What is cardiovascular endurance?

The ability of the heart and lungs to work together to provide energy during exercise.

The strength of bones in the body.

The ability to lift heavy weights.

What does muscular strength refer to?

The amount of force a muscle can produce with a single effort.

The ability to run long distances.

The effectiveness of breathing during exercise.

What does body composition refer to?

The percentage of fat, bone, water, and muscle in the body.

How can one maintain a healthy body composition?

Eating healthy foods and exercising regularly.

What does muscular endurance allow a muscle to do?

Continue to perform without getting tired.

Five Components of Physical Fitness

Identify the importance of the five components of the physical fitness teat

1 What is the Australian Qualifications Framework (AQF)? 2 Describe three nationally accredited training products. 3 Define CBA. How is CBA different from other types of assessment? Briefly describe how the four (4) principles of assessment impact the CBA process. 4 Briefly describe how the rules of evidence impact the collection of evidence. 5 Describe different methods of gathering evidence for CBA . Describe the three main assessment methods. Describe the two different assessment pathways. What is recognition of prior learning (RPL), and what are the benefits? 6 What are the three endorsed components of training packages? 7 Briefly describe the components of a unit of competency (UofC). 8 What is the ACSF? Identify the five core skills. 9 How do the dimensions of competency guide assessment practices? 10 What is the purpose and context of assessment? How can you contextualise an assessment while maintaining the integrity of the Unit of Competency? 11 What is the difference between assessment tools and assessment instruments? What are the essential parts of an assessment tool? 13 What is the purpose of mapping against units of competency? 14 Describe reasonable adjustment in competency-based assessment. What are the limitations? 15 What would you do if a candidate provided you with evidence that was not authentic? 16 How does the marker’s guide support you in objective assessment judgement? 17 Give an example of inclusive language when giving assessment instructions. 18 What are the key WHS instructions that are required prior to commencing assessment? 19 What do you look for when clustering units of competency for an assessment tool? 20 What are the key characteristics of a cohort of learners that need to be considered in the planning an assessment tool?

Tobruk, a small town on the Libyan coast, was central to much of the fighting that took place in the Western Desert during the Second World War. It had originally been developed by the Italians during their colonisation of eastern Libya during the early decades of the 20th century. With a sheltered deep water harbour it became a key naval outpost. It was fortified during the 1930s with both coastal defence batteries and a 50 kilometre-long perimeter of reinforced concrete platoon posts, and other supporting infrastructure such as gun positions, headquarters bunkers, underground supply dumps, and observation towers. When British and Commonwealth forces advanced out of Egypt and into Libya in January 1941, Tobruk was their second objective. The Italian defence perimeter was attacked by the 6th Australian Division on the morning of 22 January and the town fell the next morning. The operation resulted in approximately 27,000 Italian prisoners and the capture of over 200 artillery pieces, but cost 49 Australian lives. The 6th Division's advance pressed on beyond Tobruk and eventually they were withdrawn from Libya to be deployed to Greece.The 9th Australian Division was moved in to Libya in February 1941 to garrison the territory captured by the 6th. By this time, however, German troops had arrived in Libya to reinforce their Italian allies and they launched an offensive that the British Commonwealth forces were ill-disposed to hold back. A retreat towards Egypt commenced. The 9th Division was ordered to fall back upon Tobruk, hold it in order deny its port facilities to the Germans, and delay their advance so as to provide time for defences on the Egyptian frontier to be prepared. Tobruk and the 9th Division were subsequently encircled, beginning what became known as "the siege of Tobruk". Reinforced by the 18th Brigade of the 7th Australian Division and other British and Commonwealth troops, and resupplied by the sea, the 9th Division held Tobruk from April to September 1941. During this period it repelled two major German attacks. In September and October the 9th Division, its condition steadily declining, was relieved by the British 70th Division, which continued to defend Tobruk until the siege was finally lifted by Operation Crusader in December. The defence of Tobruk resulted in 749 Australian deaths, and another 604 became prisoners of war. Tobruk was the scene of further heavy fighting in June 1942 when the fortunes of war again saw a British Commonwealth force seeking to deny the port to the enemy. The Axis forces, however, were in no mood for another siege and launched a massive attack to capture it on 20 June. It remained in their hands until their final retreat from Libya in November 1942.John Hurst Edmondson (1914-1941), soldier, was born on 8 October 1914 at Wagga Wagga, New South Wales, only child of native-born parents Joseph William Edmondson, farmer, and his wife Maude Elizabeth, née Hurst. The family moved to a farm near Liverpool when Jack was a child. Educated at Hurlstone Agricultural High School, he worked with his father and became a champion rifle-shooter. He was a council-member of the Liverpool Agricultural Society and acted as a steward at its shows. Having served (from March 1939) in the 4th Battalion, Militia, he enlisted in the Australian Imperial Force on 20 May 1940 and was posted to the 2nd/17th Battalion. Later that month he was promoted acting corporal (substantive in November). Well built and about 5 ft 9 ins (175 cm) tall, Edmondson settled easily into army life and was known as a quiet but efficient soldier. His battalion embarked for the Middle East in October and trained in Palestine. In March 1941 the 2nd/17th moved with other components of the 9th Division to Libya and reached Marsa Brega before an Axis counter-attack forced them to retreat to Tobruk. The siege of the fortress began on 11 April. Two days later the Germans probed the perimeter, targeting a section of the line west of the El Adem Road near Post R33. This strong-point was garrisoned by the 2nd/17th's No.16 Platoon in which Edmondson was a section leader. The enemy intended to clear the post as a bridgehead for an armoured assault on Tobruk.Under cover of darkness thirty Germans infiltrated the barbed wire defences, bringing machine-guns, mortars and two light field-guns. Lieutenant Austin Mackell, commanding No.16 Platoon, led Edmondson's five-man section in an attempt to repel the intruders. Armed with rifles, fixed bayonets and grenades, the party of seven tried to outflank the Germans, but were spotted by the enemy who turned their machine-guns on them. Unknown to his mates, Edmondson was severely wounded in the neck and stomach. Covering fire from R33 ceased at the pre-arranged time of 11.45 p.m. and Mackell ordered his men to charge. Despite his wounds, Edmondson accounted for several enemy soldiers and saved Mackell's life. When the remaining Germans fled, the Australians returned to their lines. Although Edmondson was treated for his wounds, he died before dawn on 14 April 1941. The Germans' armoured attack that morning was thwarted, partly due to the earlier disruption of their plans. Edmondson was buried in Tobruk war cemetery. He had not married. His Victoria Cross, gazetted on 4 July, was the first awarded to a member of Australia's armed forces in World War II. In April 1960 Mrs Edmondson gave her son's medals to the Australian War Memorial, Canberra, where they are displayed alongside his portrait (1958) by Joshua Smith. At Liverpool a public clock commemorates Edmondson, as do the clubrooms used by the sub-branch of the Returned Services League of Australia.Perhaps my nerves will be more under control when I am by myself. There were no entries in the diary until Friday April 18 when she wrote: Fighting terrific in Greece and North Africa…. I dread the casualty list also the heaviest air raid over London to date. Account …. of heavy fighting and much use of bayonet at Tobruk. Also gives an account of a charge in which a Lieutenant and a Corporal took prominent parts on Easter Sunday night. Of course, no names. When I read it …. I was sure the Corporal was Jack…. It said no casualties but …. I know … that all is not well with Jack. ….. (and) Stuffy ….has not come home yet. On Wednesday April 23 she received a letter from Jack dated March 30 and for the first time he said the conditions were bad. The food short, water one bottle for 48 hours. It worried me terribly so I posted a parcel (of) milk tablets, chocolate milk, biscuits (and) cigarettes.Tuesday April 15 I was feeling afraid of something while I was working and packing the cake (and) had a couple of brandys to (keep going).April 26 Received the following telegram in the mail, the bus man brought it in. “It is with deep regret that I have to inform you that Corporal John Hurst Edmondson was killed in action on the 14th April and desire to convey the profound sympathy of the Ministry for the Army and the Military Board.”Her final entry

In this video we take a look at the 0:02 fetch to code 0:03 execute cycle including its effect on 0:06 the various registers we've previously 0:12 [Music] 0:14 discussed a computer is defined Definition 0:17 as an electronic device that takes an 0:20 input 0:22 processes data 0:25 and delivers output 0:29 in this simple example you can see we're 0:31 taking the input 5 0:35 we're multiplying it by 2 that's our 0:37 process 0:39 and we're outputting 10. 0:44 but this could be way more complex for 0:46 example of a game console 0:48 the input could be the buttons you press 0:50 on a controller 0:53 the processes would then be carried out 0:55 by the console itself 0:59 and the output would be some form of 1:01 update to a monitor 1:02 and sound out for a speaker possibly 1:04 vibration feedback through the 1:06 controller 1:10 to process data a computer follows a set 1:13 of instructions 1:14 known as a computer program 1:18 if we take the lid off a typical desktop 1:20 computer we can identify 1:22 two critical components the memory 1:26 that stores the program and the central 1:29 processing unit or processor 1:31 which is under this large fan and 1:33 carries out the instructions 1:37 a computer carries out its function by 1:40 fetching 1:41 instructions decoding them and then 1:43 executing them 1:44 in a continuous repetitive cycle 1:46 billions of times a second 1:48 let's look at each of these stages in a 1:50 little more detail Fetch 1:53 so let's start with the fetch stage the 1:55 very first thing that happens 1:57 is the program counter is checked as it 2:00 holds the address 2:01 of the next instruction to be executed 2:07 the address stored is then copied into 2:09 the memory address register 2:14 the address is then sent along the 2:16 address bus to main memory 2:18 where it waits to receive a signal from 2:21 the control 2:22 bus so it knows what to do 2:27 as we want to read the data that's 2:29 stored in memory address 2:30 0 0 0 0 the control unit sends 2:34 a read signal along the control bus to 2:36 main memory 2:41 now main memory knows the data needs to 2:44 be read 2:45 the content stored in memory address 000 2:49 can be sent along the data bus to the 2:51 memory data register 2:56 now as we're currently in the process of 2:58 fetching an instruction 3:00 the data received by the memory data 3:03 register gets copied 3:04 into the current instruction register 3:11 the instruction effectively has now been 3:14 fetched from memory 3:16 just before we proceed to the decode 3:18 phase we now 3:19 increment the program counter so that 3:22 the address it contains 3:24 points to the address of the next 3:26 instruction which will need to be 3:30 executed 3:32 the instruction now being held in the 3:33 current instruction register 3:35 is ready to be decoded 3:39 now as we mentioned in the previous 3:41 video the instruction is made up of two 3:43 parts 3:44 we have the op code that's what it is we 3:47 need to do 3:50 and we have the operand what are we 3:53 going to do it to 3:55 now the operand could contain the actual 3:57 data 3:58 or indeed it could contain an address of 4:01 where the data is to be found 4:06 by decoding this instruction we can see 4:08 the operation we need 4:10 is a load operation so we need to load 4:14 the contents of memory location0101 4:18 into the cpus accumulator 4:25 in the exam a simple model will be used 4:27 to describe the 4:29 structure of any given instruction 4:32 you're not going to be expected to 4:34 define how an opcode is made up 4:36 but simply to interpret opcodes in the 4:39 given context of an exam 4:40 question in the example here 4:44 you can see there's a total of 16 4:46 different opcodes available 4:48 and this is because we're using four 4:50 bits for our representation 4:56 so now we've fetched the instruction and 4:59 we've decoded it so we know what we need 5:00 to do 5:01 we're finally ready to execute it 5:05 so we now send address 0101 5:08 to the memory dress register 5:13 now we're in the memory address register 5:15 we can finally send the address 5:18 down the address bus to main memory 5:24 this time we want to read the data 5:26 that's stored in memory 5:28 and so the control unit again sends a 5:30 read signal along the control bus 5:36 so main memories now receive an address 5:38 and a read signal 5:40 so the content stored at memory location 5:43 0101 5:44 can now be sent along the data bus back 5:46 to the cpu 5:47 and into the memory data register 5:54 finally the contents of the memory data 5:56 register are copied to the accumulator 5:59 and this is one of a number of general 6:00 purpose registers found in the cpu 6:04 this first instruction is now complete Branching 6:11 so what does this program actually do 6:14 you should be able to work it through 6:16 carefully and figure it out 6:19 we're now pointing instructions zero 6:21 zero zero one in the program counter 6:23 and we're ready to fetch the second 6:25 instruction 6:27 at the end of this video we're gonna 6:29 provide you with the answer 6:34 so let's talk a second about programs 6:37 that branch 6:40 on the left here we have a very simple 6:42 piece of pseudo code 6:44 line zero says first execute this line 6:46 of code 6:47 line 1 now execute this line and then 6:50 line 2 says 6:52 if the age is greater than 18 then 6:56 we're going to execute lines 3 and 4 6:58 otherwise 6:59 we're going to execute lines six and 7:02 seven 7:03 so this program doesn't necessarily 7:05 follow strictly in sequence from line 7:07 zero through to seven there's a chance 7:10 here the program may branch and jump 7:14 around 7:16 so we're going to pretend that this 7:17 program has been loaded into memory 7:20 each line of code on the left here has 7:23 ended up 7:24 as a location in memory now this is not 7:27 strictly how this would happen in this 7:28 one-to-one way 7:29 but for the purpose of example it's 7:31 absolutely fine 7:35 so the program counter starts by 7:37 pointing to memory address zero 7:39 and we fetch the first instruction 7:41 decode it and execute it 7:44 it then updates and tells us the next 7:47 instruction 7:48 is zero zero zero one because remember 7:50 the program counter is being incremented 7:52 so we fetch it decode it and we execute 7:55 line one of our program 7:59 we then fetch line two which in binary 8:01 is one 8:02 zero 8:06 now at this point depending on what 8:10 happens during the execution 8:11 of line two the program may be required 8:15 to fetch line three from memory or 8:18 line five from memory 8:25 so let's look at how this actually works 8:27 because we've said the program counter 8:28 simply gets incremented 8:31 well in the current instruction register 8:33 we have an instruction with the op code 8:36 0 1 1 0. 8:41 now when we look this up in the decode 8:43 unit we discover that this 8:45 code means branch always 8:51 this replaces the value held in the 8:54 program counter 8:56 with the contents of the operand that's 8:58 the second part of the instruction 9:01 from the current instruction register so 9:03 this case 9:04 one zero zero one 9:09 now when the next fetch cycle begins the 9:12 program counter is obviously checked 9:14 and as its contents have been previously 9:16 updated to a new memory location 9:19 and not simply incremented the program 9:22 effectively is able to jump 9:24 around memory 9:28 so having watched this video you should 9:30 be able to answer the following key 9:32 question 9:33 how does a cpu work 9:39 okay so let's um answer the question we 9:41 posed 9:42 earlier what did that program actually 9:48 do 9:50 so this is the first fetch to code 9:53 execute cycle 9:55 and this is the one that we ran through 9:57 in detail earlier 9:58 it effectively loaded the contents of 10:01 the memory 10:02 stored at location location0101 10:05 into the accumulator in other words 10:08 the dna number 3 is moved 10:11 from memory into the cpu 10:18 we then proceed onto the second fetch 10:20 decode execute cycle 10:23 now this one adds the contents of memory 10:27 located at 0 1 1 0 10:30 to the current contents of the 10:32 accumulator 10:34 so in other words the dna number one 10:38 because that's what's stored at address 10:40 zero one one zero 10:43 is added to the number three that was in 10:45 the accumulator 10:46 the results are stored back over the 10:48 accumulator 10:49 so effectively we've done three plus one 10:53 equals four 10:58 the third fetch to code execute cycle 11:00 stores the contents which are in the 11:02 accumulator 11:03 into memory location zero one one one 11:07 and that's because the op code the first 11:09 part of this current instruction 11:10 zero zero one one is the command to 11:13 store when we look it up in the decoder 11:15 unit 11:16 so in other words the result of the 11:17 previous calculation three plus one 11:19 equals four 11:20 is now written back into main memory 11:28 the fourth fetch decode execute cycle 11:30 outputs the contents of the accumulator 11:33 remember they were copied into main 11:34 memory but they're still held in the 11:35 accumulator 11:37 so in this simple abstraction the number 11:40 four is now 11:41 output to the user so they can see the 11:43 result of the calculation 11:49 the fifth and final fetch code execute 11:51 cycle 11:52 brings a halt to the current program 11:58 so this very simple program which has 12:01 five 12:02 fetch decode execute cycles has 12:04 performed the calculation 12:06 three plus one is then stored the result 12:09 in main memory 12:10 and displayed the result four to the 12:12 user 12:13 and in a high-level language this may 12:15 look something very similar to the 12:17 following two lines of code 12:20 sum variable equals num1 plus num2 12:24 print sum to the user 12:27 so you can start to get an appreciation 12:29 here of how the high level code you 12:32 write actually ends up being fetched 12:34 decoded 12:35 and executed inside a processor 12:38 of course your processor is doing 12:40 billions and billions of these 12:42 operations a second 12:43 which when you think about it is really 12:45 very impressive 12:52 [Music] 13:03 you. make 10 questions for a standerd of a level

Lipids are large, nonpolar organic molecules. They do not dissolve in water. Lipids include triglycerides (trie-GLIS-uhr-IEDZ), phospho- lipids, steroids, waxes, and pigments. Lipid molecules have a higher ratio of carbon and hydrogen atoms to oxygen atoms than carbohydrates have. Because lipid molecules have larger numbers of carbon-hydrogen bonds per gram than other organic com- pounds do, they store more energy per gram. Fatty Acids Fatty acids are unbranched carbon chains that make up most lipids. Figure 3-10 shows that a fatty acid contains a long carbon chain (from 12 to 28 carbons) with a carboxyl group, —COOH, attached at one end. The two ends of the fatty-acid molecule have different properties. The carboxyl end is polar and is thus hydrophilic or attracted to water molecules. In contrast, the hydro- carbon end of the fatty-acid molecule is nonpolar. This end tends not to interact with water molecules and is said to be hydrophobic (HIE-droh-FOH-bik), or “water fearing.” In saturated fatty acids, such as palmitic acid, which is shown in Figure 3-10, each carbon atom is covalently bonded to four atoms. The carbon atoms are in effect full, or saturated. In contrast, linoleic acid, also shown in Figure 3-10, has carbon atoms that are not bonded to the maximum number of atoms to which they can bond. Instead, they have formed double bonds within the carbon chain. This type of fatty acid is said to be unsaturated. Triglycerides Three classes of lipids important to living things contain fatty acids: triglycerides (fats), phospholipids, and waxes. A triglyceride is composed of three molecules of fatty acid joined to one molecule of the alcohol glycerol. Saturated triglycerides are composed of saturated fatty acids. They typically have high melting points and tend to be hard at room temperature. Common dietary saturated triglycerides include butter and fats in red meat. In contrast, unsaturated triglycerides are composed of unsaturated fatty acids and are usually soft or liquid at room temperature. Unsaturated triglycerides are found primarily in plant seeds where they serve as an energy and carbon source for germinating plants. Phospholipids Phospholipids have two, rather than three, fatty acids attached to a molecule of glycerol. They have a phosphate group attached to the third carbon of the glycerol. As shown in Figure 3-11, the cell membrane is made of two layers of phospholipids, called the lipid bilayer. The inability of lipids to dissolve in water allows the mem- brane to form a barrier between the inside and outside of the cell. Hydrophilic “head” Phospholipids Hydrophobic “tail” Phospholipids Water Water The lipid bilayer of a cell membrane is a double row of phospholipids.The “tails” face each other.The “head” of a phospholipid, which contains a phosphate group, is polar and hydrophilic.The two tails are two fatty acids and are nonpolar and hydrophobic. FIGURE 3-11 H C H C O OH H C H H C H H C H H C H H C H H C H H C H H C H H C H H C H H C H H C H C H H H H C H H C H H C H H C H H C H C O OH H C H H C H H C H C H C H C H H C H H C H C H H C H H C H C H H H H C H Fatty acids have a polar carboxyl head, highlighted in purple, and a nonpolar hydrocarbon tail, highlighted in green. FIGURE 3-10 Palmitic acid Linoleic acid mb06se_bchs02.qxd 5/18/07 10:49 AM Page 59 60 CHAPTER 3 1. Compare the structure of monosaccharides, dis- accharides, and polysaccharides. 2. How are proteins constructed from amino acids? 3. How do amino acids differ from one another? 4. Describe a model of enzyme action. 5. Why do phospholipids orient in a bilayer when in a watery environment, such as a cell? 6. Describe how the three major types of lipids differ in structure from one another. 7. What are the functions of the two types of nucleic acids? CRITICAL THINKING 8. Applying Information Before a long race, run- ners often “carbo load.” This means that they eat substantial quantities of carbohydrates. How might this help their performance? 9. Recognizing Relationships High temperatures can weaken bonds within a protein molecule. How might this explain the effects of using a hot curling iron or rollers in one’s hair? 10. Applying Information You want to eat more unsaturated than saturated fats. Name examples of foods you would eat more of and less of. SECTION 2 REVIEW Waxes A wax is a type of structural lipid consisting of a long fatty-acid chain joined to a long alcohol chain. Waxes are waterproof, and in plants, form a protective coating on the outer surfaces. Waxes also form protective layers in animals. For example, earwax helps pre- vent microorganisms from entering the ear canal. Steroids Unlike most other lipids, which are composed of fatty acids, steroid molecules are composed of four fused carbon rings with various functional groups attached to them. Many animal hor- mones, such as the male hormone testosterone, are steroid com- pounds. One of the most familiar steroids in humans is cholesterol. Cholesterol is needed by the body for nerve and other cells to func- tion normally. It is also a component of the cell membrane. NUCLEIC ACIDS Nucleic acids are very large and complex organic molecules that store and transfer important information in the cell. There are two major types of nucleic acids: deoxyribonucleic acid and ribonucleic acid. Deoxyribonucleic acid, or DNA, contains information that deter- mines the characteristics of an organism and directs its cell activi- ties. Ribonucleic (RIE-boh-noo-KLEE-ik) acid, or RNA, stores and transfers information from DNA that is essential for the manufactur- ing of proteins. Some RNA molecules can also act as enzymes. Both DNA and RNA are polymers, composed of thousands of linked monomers called nucleotides (NOO-klee-uh-TIEDS). As shown in Figure 3- 12, each nucleotide is made of three main components: a phosphate group, a five-carbon sugar, and a ring-shaped nitrogenous base.

Understanding Quantum Theory of Electrons in Atoms The goal of this section is to understand the electron orbitals (location of electrons in atoms), their different energies, and other properties. The use of quantum theory provides the best understanding to these topics. This knowledge is a precursor to chemical bonding. As was described previously, electrons in atoms can exist only on discrete energy levels but not between them. It is said that the energy of an electron in an atom is quantized, that is, it can be equal only to certain specific values and can jump from one energy level to another but not transition smoothly or stay between these levels. The energy levels are labeled with an n value, where n = 1, 2, 3, …. Generally speaking, the energy of an electron in an atom is greater for greater values of n. This number, n, is referred to as the principal quantum number. The principal quantum number defines the location of the energy level. It is essentially the same concept as the n in the Bohr atom description. Another name for the principal quantum number is the shell number. The shells of an atom can be thought of concentric circles radiating out from the nucleus. The electrons that belong to a specific shell are most likely to be found within the corresponding circular area. The further we proceed from the nucleus, the higher the shell number, and so the higher the energy level (Figure 9.4.1). The positively charged protons in the nucleus stabilize the electronic orbitals by electrostatic attraction between the positive charges of the protons and the negative charges of the electrons. So the further away the electron is from the nucleus, the greater the energy it has. This quantum mechanical model for where electrons reside in an atom can be used to look at electronic transitions, the events when an electron moves from one energy level to another. If the transition is to a higher energy level, energy is absorbed, and the energy change has a positive value. To obtain the amount of energy necessary for the transition to a higher energy level, a photon is absorbed by the atom. A transition to a lower energy level involves a release of energy, and the energy change is negative. This process is accompanied by emission of a photon by the atom. The following equation summarizes these relationships and is based on the hydrogen atom: The values nf and ni are the final and initial energy states of the electron. The principal quantum number is one of three quantum numbers used to characterize an orbital. An atomic orbital, which is distinct from an orbit, is a general region in an atom within which an electron is most probable to reside. The quantum mechanical model specifies the probability of finding an electron in the three-dimensional space around the nucleus and is based on solutions of the Schrödinger equation. In addition, the principal quantum number defines the energy of an electron in a hydrogen or hydrogen-like atom or an ion (an atom or an ion with only one electron) and the general region in which discrete energy levels of electrons in a multi-electron atoms and ions are located. Another quantum number is l, the angular momentum quantum number. It is an integer that defines the shape of the orbital, and takes on the values, l = 0, 1, 2, …, n – 1. This means that an orbital with n = 1 can have only one value of l, l = 0, whereas n = 2 permits l = 0 and l = 1, and so on. The principal quantum number defines the general size and energy of the orbital. The l value specifies the shape of the orbital. Orbitals with the same value of l form a subshell. In addition, the greater the angular momentum quantum number, the greater is the angular momentum of an electron at this orbital. Orbitals with l = 0 are called s orbitals (or the s subshells). The value l = 1 corresponds to the p orbitals. For a given n, p orbitals constitute a p subshell (e.g., 3p if n = 3). The orbitals with l = 2 are called the d orbitals, followed by the f-, g-, and h-orbitals for l = 3, 4, 5, and there are higher values we will not consider. There are certain distances from the nucleus at which the probability density of finding an electron located at a particular orbital is zero. In other words, the value of the wavefunction ψ is zero at this distance for this orbital. Such a value of radius r is called a radial node. The number of radial nodes in an orbital is n – l – 1. Consider the examples in Figure 9.4.2. The orbitals depicted are of the s type, thus l = 0 for all of them. It can be seen from the graphs of the probability densities that there are 1 – 0 – 1 = 0 places where the density is zero (nodes) for 1s (n = 1), 2 – 0 – 1 = 1 node for 2s, and 3 – 0 – 1 = 2 nodes for the 3s orbitals. The s subshell electron density distribution is spherical and the p subshell has a dumbbell shape. The d and f orbitals are more complex. These shapes represent the three-dimensional regions within which the electron is likely to be found. Principal quantum number (n) & Orbital angular momentum (l): The Orbital Subshell: https://youtu.be/ms7WR149fAY If an electron has an angular momentum (l ≠ 0), then this vector can point in different directions. In addition, the z component of the angular momentum can have more than one value. This means that if a magnetic field is applied in the z direction, orbitals with different values of the z component of the angular momentum will have different energies resulting from interacting with the field. The magnetic quantum number, called ml, specifies the z component of the angular momentum for a particular orbital. For example, for an s orbital, l = 0, and the only value of ml is zero. For p orbitals, l = 1, and ml can be equal to –1, 0, or +1. Generally speaking, ml can be equal to –l, –(l – 1), …, –1, 0, +1, …, (l – 1), l. The total number of possible orbitals with the same value of l (a subshell) is 2l + 1. Thus, there is one s-orbital for ml = 0, there are three p-orbitals for ml = 1, five d-orbitals for ml = 2, seven f-orbitals for ml = 3, and so forth. The principal quantum number defines the general value of the electronic energy. The angular momentum quantum number determines the shape of the orbital. And the magnetic quantum number specifies orientation of the orbital in space, as can be seen in Figure 9.4.3. Figure 9.4.4 illustrates the energy levels for various orbitals. The number before the orbital name (such as 2s, 3p, and so forth) stands for the principal quantum number, n. The letter in the orbital name defines the subshell with a specific angular momentum quantum number l = 0 for s orbitals, 1 for p orbitals, 2 for d orbitals. Finally, there are more than one possible orbitals for l ≥ 1, each corresponding to a specific value of ml. In the case of a hydrogen atom or a one-electron ion (such as He+, Li2+, and so on), energies of all the orbitals with the same n are the same. This is called a degeneracy, and the energy levels for the same principal quantum number, n, are called degenerate energy levels. However, in atoms with more than one electron, this degeneracy is eliminated by the electron–electron interactions, and orbitals that belong to different subshells have different energies. Orbitals within the same subshell (for example ns, np, nd, nf, such as 2p, 3s) are still degenerate and have the same energy. While the three quantum numbers discussed in the previous paragraphs work well for describing electron orbitals, some experiments showed that they were not sufficient to explain all observed results. It was demonstrated in the 1920s that when hydrogen-line spectra are examined at extremely high resolution, some lines are actually not single peaks but, rather, pairs of closely spaced lines. This is the so-called fine structure of the spectrum, and it implies that there are additional small differences in energies of electrons even when they are located in the same orbital. These observations led Samuel Goudsmit and George Uhlenbeck to propose that electrons have a fourth quantum number. They called this the spin quantum number, or ms. The other three quantum numbers, n, l, and ml, are properties of specific atomic orbitals that also define in what part of the space an electron is most likely to be located. Orbitals are a result of solving the Schrödinger equation for electrons in atoms. The electron spin is a different kind of property. It is a completely quantum phenomenon with no analogues in the classical realm. In addition, it cannot be derived from solving the Schrödinger equation and is not related to the normal spatial coordinates (such as the Cartesian x, y, and z). Electron spin describes an intrinsic electron “rotation” or “spinning.” Each electron acts as a tiny magnet or a tiny rotating object with an angular momentum, even though this rotation cannot be observed in terms of the spatial coordinates. The magnitude of the overall electron spin can only have one value, and an electron can only “spin” in one of two quantized states. One is termed the α state, with the z component of the spin being in the positive direction of the z axis. This corresponds to the spin quantum number ms=12. The other is called the β state, with the z component of the spin being negative and ms=−12. Any electron, regardless of the atomic orbital it is located in, can only have one of those two values of the spin quantum number. The energies of electrons having ms=−12 and ms=12 are different if an external magnetic field is applied. Figure 9.4.5 illustrates this phenomenon. An electron acts like a tiny magnet. Its moment is directed up (in the positive direction of the z axis) for the 12 spin quantum number and down (in the negative z direction) for the spin quantum number of −12. A magnet has a lower energy if its magnetic moment is aligned with the external magnetic field (the left electron) and a higher energy for the magnetic moment being opposite to the applied field. This is why an electron with ms=12 has a slightly lower energy in an external field in the positive z direction, and an electron with ms=−12 has a slightly higher energy in the same field. This is true even for an electron occupying the same orbital in an atom. A spectral line corresponding to a transition for electrons from the same orbital but with different spin quantum numbers has two possible values of energy; thus, the line in the spectrum will show a fine structure splitting. The Pauli Exclusion Principle An electron in an atom is completely described by four quantum numbers: n, l, ml, and ms. The first three quantum numbers define the orbital and the fourth quantum number describes the intrinsic electron property called spin. An Austrian physicist Wolfgang Pauli formulated a general principle that gives the last piece of information that we need to understand the general behavior of electrons in atoms. The Pauli exclusion principle can be formulated as follows: No two electrons in the same atom can have exactly the same set of all the four quantum numbers. What this means is that electrons can share the same orbital (the same set of the quantum numbers n, l, and ml), but only if their spin quantum numbers ms have different values. Since the spin quantum number can only have two values (±12), no more than two electrons can occupy the same orbital (and if two electrons are located in the same orbital, they must have opposite spins). Therefore, any atomic orbital can be populated by only zero, one, or two electrons. The properties and meaning of the quantum numbers of electrons in atoms are briefly

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