m}}\right)\). For example \(\text{30}\text{ m·s$^{-1}$} \times \text{3,6} = \text{108}\text{ km·h$^{-1}$}\)

  2. Converting temperature

    Converting between the Kelvin and Celsius temperature scales is simple. To convert from Celsius to Kelvin add \(\text{273}\). To convert from Kelvin to Celsius subtract \(\text{273}\). Representing the Kelvin temperature by \({T}_{K}\) and the Celcius temperature by \({T}_{℃}\):

    \[{T}_{K} = {T}_{℃} + 273\]

Changing the subject of a formula (ESAL)

Very often in science you will have to change the subject of a formula. We will look at two examples. (Do not worry if you do not yet know what the terms and symbols mean, these formulae will be covered later in the book.)

  1. Moles

    The equation to calculate moles from molar mass is: \(n=\dfrac{m}{M}\), where \(n\) is the number of moles, \(m\) is the mass and \(M\) is the molar mass. As it is written we can easily find the number of moles of a substance. But what if we have the number of moles and want to find the molar mass? We note that we can simply multiply both sides of the equation by the molar mass and then divide both sides by the number of moles.

    \begin{align*} n& = \frac{m}{M} \\ nM & = m \\ M & = \frac{m}{n} \end{align*}

    And if we wanted the mass we would use: \(m=nM\).

  2. Energy of a photon

    The equation for the energy of a photon is \(E=h\dfrac{c}{\lambda }\), where \(E\) is the energy, \(h\) is Planck's constant, \(c\) is the speed of light and \(\lambda\) is the wavelength. To get \(c\) we can do the following:

    \begin{align*} E & = h\frac{c}{\lambda } \\ E\lambda & = hc \\ c & = \frac{E\lambda }{h} \end{align*}

    Similarly we can find the wavelength we use: \(\lambda =\dfrac{hc}{E}\) and to find Planck's constant we use: \(h=\dfrac{E\lambda }{c}\).

Rate, proportion and ratios (ESAM)

In science we often want to know how a quantity relates to another quantity or how something changes over a period of time. To do this we need to know about rate, proportion and ratios.


The rate is always a change in a quantity per unit time. The average rate is the amount by which some quantity changes divided by the time interval during which the change takes place. So, for example, we can calculate the rate of change of velocity per unit time \(\left(\frac{\Delta \vec{v}}{\Delta{t}}\right)\) or the rate of change in concentration per unit time \(\left(\frac{\Delta c}{\Delta{t}}\right)\). Note that \(\Delta\) represents "the change in" a given quantity, so \(\Delta \vec{v}\) is the change in velocity, \(\Delta{c}\) is the change in concentration, and \(\Delta{t}\) is the size of the time interval during which the change takes place.

Ratios and fractions:

A fraction is a number which represents a part of a whole and is written as \(\dfrac{a}{b}\), where \(a\) is the numerator and \(b\) is the denominator. A ratio tells us the relative size of one quantity (e.g. the number of moles of reactants) compared to another quantity (e.g. the number of moles of product). \(\text{2}:\text{1}\) ; \(\text{4}:\text{3}\), etc. Ratios can also be written as fractions as percentages (fractions with a denominator of \(\text{100}\)).


Proportion is a way of describing relationships between values or between constants. We can say that \(x\) is directly proportional to \(y\) (\(x\propto y\)) or that \(a\) is inversely proportional to \(b\) (\(a\propto \frac{1}{b}\)). It is important to understand the difference between directly and inversely proportional.

  • Directly proportional

    Two values or constants are directly proportional when a change in one leads to the same change in the other. This is a more-more relationship. We can represent this as \(y\propto x\) or \(y=kx\) where \(k\) is the proportionality constant. We have to include \(k\) since we do not know by how much \(x\) changes when \(y\) changes. \(x\) could change by \(\text{2}\) for every change of \(\text{1}\) in \(y\). If we plot two directly proportional variables on a graph, then we get a straight line graph that goes through the origin \(\left(0;0\right)\):

  • Inversely proportional

    Two values or constants are inversely proportional when a change in one leads to the opposite change in the other. We can represent this as \(y=\frac{k}{x}\). This is a more-less relationship. If we plot two inversely proportional variables we get a curve that never cuts the axis:


Constants in equations (ESAN)

A constant in an equation always has the same value. For example the speed of light (\(c=\text{2,99} \times \text{10}^{\text{8}}\text{ m·s$^{-1}$}\), Planck's constant (\(h\)) and Avogadro's number (\({N}_{A}\)) are all examples of constants that are used in science. The following table lists all the contants you will encounter in this book.



Value and units

SI Units

Atomic mass unit


\(\text{1,67} \times \text{10}^{-\text{24}}\) \(\text{g}\)

\(\text{1,67} \times \text{10}^{-\text{27}}\) \(\text{kg}\)

Charge on an electron


\(-\text{1,6} \times \text{10}^{-\text{19}}\) \(\text{C}\)

\(-\text{1,6} \times \text{10}^{-\text{19}}\) \(\text{s·A}\)

Speed of sound (in air, at \(\text{25}\) \(\text{℃}\))

\(\text{344}\) \(\text{m·s$^{-1}$}\)

Speed of light


\(\text{3} \times \text{10}^{\text{8}}\) \(\text{m·s$^{-1}$}\)

Planck's constant


\(\text{6,626} \times \text{10}^{-\text{34}}\) \(\text{J}\)

\(\text{6,626} \times \text{10}^{-\text{34}}\) \(\text{kg·m$^{2}$·s$^{-1}$}\)

Avogadro's number


\(\text{6,022} \times \text{10}^{\text{23}}\)

Gravitational acceleration


\(\text{9,8}\) \(\text{m·s$^{-2}$}\)

Trigonometry (ESAO)

Trigonometry is the relationship between the angles and sides of right angled triangles. Trigonometrical relationships are ratios and therefore have no units. You should know the following trigonometric ratios:

  • Sine

    This is defined as \(\sin A = \frac{\text{opposite}}{\text{hypotenuse}}\)

  • Cosine

    This is defined as \(\cos A = \frac{\text{adjacent}}{\text{hypotenuse}}\)

  • Tangent

    This is defined as \(\tan A = \frac{\text{opposite}}{\text{adjacent}}\)

Using significant figures

Textbook Exercise 1.2

Write the following quantities in scientific notation:

  1. \(\text{10 130}\) \(\text{Pa}\) to \(\text{2}\) decimal places

  2. \(\text{978,15}\) \(\text{m·s$^{-2}$}\) to one decimal place

  3. \(\text{0,000001256}\) \(\text{A}\) to \(\text{3}\) decimal places

Solution not available at present.

For each of the following symbols, write out the unit in full and write what power of \(\text{10}\) it represents:

  1. \(\text{μg}\)

  2. \(\text{mg}\)

  3. \(\text{kg}\)

  4. \(\text{Mg}\)

Solution not available at present.

Write each of the following in scientific notation, correct to \(\text{2}\) decimal places:

  1. \(\text{0,00000123}\) \(\text{N}\)

  2. \(\text{417 000 000}\) \(\text{kg}\)

  3. \(\text{246 800}\) \(\text{A}\)

  4. \(\text{0,00088}\) \(\text{mm}\)

Solution not available at present.

For each of the following, write the measurement using the correct symbol for the prefix and the base unit:

  1. \(\text{1,01}\) microseconds

  2. \(\text{1 000}\) milligrams

  3. \(\text{7,2}\) megametres

  4. \(\text{11}\) nanolitre

Solution not available at present.

The Concorde is a type of aeroplane that flies very fast. The top speed of the Concorde is \(\text{844}\) \(\text{km·hr$^{-1}$}\). Convert the Concorde's top speed to \(\text{m·s$^{-1}$}\).

Solution not available at present.

The boiling point of water is \(\text{100}\) \(\text{℃}\). What is the boiling point of water in Kelvin?

To convert from Celcius to Kelvin we add \(\text{273}\):

\(\text{100}\text{ ℃} = (\text{100}+\text{273}) \text{K} = \text{373}\text{ K}\)