S112_1 Scales in space and time About this free course This free course is an adapted extract from the Open University course S112 Science: concepts and practice www.open.ac.uk/courses/modules/s112 . This version of the content may include video, images and interactive content that may not be optimised for your device. You can experience this free course as it was originally designed on OpenLearn, the home of free learning from The Open University – www.open.edu/openlearn/science-maths-technology/scales-space-and-time/content-section-0 There you’ll also be able to track your progress via your activity record, which you can use to demonstrate your learning.
1 Working with large and small numbers When exploring different scales, you will inevitably encounter very big and very small numbers, and need to be able to express these and use them in calculations. Values will also be measured to varying degrees of precision and sometimes presented in different units from those you are familiar with. The size–time explorer and associated questions rely on some knowledge of some fundamental mathematical concepts that underpin scientific study and practice. Before you explore the tool it is worth spending some time (re-)visiting these mathematical ideas. This is a brief look at some principles before you have the opportunity to apply your skills to exploring scales, but links are provided to more in-depth maths support, if you need it.
1.1 Scientific notation Large and small numbers are best expressed by using scientific notation. The convention of scientific notation is that a quantity is presented as a number, equal to or greater than 1 but less than 10, multiplied by a power of ten. For example, 3500 000, or 3.5 million, is written as 3.5 × 106. Here, the number part is 3.5, which is clearly greater than 1 but less than 10, and the power of ten is 6. Similarly, 0.0095 is written as 9.5 × 10−3 in scientific notation. Table 1 shows some examples of powers of ten and their meanings. Table 1  Translating values into powers of ten
Thousands Hundreds Tens Units Tenths Hundredths Thousandths
Value 1000 100 10 1 0.1 0.01 0.001
Power of 10 103 102 101 100 10−1 10−2 10−3
You might already be familiar with these concepts. Have a go at the following questions to test your knowledge. Write 365 000 000 in scientific notation. 365 000 000 is written in scientific notation as 3.65 × 108. Write 0.0465 in scientific notation. 0.0465 in scientific notation is written as 4.65 ×10−2. How confident did you feel completing those questions? If you were able to answer the questions correctly then move straight to the next section. If you found some of the questions challenging then look at the additional maths resources that are available from the link below. If you need any guidance on the maths content, take a look at the badged open course, Mathematics for science and technology.
1.2 Units When exploring different scales, you will inevitably encounter a number of different units. There is an International System of Units that gives a preferred unit (known as an SI unit) for each type of measurement. For time, this is seconds (abbreviated to s) and for distance this is metres (m). These are both known as base units. There are seven base units in the SI system and every other SI unit can be expressed as a combination of these. As you will see, the main axes of the size–time explorer uses SI units, but you will come across other units that are also used for time and length. Note that when the symbols for units are used, do not make them plural. For example, write ‘4 cm’, not ‘4 cms’.  <b>Units with prefixes</b> The SI system also includes standard prefixes that provide another way to write large and small values more efficiently. You will be familiar with a kilo, meaning a thousand, from a kilometre (1000 m), and also centi, meaning a hundredth, from centimetres (0.01 m, or one-hundredth of a metre). So, instead of writing 3000 metres, write 3 kilometres, or 3000 m = 3 km. Similarly, 0.04 m = 4 cm. Table 2 contains a selection of prefixes, from the large to the small. Note that the case, or capitalisation, of the symbol is important; for example, a mm is a thousand million (i.e. a billion) times smaller than a Mm. Also, there is not a prefix for every power of ten. Some prefixes, such as deca (da), are rarely used. Table 2  Examples of prefixes
Name Symbol 10n Meaning
giga G 109 1000 000 000
mega M 106 1000 000
kilo k 103 1000
hecto h 102 100
deca da 101 10
100 1
deci d 10−1 0.1
centi c 10−2 0.01
milli m 10−3 0.001
micro µ 10−6 0.000 001
nano n 10−9 0.000 000 001
Apart from centi, which is commonly used in centimetres, and deci, which is sometimes used to measure volume in decilitres, the major prefixes increase or decrease by three powers of 10: kilo (103), mega (106), giga (109), for example. How would you describe 0.000 003 seconds using a SI prefix? 0.000 003 seconds can be written as 3 microseconds, or 3 µs. How would you describe 0.045 metres using a SI prefix? This can be written as 45 millimetres or 45 mm, or alternatively as 4.5 centimetres or 4.5 cm. Both are correct, but one may be more appropriate than the other in a given context. Although metres and seconds are the standard SI units for length and time, other units are sometimes used that are appropriate to scales frequently encountered in a particular field. For example, the age of a person might be given in years (y) rather than in seconds as it is more meaningful in that context. However, it can be easily converted from one unit to another, and during this course you will become familiar with that process. Note that you will encounter both SI units and other commonly used units in the size–time explorer. Don’t forget, if you need any guidance on the maths content, take a look at the badged open course, Mathematics for science and technology.
<b>Compound units and rates</b> Some quantities need multiple units (known as compound units) to describe them. One such quantity is speed, which is a measure of distance travelled over time taken. For example, at the time of writing (2017), the women’s world record for the 100.0 m sprint is 10.49 s, as set by Florence Griffith-Joyner in 1988. To calculate the (average) speed at which she ran, you need to divide the distance over the time: $\begin{array}{l}\frac{100\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}}{10.49\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}}=9.532\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\frac{\text{m}}{\text{s}}\\ \phantom{\frac{100\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}}{10.49\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}}}=9.532\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}\end{array}$ Her average speed was therefore 9.532 metres per second (m s−1). This value describes the average rate at which she covered distance over the course of the race, namely 9.532 metres every second. Note the way the units are written in the example above. The result of the calculation could be reported as ‘metres per second’, ‘m per s’ or ‘m/s’, but ‘m s−1’ is most appropriate for scientific reporting as it is the most concise. The notation of ‘negative exponents’ is commonly used for units. So, for example, just as $\frac{1}{25}=\frac{1}{{5}^{2}}={5}^{-2}$, can be expressed $\frac{1}{\text{s}}$ as s−1 and $\frac{1}{{\text{m}}^{3}}$ as m−3. Writing compound units using negative exponents is generally good scientific practice. You should also note that care is needed when writing and interpreting compound units. A space is required between the ‘m’ and the ‘s’ in m s−1. The unit ms−1 has a completely different meaning, namely ‘per millisecond’. Any quantity that changes with time can be expressed as a rate. For example, it takes 5 minutes for an oven to heat up from room temperature, 20 °C, to 200 °C. The change in temperature is 180 °C, so the (average) rate at which the temperature changes is: $\begin{array}{l}\frac{200\phantom{\rule[-0.0ex]{.3em}{0.0ex}}°\text{C}-20\phantom{\rule[-0.0ex]{.3em}{0.0ex}}°\text{C}}{5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{min}}=\frac{180\phantom{\rule[-0.0ex]{.3em}{0.0ex}}°\text{C}}{300\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}}\\ \phantom{\frac{200\phantom{\rule[-0.0ex]{.3em}{0.0ex}}°\text{C}-20\phantom{\rule[-0.0ex]{.3em}{0.0ex}}°\text{C}}{5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{min}}}=0.6\phantom{\rule[-0.0ex]{.3em}{0.0ex}}°\text{C}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}\end{array}$ Usually SI units should be used for calculations, which is why minutes were converted to seconds in the example above. However, at times, other non-SI units might be more meaningful. For example, reporting the oven temperature in degrees Celsius (°C) is far more familiar than using the SI unit of temperature – kelvin, K. It is important to include units throughout a calculation, as well as in a final answer. It makes it clear to the reader (and yourself!) what units are used, and where units have been converted. Don’t forget, if you need any guidance on the maths content, take a look at the badged open course, Mathematics for science and technology.
1.3 Precision and magnitude When working with quantities at any scale, and when combining numbers through addition, multiplication, etc., it is important to report the answers with an appropriate level of precision. To do this you need to consider the number of significant figures, which is the number of digits that carry a ‘meaning’ for the measurement. The cardinal rule here is shown below: Key point An answer in a calculation cannot be more precise than the least precise number used in that calculation. For example, 361 has three significant figures and 31.01 has four significant figures. When reporting an answer to a calculation, you should use the number of significant figures from the least precise value (or value with the smallest number of significant figures) in the calculation. If you multiply 361 by 31.01, the answer provided by a calculator is 11 194.61. This value has seven significant figures, but the least precise value in the calculation (361) has just three significant figures. The answer should therefore be quoted to the same level of precision, which is 11 200. In this example, the part of the number that affects the third significant figure, namely 194.61, is rounded to 200 rather than 100. (More information on rounding is provided in Box 1.) Box 1 A reminder about rounding When rounding values, look only at the digit immediately to the right of where you want to stop the number of significant figures. If it is less than 5, round it down, if it is equal to or greater than 5, round it up. For example, if rounding 11 194.61 to three significant figures, stop after the third significant figure. Therefore, look at the digit to the right of this third significant figure, and see it is a 9. As 9 is greater than or equal to 5, round up by increasing the third significant figure by 1. In effect, 11 194.61 is nearer to 11 200 than it is to 11 100. When giving an answer to a calculation you should also quote the number of significant figures used. Note that you can do this by stating ‘to x significant figures’ in your final answer, or abbreviate it as ‘to x sig figs’ or even ‘to x s.f. (where x is the number of digits in question). Precision and exact integers There are some special cases where numbers do not affect the number of significant figures, such as exact integers. Sometimes this is encountered this when applying a formula. For example, the perimeter of a square = 4 × edge length, because, by definition, a square has four sides of equal length. Here the integer 4 is exactly four, and not more or less. A square has exactly four sides so this number is a multiplier and not an amount measured. The edge length, in contrast, is a measure, and therefore determines how precisely the perimeter can be reported. If the side of a square was measured be 2.62 cm, which has three significant figures, then: $\begin{array}{l}\text{perimeter}=4×2.62\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{cm}\\ \phantom{\text{perimeter}}=10.48\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{cm}\\ \phantom{\text{perimeter}}=10.5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{cm}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{to 3 sig fig}\end{array}$ Note, again, that the answer given by the calculator has been provided as well as the answer rounded to the appropriate number of significant figures to make work flow clear. This approach is good practice when it is appropriate to show working in calculations. <b>Significant figures and decimal places</b> When using numbers with decimal places, zeros that are after the decimal point but precede non-zero numbers are not included in the count of significant figures. Such digits are just place-holders indicating number size, and they are not giving you information about how precisely you know the value. For example, 0.0034 has two significant figures, meaning you are confident it is closer to 0.0034 than it is to 0.0033 or 0.0035. In contrast, when dealing with zeros that fall after the decimal point and follow non-zero numbers these digits are significant, because they are giving you information about the precision. So 0.00340 has three significant figures, meaning you are confident it is closer to 0.00340 than it is to 0.00339 or 0.00341. This is a special case that only affects decimal numbers. So, 3000 has one significant figure, but 3000.0 has five significant figures because it indicates that it has been measured to one decimal place. Another benefit of using scientific notation is that it is an efficient way to show the number of significant figures. For example, 3000 would be written in scientific notation as 3 × 103 but 3000.0 would be written as 3.0000 × 103, where it is clear that the second number has more (five) significant figures. You might already be familiar with the concepts of precision and significant figures. A selection of questions is provided below for you to test your knowledge. If you multiply 3.01 by 2.1 (3.01 × 2.1) to how many significant figures should you report the answer? The answer should be reported to two significant figures because 2.1 is the least precise number and is quoted to two significant figures. What is the result of multiplying 3.01 by 2.1, to the correct level of precision? 3.01 × 2.1 = 6.321, which to two significant figures is 6.3 (because 6.321 is nearer to 6.300 than it is to 6.400). Express 2052 × 0.033 with the appropriate precision. The exact answer is 67.716, which has five significant figures. However, 2052 is significant to four figures and 0.033 is significant to two figures. The answer should therefore not be quoted to more than two significant figures either, so 67.716 = 68 to 2 sig fig. Express the number 9.2499 × 103 to two significant figures. 9.2499 × 103 is expressed as 9.2 × 103 to two significant figures. (The first digit after the final significant figure in question is 4, which is less than 5 so the number is rounded down to 9.2 rather than up to 9.3.) <b>Magnitude</b> It can be just as important to appreciate that an answer is in the tens versus in the millions, as it is to know the exact amount. Alongside precision you need to develop an appreciation of the magnitude of numbers. You do this by considering the nearest power of ten. For example, 340 m is closer to 100 m (102 m) than it is to 1000 m (103 m). Hence its nearest order of magnitude is 102. An example of when the magnitude of a number is useful is if someone has over 10 million followers on Twitter. Knowing there are roughly 107 people potentially reading their tweets may be more important than knowing exactly how many there are at a given minute. Using another Twitter example, someone with 83 839 followers has closer to 100 000 followers than 10 000 followers, hence the magnitude or power of ten that best describes their following is 105. Using the magnitude, you can then easily compare between these two examples: Note that ~ means ‘approximately equal to’, and applies only to the first step in the expression above because the exact numbers of followers has been rounded to provide these values. The three parts thereafter are exactly equal to one another, which is why the equals symbol (=) has been used. Ten million (107) is one hundred (102) times larger than one hundred thousand (105), or two orders of magnitude greater. It is therefore easy to see that Person 1 has 100 times more followers than Person 2. Note also that the order of magnitude quoted is the power of ten difference between the two numbers, in this example, 2. It is important to appreciate that if A is two orders of magnitude larger than B, then A is 100 times larger, not twice as large. You might already be familiar with the concepts of magnitude. A question is provided below for you to test your knowledge. Express the number 9.2499 × 103 to the nearest order of magnitude. 9.2499 × 103 is approximately 104 to the nearest order of magnitude, because 9.2499 × 103 is closer to 104 than 103. That is, 9249.9 is closer to 10 000 than 1000. Don’t forget, if you need any guidance on the maths content, take a look at the badged open course, Mathematics for science and technology.
Event Temperature/K Time since big bang/s
big bang unknown 0
neutrons and protons formed ~100 000 10−6
quarks exist in a plasma 100 000 000 000 10−12
quarks and other fundamental particles form stable particles 10 000 000 000 000 10–1000
today 2.7 × 1017
Table 3B
Event Temperature/K Time since big bang/s
big bang unknown 0
quarks exist in a plasma 10 000 000 000 000 10−12
quarks and other fundamental particles form stable particles 100 000 000 000 10−6
neutrons and protons formed ~100 000 10–1000
today 2.7 × 1017
Table 3B is correct. It shows the correct combination of events, temperatures and times, with the universe cooling over time, and the times listed in a logical order.
Question 4 Use the information from Question 3 to consolidate your learning and fill in the missing gaps in the paragraph below. Summary Here, you learned about the current size of the Universe and its age, as well as it origin by working backward through time. The Universe is the biggest thing known, but considering how it began requires you to think about some of the very smallest things known, including subatomic particles such as protons and neutrons. Next: That’s as big as possible! Now return to the size–time explorer to choose your next level of scale, or use the ‘Next >’ button to go to the next level down on the size scale.
2.3 The Earth In the slide show below, the dimensions of the Earth and its trajectory around the Sun are given, together with the time it takes the Earth to spin on its axis and complete an orbit of the Sun. Using these data, you can consider the speed at which the Earth is moving. The way to describe a unique location on Earth is covered, along with the implications for life of the Earth’s motion over different timescales. While working through the slides, record the following information to use in the questions that follow: the average radius of the Earth the average distance and duration of the Earth’s orbit around the Sun the time it takes for the Earth to spin once on its axis. Implications of the spin and orbit of Earth Question 1 What is the average circumference of the Earth? Use the value of the radius you recorded from the video and the following equation: circumference = 2 × π × r where r is the radius and π has a value of 3.14. Write your answer without spaces between consecutive digits, and don’t forget to consider significant figures. You may have wondered about what number of significant figures was appropriate for your answer. Let’s look at the data provided. The radius was given to 4 significant figures and the value for π was given to three significant figures. However, the 2 might appear to have only 1 significant figure, but in fact it is a special case. It is an integer, and so is viewed as being infinitely precise – it isn’t something measured, it is an absolute value. So, the least precise value here is π, so the answer should be quoted to the same precision, and so the answer is given as 40 000 km (to three significant figures). It is very important to quote the level of precision here as, without it, 40 000 km has an ambiguous number of significant figures. It could be understood to be an answer given to the precision of 1 significant figure or 5 significant figures, when actually the value is calculated to 3 significant figures. Question 2 The circumference at the Equator is 40 030 km. (Remember the Earth is widest at the Equator so this value is slightly different from that given in Question 1.) Ignoring the fact that the Earth is travelling around the Sun, how fast would you be moving if you were standing at the Equator? (Note that speed is calculated as distance travelled over (divided by) time taken, and you’ll need to use the time it takes for the Earth to rotate once in your calculation.) Give your answer to the nearest km h−1, remembering to quote your answer to the correct number of significant figures. Question 3 When standing in the Equator, or indeed anywhere on Earth, you don’t feel as though you are moving at such high speeds as you have just calculated (or indeed any speed). The following is an explanation given by one of the physicists at The Open University. Use your developing knowledge of units, and rates, to complete the paragraph by selecting the correct options from the drop-down lists. There was a tricky conversion in this question, where 450 km h−2 was converted to 0.035 m s−2. Here’s some additional information about how this was done. The value 450 km h−2 is a measure of acceleration (not speed), which indicates that the Earth has an acceleration of 450 km per hour, each hour. To convert km h−2 to m s−2 you first convert km to m (by multiplying by 1000), which gives 450 km h−2 × 1000 m km−1 = 450 000 m h−2 Then you need to convert hours into seconds, twice, because the acceleration is ‘m per hour per hour’. As there are 3600 seconds in an hour, you should divide by 3600, but because you are working with h−2, you need to do apply to conversion twice. Hence: $\begin{array}{l}\frac{450\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}000\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{h}}^{-2}}{3600\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{h}}^{-1}×3600\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{h}}^{-1}}=0.03472\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-2}\\ \phantom{\frac{450\phantom{\rule[-0.0ex]{0.3em}{0.0ex}}000\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{h}}^{-2}}{3600\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{h}}^{-1}×3600\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{h}}^{-1}}}=0.035\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-2}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{to 3 sig fig}\end{array}$ Notice how dividing by 3600 s h−1 × 3600 s h−1 results in the h−2 on the top cancelling with the h−1 × h−1 on the bottom. As described in the slides, locations at the same latitude experience the same lengths of day. Locations at the same longitude are (generally) in the same time zone, but experience differences in day length across the year due to the tilt of the axis on which the Earth spins. you can compare two locations: High Park in England (51.84 °N, 1.4 °W), which is marked on Figure 1, and the city of Accra (5.6 °N, 0.2 °W). Question 4 Question 5 To explore the variation in the time that locations have daylight, Table 4 below gives the sunrise and sunset times for two locations at the same longitude but different latitudes. Some of the values are missing (see X–Z). Select the correct daylight length times for these values from the options below. Table 4 Sunrise and sunset times (nearest 15 minutes), daylight length and long-term minimum and maximum daily temperatures for two locations on the same longitude: High Park, England (Europe) and Accra, Ghana (Africa).
High Park, United KingdomLatitude: 51.84 °N Accra, Ghana Latitude: 5.6 °N
Sunrise Sunset Daylight length/hours:min Min–max temp/°C Sunrise Sunset Daylight length/ hours: min Min–max temp/°C
Mar 20 6:00 18:15 W 3.7–10.0 6:00 18:15 12:15 24.5–32.5
Jun 20 4:45 21:30 X 10.9–20.3 5:45 18:15 12:30 22.8–29.6
Sep 20 6:45 19:00 Y 10.7–19.1 5:45 18:00 12:15 22.5–29.1
Dec 20 8:15 15:45 Z 2.3–7.7 6:00 18:00 12:00 22.5–31.6
Main species/forest type Woodland area/km2 Total area km2
Areas 0.02 km2 and greater Areas 0.001–0.02 km2
Oak 1478 X 1586
Beech 606 35 641
Sycamore 450 38 488
Ash 963 86 Y
Birch 685 11 696
Total 5853 622 6475
Question 7 Choose the correct words from the drop-down lists to complete the paragraph below to consider the significance of the preservation of the oak ecosystem in the video, using information from Table 5. Summary In this section, you learned about a remnant of oak woodland in the grounds of Blenheim Palace that contains ancient oaks. It is a valuable site because oaks have long lifespans and so the ecosystems in which they occur need long-term protection to persist and allow oaks to reach the large size that supports many other species. Next: Now look at how oaks colonised the British Isles, or you can return to the size–time explorer and choose for yourself, or use the ‘Next >’ button to go to the next level down on the size scale.
2.6 An oak tree The following slide show describes the life of an oak, using an ancient oak from High Park at Blenheim Palace as a particular example. To better appreciate the age and size of the oak, and the link between them, some calculations are needed. While working through the slides, record the following information to use in the questions that follow: the age and girth of the oldest High Park oak the period over which an oak grows fastest the approximate age at which an oak becomes mature and produces acorns. The life of an ancient oak Question 1 Use the conservative age of the ancient oak, estimated in 2017, to calculate the year in which it would it have germinated if it was exactly this age. Question 2 Using some of the other information that you recorded, in which year did the High Park oak’s ‘growth spurt’ end if it was exactly it’s conservatively estimated age? Question 3 The time axis in the size–time explorer is in seconds, and the size axis is in metres. Nevertheless, you can use many other units for time (minutes, days, weeks, millennia) depending on the context, so you need to be able to convert between them. In this question, you will calculate, step by step, the age in seconds of the ancient High Park oak – assuming it is exactly 800 years old. Use the interactive Figure 5 to help you work through the calculations from converting years to days (assuming 1 year = 365.25 days to include leap years), then to hours, to minutes and finally to seconds. At this stage don’t worry about significant figures or scientific notation. Figure 5 Convert the age of the ancient oak into SI units of seconds. Question 4 Give the age of the oak in standard scientific notation to three significant figures. That is a lot of seconds! You can check that the number you calculated fits with where the oak tree is placed on the size–time explorer. Now let’s look now at how the oak changed in size over that time. The tree heights over time from the slides are plotted in Figure 6. The first seven data points from the left were calculated from an approximately 220 year-old tree felled at Blenheim Palace, and the data point on the extreme right is from a 800 year old oak that is still standing there. Question 5 Select the terms from the drop-down options below that best describe the graph in Figure 6 Question 6 What is the girth, or distance around the trunk, of the ancient High Park oak?  Summary Oaks can live for over one thousand years, but ancient oaks are rare. Survival to this age requires successful germination, growth through seedling and sapling stages, withstanding damage from herbivores and weather and, across the British Isles, being protected from being cut down for timber. Oaks grow very quickly for about 120 years, rapidly increasing in height in this time, in comparison to later years. Many other species depend on oaks, making them a critical part of oak woodland ecosystems. Next: It is recommended that you keep zooming in, and looking at parts of an organism or living thing – in this case a leaf on the oak, which you can do with the ‘Next >’ button. Alternatively, you can return to the size–time explorer and choose a level yourself.
Age bracket/years Number of years spent in
this age bracket
Number of leaves produced each year Total number of leaves produced during age bracket
1–20 20 × 10 000 = 200 000
21–100 × =
101–500 × =
501+ × =
Total:
The completed Table 6 is shown below: Table 6 (Completed)
Age bracket/years Number of years spent in
this age bracket
Number of leaves produced each year Total number of leaves produced during age bracket
1–20 20 × 10 000 = 200 000
21–100 80 × 200 000 = 16 000 000
101–500 400 × 400 000 = 160 000 000
501+ 300 × 300 000 = 90 000 000
Total: 266 200 000
Using the data in the video, you can calculate that an 800 year old oak could have produced 266 200 000 leaves in its lifetime. That’s nearly 300 million leaves!
Are you surprised at this number? Did you ever consider that a tree might produce this many leaves over its lifetime? Scientists estimate the number of leaves on a tree for multiple reasons, but some examples are to: estimate the amount of whole-tree photosynthesis or carbon fixation determine how much of a tree’s resources are allocated to leaves compared to other parts of the plant consider how much light is intercepted or if self-shading occurs, and estimate nitrogen use.
Question 6 In Question 5, you estimated the number of leaves an 800 year old oak could produce in its lifetime. Give a reason why it is useful to estimate a number like this instead of determining the number exactly. One reason is that you can get a good estimate using techniques including digital imaging without damaging the tree. Cutting down a tree and counting the leaves is one way to determine leaf number, but clearly that’s not possible to do multiple times in its lifetime. Another reason is it is clearly extremely difficult, if not impractical, to count all of the leaves on a tree every year for 800 years, and it is very unlikely that you could provide an accurate number. (It would take generations of people collecting every single fallen leaf and counting them!) Question 7 Based on the information in Question 5 suggest two limitations, or sources of error, with approximating the number of leaves in this way. There are several possible limitations; here are two. The age brackets (time spans) for each of the growth periods, particularly the latter two, are very large (100 years and 500 years, respectively) so any error in the average annual number of leaves will be magnified when multiplied by such large numbers and will render the estimate inaccurate. The leaf numbers for each age range are averages determined from other plants of the same species, but likely to be growing in a different area. The actual oak tree under scrutiny will almost certainly have produced a different numbers of leaves. Question 8 Using the details you recorded regarding how thick oak leaves are, and how long it takes them to grow, add the bar ‘Oak leaf expansion’ to the size–time explorer. Summary Here you determined the order of magnitude that leaf sizes can vary across plant species and considered the dimensions of an oak leaf. In addition, you used a table to help organise numbers in a multi-part calculation when you estimated the number of leaves an ancient oak produces in its lifetime. By making calculations such as these, you can better appreciate the implications of the deciduous nature of oak leaves. Next: Now look at the components of a leaf at even smaller scale by examining a stomata (the pores on a leaf), which you can do with the ‘Next >’ button, or you can return to the size–time explorer and pick a new level yourself.
2.8 Stomata The following video explains what stomata are and their function in gas exchange between the inside of a leaf and the air. To appreciate the size, abundance and rate of movement of the guard cells that form stomata, some measurements and calculations are needed. Using information from the video and some additional details you can calculate the number of stomata on a leaf and a whole tree! You should also consider why they might not move more quickly. There are quite a few calculations in this section, but you will be stepped through them. While watching the video, record the following information to use in the questions that follow: the length of a guard cell to the nearest micrometre (μm) the density of stomata (per square millimetre) on the upper surface of an oak leaf how long it takes stomata to open or close, to the nearest 5 minutes. (You will need to use the clock in the top left-hand corner of the video to estimate this.) Stomata: specialised leaf cells JULIA COOKE Plants make food through photosynthesis, by converting carbon dioxide and water into sugars and other carbohydrates, using sunlight as energy. The first part of the process is to allow carbon dioxide into the leaves through little pores on the leaf surfaces. On this oak leaf, there are roughly 500 pores in each one millimetre by one millimetre square of leaf on the lower side, and 100 pores in each square millimetre of leaf on the upper side. The pores are called stomata, or stomates. As well as the entry of carbon dioxide into the leaf, water also evaporates from the plant through these pores, known as transpiration, and oxygen is released, which is critical to our own survival. Plants take up carbon dioxide and release oxygen and water vapour. We call this gas exchange. Plants carefully balance gas exchange to maximise photosynthesis but avoid too much water loss by opening and closing the stomata. To open the pores, water is pumped into the two guard cells that border each stomate, and water is pumped out of the cells to close them. Opening and closing pores isn’t instantaneous. It takes time and occurs in response to environmental cues. In oak leaves, stomata respond to light levels, because light is needed as the energy source for photosynthesis. They can also respond to water availability in the air and soil. Question 1 What is the length of a guard cell, to the nearest micrometre?  Question 2 The density of stomata is the number of stomata in a given area. In the video, stomatal density is stated as the number of stomata per square millimetre (stomata per mm2, or stomata mm−2). If the average size of an oak leaf is 15 cm2, you can use the density of stomata to work out roughly how many stomata are on the top side of a leaf. Let’s go through the calculations step by step. You may find Figure 7 will help you to visualise the scales. Question 3 First, how many square millimetres are there in a square cm? Question 4 Note the way that the units were derived for the final answer. In multiplying ‘stomata mm−2’ by ‘mm2 cm−2’ the ‘mm’ terms cancelled each other out, leaving just ‘stomata cm−2’. Question 5 Finally, use your answer to Question 4 to work out how many stomata there are on the upper surface of a leaf which has an area of 15 cm2. Provide your answer using standard scientific notation.  Question 6 The video indicated that there is a higher density of stomata on the lower surface of the leaf (500 stomata mm−2). Using this value and the answer to Question 5, work out the total number of stomata on both sides of the leaf.  Another way to work out the answer to this question is to consider the ratio of stomatal density between the upper and lower side of the leaf. Using the stomata densities on either side of the leaf – 500 mm−2 on the lower side and 100 mm−2 on the upper side – it is possible to calculate that there are $\frac{500\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{stomata}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}{\text{m}}^{-2}}{100\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{stomata}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}{\text{m}}^{-2}}=5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{times as many stomata on the lower side}$ times as many stomata on the lower side. Using this ratio, if there are 1.5 × 105 stomata on the upper surface there must be 7.5 × 105 on the lower side. Adding the two numbers together gives the total number of stomata on the leaf as 9.0 × 105. Question 7 A large oak might have 400 000 leaves in a single year. How many stomata would there be on a tree with this many leaves? Clearly these are estimates, as the actual number of stomata on every leaf, or even exactly how many leaves are on the tree haven’t been counted. However, these values are based on numbers published in scientific reports for Quercus robur and should provide a reasonable estimate. Next time you walk past a large tree, take a minute to think about how many stomata their might be on the whole tree… and then think about how many stomata there might be in a forest! Researchers use high precision equipment, such as leaf chambers and infrared gas analysers to measure the amount of CO2 taken up by a leaf over time (Figure 8). You can use the information from such measurements together with your earlier calculations to estimate the number of CO2 molecules taken up by stomata. Question 8 Experiments have shown that an oak leaf in sunlight can take up CO2 at a net rate of 4.4 × 10−9 grams per cm2 of leaf per second (or 4.4 × 10−9 g CO2 cm−2 s−1). Calculate the mass of carbon dioxide taken up by one stomata in one second by dividing this value by the number of stomates in a cm2 of leaf. You can turn this mass of CO2 into a number of molecules of CO2. It turns out that 44 g of CO2 contain 6.023 × 1023 molecules (this is called Avogadro’s number). So if there are 6.023 × 1023 molecules in 44 g of CO2 there must be To calculate the number of molecules of CO2 taken in by each stomata per second you now need to multiply the mass of CO2 per stomata per second by the number of molecules of CO2 per gram of CO2. $\begin{array}{l}7.3×{10}^{-14}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{g}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{stomata}}^{-1}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}×1.4×{10}^{22}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{molecules}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{g}}^{-1}\\ \phantom{7.3×{10}^{-14}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{g}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{stomata}}^{-1}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}}=10.22×{10}^{8}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{molecules}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{stomata}}^{-1}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}\\ \phantom{7.3×{10}^{-14}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{g}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{stomata}}^{-1}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}}=1.0×{10}^{9}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{molecules}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{stomata}}^{-1}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{(to 2 sig fig)}\end{array}$ This gives a value of 1.0 × 109 molecules CO2 stomata−1 s−1. That is 1 billion molecules of CO2 per stomata per second! Each of the stomata will open and close during the day to regulate CO2 uptake and water loss. Now it is time to consider the time it takes stomata to open and close. Question 9 How long does it take a stomata in the animation to open and close (to the nearest 5 minutes)?  Question 10 In oaks, light stimulates stomata to open and lack of light stimulates closure. Why might stomatal closure take some minutes rather than being instantaneous? Suggest possible reasons. Note: this question is to encourage you to consider possibilities rather than necessarily knowing the ‘right’ answer. (Your answer should be between 50 and 100 words.) Light levels can change very quickly, as a leaf can be shaded by others intermittently in windy conditions or as clouds roll by. The plant uses energy to open and close stomata, which could be wasted if the stomata were to respond too quickly to environmental conditions. However, if the weather clouds over for a long time, or the leaf moves into long-term shade and low light levels prohibit photosynthesis, stomata need to close to reduce water loss. Summary In this section you learned that microscopic guard cells form stomata or pores, which allow billions of molecules of CO2 to be exchanged between the leaf and the atmosphere every second. The stomata open and close to regulate the movement of gases. Through a series of calculations, you were able to consider just how many stomata there are on a leaf and a whole tree. There were some really massive numbers there! Although this section focuses on the scale of cells (micrometres) your calculations showed how process at this scale affect bigger (leaves and trees) and smaller (molecules) size scales. Next: Now look at the components of a leaf at even smaller scales by examining a chloroplast within a stomatal cell, which you can do with the ‘Next >’ button, or you can return to the size–time explorer to choose a level for yourself.
2.10 Glucose The following video introduces glucose molecules and their size. It describes how glucose stores energy, in some cases by forming more complex molecules that can be stored for longer. Plants and animals obtain glucose in different ways, but use it in the same way. Here you will consider the size of a glucose molecule and the timescales over which plants and animals use it. While watching the video, record the following information to use in the questions that follow: the length of the linear form of the glucose molecule the amounts of glucose used by an oak leaf for growth and maintenance per day definition of autotroph and heterotroph a description of the solubility of glucose and starch the uses of energy listed (e.g. growth). Glucose: a molecule that stores energy JULIA COOKE Glucose, C6H12O6, is a sugar ubiquitous in nature as a way of storing energy. It takes energy to assemble the molecule, and this stored energy is what is useful to living things. Glucose is a carbohydrate because it contains only carbon, hydrogen and oxygen. And the hydrogen and oxygen are in a ratio of two to one. Glucose can occur in a linear or cyclic form. Autotrophs, including plants, make their own food using energy from sunlight to assemble glucose. They make glucose from non-living ingredients, water, and carbon dioxide. Autotrophs are also called ‘producers’. Heterotrophs consume other living things as food sources. A caterpillar, for example, eats a leaf to get energy, including glucose, and a bird will eat a caterpillar is its energy source. Heterotrophs are also called ‘consumers’. Plants store glucose as starch, and animals store it as glycogen, both by chemically linking glucose molecules together. Starch and glycogen are similar but have different branching patterns. They are useful ways to store glucose because, unlike glucose, theyre not soluble. When needed, both are broken down by breaking the chemical bonds to form glucose. Regardless of where the glucose came from, all living things obtain energy by breaking glucose back into carbon dioxide and water again in a process called ‘cellular respiration’. This releases the energy used to assemble glucose initially, and the energy is used to sustain life. Respiration occurs in heterotrophs such as the caterpillar and us. We take in oxygen and breathe out carbon dioxide. Autotrophic plants both respire and photosynthesise. Living things use energy to grow and reproduce and some is lost in waste, not able to be used by the organism. Living things use energy to do work, such as moving and lifting, and use energy to produce heat, and in maintenance. It takes energy to assemble the molecules of life. We can look at how much glucose an oak leaf uses for growth and maintenance. Research has determined that, in a single day, a young leaf of an oak tree used 1.3 grams of glucose for each gram of growth. It used 27 milligrams of glucose to maintain each gram of leaf. Thats nearly 50 times as much glucose to use for growth compared to maintaining the tissue. Getting enough energy or glucose contributes to some of the patterns we see in nature. Plants can only grow where they can get enough sunlight to power photosynthesis. Animals can only survive where there is enough of the right kind of food to sustain them. A butterfly, for example, will lay its eggs away from other eggs to increase chances that there is enough food for the larvae and food which locations are sought after. Many distribution patterns and behaviours are driven by the need for glucose, the molecule that stores energy. Question 1 How big is a glucose molecule? Give your answer in scientific notation. Question 2 How much glucose is needed to maintain a gram of oak leaf each day? Give your answer in scientific notation. The English oak, Quercus robur, is a deciduous species that loses its leaves over winter. The following graph (Figure 11) shows the amount of sugars (including glucose) and starch – both of which are carbohydrates – in English oak twigs during the year. Three specific time periods are shown: just before leaves are lost, when the tree has no leaves, and just before bud break (which is when the leaf buds begin expanding into leaves). The columns are the mean values of 10 twigs, each from a different plant. Question 3 Use your notes from the video together with Figure 11 to describe the graph and interpret its significance in terms of how a plant uses glucose and starch to store energy. Select the most appropriate terms from the drop-down lists to complete the following paragraph. A very simple food chain was shown in the video. It involved a bird (great tit) obtaining its energy (and protein) by eating the caterpillar (hairstreak butterfly larvae), which in turn obtained glucose by eating the oak leaf. You can measure the blood glucose concentrations in birds in order to compare them with other species. Table 7 shows the average blood sugar levels of 14-day-old blue tit and great tit chicks (both of which can feed on caterpillars on oak leaves). For comparison, it also contains the levels of a healthy adult human both before and after eating. Table 7 Average blood glucose concentrations in blue tit and great tit chicks in an oak woodland in Poland (Kaliński et al., 2015) and for an adult human after fasting and two hours after eating (Diabetes UK, n.d.).
Animal Average blood glucose concentration/mg ml−1
Blue tit chick 2.503
Great tit chick 2.684
Adult human 2 h after eating 1.400
Question 4 Use your list of the functions of glucose in an animal from the video to comment on the relative amounts of glucose in the bloodstream of tits and humans. Specifically, consider the age of the animals and what types of activity (work) the adult animals listed in the table do that might account for the differences. (Your answer should be between 50 and 100 words.) The tit chicks have two to three times the concentration of glucose in their blood compared to adult humans. Glucose is used to power growth, reproduction, work, generate heat, maintain cells and some is lost in waste. As the tits are just 14 days old they are likely to be growing quickly. Growth requires lots of energy in comparison to maintaining cells, so this could be one reason for the high blood glucose levels in the chicks, compared to the human. Alternatively, flying uses a lot of energy, so perhaps birds have higher blood glucose levels in general than animals that do not fly. Summary Here you learned about a molecule, glucose, which stores energy in both plants and animals. It takes energy to synthesise glucose and when glucose is broken down into CO2 and water again, that energy is released (through cellular respiration). Your calculations and interpretation of graphs and tables of data allowed you to consider how plants and animals use glucose over time and at different stages in their development. Next: Now look at even smaller scales by examining a carbon atom, which you can do with the ‘Next >’ button, or you can return to the size–time explorer to choose a level for yourself.
2.11 Carbon The following video looks at a carbon atom, exploring its size and inclusion in diverse organic molecules. Carbon is ‘fixed’ by plants through photosynthesis – by which you mean incorporated into molecules used by living organisms. Here you can explore the length of time the carbon is stored in various forms after a plant dies before being released into the atmosphere again as CO2. Humans use fossil fuels to power many of the things used in your daily lives. These fossils fuels are energy stored in carbon molecules fixed through photosynthesis that occurred hundreds of millions of years ago, during the Carboniferous and Jurassic periods. While watching the video, record the following information to use in the questions that follow: the diameter of a carbon atom how long it takes an oak leaf and an acorn to grow the dates of the Carboniferous and Jurassic periods, when much of the carbon forming today’s coal, oil and gas reserves was fixed though photosynthesis. Carbon in energy sources JULIA COOKE Carbon is a chemical element, often abbreviated to the letter C. Individual carbon atoms are too small to be seen by eye or even through a light microscope because they’re smaller than the wavelengths of visible light. Carbon is frequently depicted as a black sphere, but is represented in other ways depending on the type of information we want to show, such as chemical bonds or molecule structure. Carbon is a component of an impressive diversity of molecules, including carbon dioxide and glucose and many biological molecules, including DNA, proteins, fibres, and cellulose, as well as synthetic products such as plastics. We obtain carbon compounds from our food to fuel our bodies and as building blocks to make our cells. Energy from the sun as light and carbon as carbon dioxide from the air can be captured by plants and turned into something ready to eat within a few weeks. The term ‘organic’ is used in science to describe compounds containing carbon. Some organic energy sources take a long time to form but can store large amounts of energy. We burn wood for heating, but that wood may have taken decades for the tree to accumulate. When plants and animals die, much of their organic matter decomposes and is broken down into simpler molecules, and eventually, the carbon is released back into the atmosphere as carbon dioxide. But in some conditions, decomposition doesnt occur, and the carbon is trapped and stored. Coal, a fossil fuel, began forming between 360 and 280 million years ago when huge areas of peat bogs and swamps grew and then were buried. Over time, they were subjected to very high pressures and heat, which transformed the plant material into coal. Crude oil is formed when large quantities of tiny plants and animals, zooplankton and algae, are buried under sedimentary rock and subjected to high pressures and heat. Between 180 and 140 million years ago, during the Jurassic period, the carbon for much of today’s crude oil was deposited. When we burn fossil fuels in power plants and cars, we’e using energy from the sun and carbon fixed through photosynthesis hundreds of millions of years ago. They can’t be replaced quickly, as hundreds of thousands to millions of years are needed to accumulate the solar energy and carbon to form coal and oil. Question 1 What is the diameter of a carbon atom? Question 2 The video explained that you cannot see an individual carbon atom because they are shorter than the wavelengths of visible light. The shortest wavelength of light that you can see is about 390 nm (which is violet light). How many orders of magnitude longer is this than a carbon atom? Question 3 Carbon is part of glucose molecules. Choose the appropriate words below to compare the sizes of a single carbon atom to a glucose molecule. Many of our fuel sources use carbon-based compounds as sources of energy, from foods that you metabolise to wood, coal, crude oil and natural gas that you burn. Different sources vary in the amounts of energy released per unit mass. Table 8 lists different fuel sources, the amount of energy released per kg, and the amount of CO2 emissions for the energy produced. Joules (J) is an SI unit that describes energy; you will have seen it listed on food wrappers. Table 8 The age, amount of energy per kg, and CO2 emissions and % of carbon in different carbon sources exploited for energy
Energy source Net calorific value/MJ kg−1 CO2 emissions/g MJ−1 % Carbon
Wood and leaves 16 4 42
Coal 29 115 67
Natural gas 38 50 76
Crude oil 42 72 89
Question 4 Use information from the video to determine the age of the different energy sources in Table 8. (Note that for coal, natural gas and crude oil you should express your answer using scientific notation.) Question 5 Tabulated information, such as Table 8, facilitates comparisons between values. Select the correct words in the sentences below to describe the patterns in energy and carbon content and emissions of the different energy sources. The amount of CO2 emissions depends on the carbon content and the ratio of carbon: hydrogen in fuel. Although all energy sources are formed from the fixation of carbon dioxide through photosynthesis, wood, coal and gas are made of different compounds. Gas, often methane (CH4), has a higher ratio of hydrogen to carbon in comparison to coal and wood, and therefore lower CO2 emissions than coal for a given energy release. Question 6 Returning to the oak tree, how long does it take for an oak to produce an oak leaf? Use the equation below to aid your conversion from weeks to seconds. You can identify pairs of units that occur both in the numerator and denominator, which will cancel each other and give the final units for the answer. (The cancelling of weeks are shown as an example.) Leave no spaces between digits in your answer. $5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\overline{)\text{week}}×7\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\frac{\text{day}}{\overline{)\text{week}}}×24\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\frac{\text{hour}}{\text{day}}×60\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\frac{\text{minute}}{\text{hour}}×60\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\frac{\text{second}}{\text{minute}}=\phantom{\rule[-0.0ex]{0.2em}{0.0ex}}?$ Note that this is a special case where each number is an exact integer. Each day has exactly 24 hours, each minute has exactly 60 seconds, etc, so you can report the answer as 3024 000 seconds. Question 7 Organisms including oaks use carbon to make many different types of molecules. Acorns are full of carbon-based sugars and starches to fuel the embryo when an acorn germinates; hence acorns are an attractive energy source for animals. Oaks also use carbon to make defensive compounds to deter animals. The English oak makes compounds called sesquiterpenes which have been shown to deter insects. One of these compounds, farnesene, contains 15 carbon atoms of which 13 form a long carbon chain. Hence the molecule has a length of roughly 1.8 × 10−9 m (calculated as 13 × 1.4 × 10−10 m). Use this size and the time it takes an acorn to grow to add the bar ‘Acorn and sesquiterpenes production’ to the size–time explorer. You can use the same calculation as you used in Question 6 to calculate the length of time it takes for the oak to produce an acorn. Summary After learning about carbon atoms and how small they are, you learned that carbon is a component of many of our fuel sources. The food you eat is produced relatively quickly – as is the wood you burn – but fossil fuels are ancient sources of carbon that were fixed by plants many millions of years ago. They cannot be replaced quickly as it takes time to build up large reserves of these compounds, of which tiny carbon atoms are a significant component. Next: Now look at even smaller scales by examining the components of an atom, which you can do with the ‘Next >’ button, or you can return to the size–time explorer to choose a level for yourself.
2.12 Protons and neutrons The following slide show considers the size of subatomic particles, protons and neutrons, and their role in determining the identity of an element. You will consider the size of protons and neutrons relative to an atom, and how to use isotopes of carbon to date samples for a specific period. While working through the slides, record the following information to use in the questions that follow: the diameter of a neutron the half-life of carbon-14 (14C). Subatomic particles: protons and neutrons Question 1 Question 2 In another part of this topic you learn that the diameter of a carbon atom is about 1.4 × 10−10 m. How many times larger is a carbon atom than a neutron? Report your answer in full, not using scientific notation, and leaving no spaces between digits. Question 3 Radioactive 14C (read aloud as ‘14-C’) decays to 14N in a predictable way. The half-life of 14C is the number of years it takes half the amount of 14C remaining in a sample to decay. Table 9 shows the percentage of 14C remaining against years since plant death for a sample. Using the value of the half-life you noted from the slide show, select the correct values for the list below to replace W, X, Y and Z. Table 9 Decay of 14C in dead plant tissue.
Years since plant death % 14C remaining
0 100
5730 50
11 460 25
W 12.5
X 6.3
28 650 Y
34 380 Z
40 110 0.8
45 840 0.4
Question 4 There are equations describing the curve of the decay graph that was shown in the slide show, and these allow an organic sample to be dated accurately. However, you can use Table 9 to estimate the age of sample. Question 5 Different radioactive isotopes have different half-lives. The time frame varies substantially across the elements. For uranium-232, the half-life is just 68.9 years. Use the diameter of a uranium atom (312 pm or picometres), together with the half-life of 232U to add the ‘Half-life of a uranium-232’ bar to the size–time explorer. (Note that 1 pm = 1 × 10–12 m). Summary In this section you have learned that atoms consist of subatomic particles including protons and neutrons. The number of protons determines the element. The number of neutrons in atoms can vary, which results in different isotopes of the same element (e.g. 12C, 13C and 14C). Neutrons are lost from atoms at predictable rates. This phenomenon has various practical scientific applications, including radiocarbon dating. Next: Look at the smallest scale now: quarks, which you can do with the ‘Next >’ button. Alternatively, you can return to the size–time explorer and choose a level for yourself.
2.13 Quarks and photons The following slide show describes elementary particles including quarks and photons. While quarks are tiny – smaller than neutrons – they do not have a defined size. However, they are placed representing the smallest size in the size–time explorer. Photons are packets of energy, which have properties of both waves and particles. They have wavelengths and frequencies, some of which you can see as different coloured light. The activities for this section focus on photons. While working through the slides, record the following information to in the questions that follow: the size, or description of the size, of an elementary particle the speed of light in a vacuum the wavelength of red photons absorbed by plants (you may need information from multiple slides to work this out). Elementary particles: quarks and photons This video contains these scenes: 1. Image of a carbon atom, with a bundle of red and green spheres in the centre, surrounded by blue spheres. Each red sphere represents a proton, containing 2 purple spheres and 1 yellow sphere (3 quarks). Each green sphere represents a neutron, containing 1 purple sphere and two yellow spheres (3 quarks). Each blue sphere represents and electron; 2. Image showing quark flavours – up and down, charm and strange, top and bottom. They are shown as spheres labelled with the first letter of the flavour, with u, c and t coloured purple, and d, s and b coloured orange; 3. Drawing of the electromagnetic spectrum, in order of increasing wavelength, decreasing frequency: a. Gamma rays (lowest wavelength) b. X-Rays c. Ultraviolet d. Visible (expanded into colours violet, blue, green, yellow, orange, red) e. Infrared f. Microwaves g. Radio waves The final sentence should read ‘Photons with a wavelength of around 400–700 nm are visible to humans, as shown here.’; 4. Line graph of photons absorbed / % (percent) against wavelength / nm, over the range 400 – 700 nm). The percentage absorbed is close to 100 percent for around 400 – 500 nm, falls to around 20 % between 500 and 570 nm (the range of wavelengths of light absorbed by chlorophyll), then rises to over 90% at around 660 nm, then falls to almost zero by 700 nm. The line is labelled ‘pigment extract’; 5. Drawing of a green leaf. Photons of different wavelengths are shown as coloured arrows pointing downwards towards the leaf. A green arrow shown pointing upwards from the top of the leaf, is labelled ‘some green light is not absorbed but reflected.’ A green arrow shown pointing downwards from under the leaf is labelled ‘green light can also be transmitted through the leaf’. The leaf is also labelled – ‘photons of many wavelengths are absorbed by leaves and power photosynthesis’. Erratum Please note that there is a small error in the voice over for Slide 3 of this slide show. The final sentence states that ‘Photons with a frequency of around 400–700 nm are visible to humans...’, but this should be ‘Photons with a wavelength of around 400–700 nm are visible to humans...’. The value of the speed of light quoted in the slides gives a value in m s−1. In this question you will calculate how many seconds it takes a photon to travel a metre, which can be expressed in units of s m−1. To do this you need to invert the relationship, so n m s−1 (where n could be any number) becomes: As an example, if a ball travels 5 m s−1, you can write this as: $\begin{array}{l}5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}=\frac{5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}}{\text{s}}\text{}\\ \phantom{5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{s}}^{-1}}=\frac{5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}}{1\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}}\end{array}$ The inverse of this is therefore: $\begin{array}{l}\frac{1\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}}{5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}}=\frac{1}{5}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}\phantom{\rule[-0.0ex]{.3em}{0.0ex}}{\text{m}}^{-1}\text{}\\ \phantom{\frac{1\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}}{5\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}}}=0.2\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{s}{\phantom{\rule[-0.0ex]{.3em}{0.0ex}}\text{m}}^{-1}\end{array}$ Which means the ball takes 0.2 seconds to travel 1 m. Question 1 Question 2 The Earth is about 1.5 × 1011 metres from the Sun, our main source of light. Using the answer from Question 1, which of the following correctly calculates how long it takes light from the Sun to reach the Earth. Question 3 Using the answer to Question 2, how many minutes does it take photons from the Sun to reach Earth? Round your answer to one significant figure. Figures in the slides showed the visible light spectrum of photons with different wavelengths, and the proportion of these absorbed by chlorophyll and proteins in photosynthetic plants. There were two main peaks in the absorption graph. In addition, the equation describing the relationship between photon wavelength and frequency was given as: $\lambda f=c$ where λ represents the wavelength of light, f is the frequency (in cycles per second) and c is the speed of light in a vacuum. This equation can be rearranged to calculate the frequency of waves, such that: $f=\frac{c}{\lambda }$ Question 4 Use values for the speed of light and the wavelength of red light corresponding to a peak in absorption to determine the frequency of the photons absorbed by plants. Question 5 Use the time it takes photons to travel from the Sun to the Earth, and the size of the wavelength of red light to add the bar ‘Red light travelling from the Sun to the Earth’ to the size–time explorer. Summary From your calculations, you now know how long it takes photon from the Sun to reach the Earth, and which photons are absorbed in photosynthesis. The slide show finished with the comment that the efficiency of photosynthesis is related to the way elementary particles move through leaves, which forms a nice link to other levels in the size–time explorer. Study note Don’t forget to answer the questions and fill in the entries associated with this level in the scales data table that you downloaded or printed in Activity 1. Next: That’s as small as you can go. It is recommended that you now go to ‘The oak woodland’ section and start working towards the bigger size scales, or you can return to the size–time explorer and choose your own level.
3 Comparing and connecting scales Now that you have worked through the resources and questions for each scale level, you should have an appreciation to of the huge range of scales encountered across the sciences. To consolidate your learning the next task is to compare the time and size scales across two levels. Write a paragraph describing the relationship between any two levels that you found interesting, and what gave you an appreciation of both the time and size scales between those levels. For example, you might choose to compare the oak tree and oak ecosystem levels, and describe the implications of the duration of the lifecycle of an oak on the history of oak ecosystem conservation. You can revisit the size–time explorer if you want to. Your paragraph should be no more than 250 words in length. Be sure to clearly identify the two levels and then explain how the connection between them contributed to your understanding of both size and time scales in science, specifically referring to the numbers and scales involved. Indeed, you could use some of the maths skills you have been practising to compare the scales between levels. Note that the two levels do not have to be adjacent in the size–time explorer, but they can be if you choose. Conclusion Well done! Using the size–time explorer, you have explored many different scales in space and time, centred around an oak tree. Through this you have strengthened your ability to manipulate small and large numbers as well as developed or deepened an appreciation of the huge range of scales encountered in science, across multiple disciplines.  The activities encouraged you to make connections between the various scale levels, and therefore appreciate the implication of fast and slow processes as well as the influence of big and small objects on each other. Hopefully you also have appreciation for how knowledge from different scientific disciplines contribute to each other. At the beginning of the course you watched a video of physicists, chemists, biologists and geologists talking about how they saw the oak tree, where they spoke about massive forces and miniscule molecules, and processes that occur over milliseconds and millennia. Having completed this course, you should now have a new appreciation of scales at which processes are happening all around you too. This OpenLearn course is an adapted extract from the Open University course S112 Science: concepts and practice.