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Before beginning this section, you may wish to review the section entitled The Nuclear Atom in an earlier lesson. There you will find a discussion of the general nature of the atomic nucleus, the definitions of isotopes, atomic number, mass number, nuclides, and nuclide symbols — all of which are extensively used in the material that follows.
Atomic nuclei are composed of two basic kinds of particles: protons and neutrons (the two are known generically as nucleons.) These two particles have almost identical masses, and differ only in their electrical charges; their properties, along with those of the electron which balances the electric charge of the nucleus in the complete atom, are shown below.
Particle | Symbol | Charge | Mass (10–27 kg) |
Molar mass grams/mol |
Mass number and spin |
---|---|---|---|---|---|
proton | p | +1 | 1.672622 | 1.007276 | 1, ½ |
neutron | n | 0 | 1.674927 | 1.008665 | 1, ½ |
electron | e | –1 | 0.000911 | 0.000546 | 0, ½ |
From the standpoint of Chemistry, we usually think of protons as components of the atomic nucleus, whose number Z defines a particlar element. The chemical symbol of the proton is H+.
By far the most important chemistry-related use of the proton is in proton nuclear magnetic resonance ("proton NMR"), in which observations are made of the extent to which the nuclear spins of the protons within hydrogen atoms (usually in organic molecules) are affected by the spins of nearby electrons. This is not only an important tool for investigating the chemical environments of different hydrogen atoms (and thus deducing the structures of molecules), but it also serves as the basis of the medical imaging technique known as MRI.
Protons in Chemistry Although the H+ symbol is widely used to denote the hydronium ion H3O+(aq) in acidic solutions, the bare proton cannot exist in aqueous solutions owing to its high charge density. For the same reason, protons are highly reactive chemically; when injected into matter, they combine with the electron cloud of any molecule, usually creating a "free radical" in the process. This property of protons makes them useful in cancer treatment, as described below.
Protons in medicine Protons, produced in accelerators, can be focussed electromagnetically into very narrow beams that can destroy cancer tissue in a much more controlled way than X-rays. Unlike X-ray photons, protons possess mass that deliver kinetic energy that can concentrated at the site of the tumor, while causing minimal damage to tissues ahead of or beyond it.
Protons in nature Free protons are formed in stellar atmospheres ("plasmas") where the temperature is sufficiently high (over 106 K) to shift the endothermic process H = H+ + e– well to the right. Thus protons are a major component of the solar wind that bathes the earth. The other significant source is cosmic radiation ("cosmic rays"), a stream of electrons, protons and other nuclei having extremely high kinetic energies. Protons account for about 90% of these particles, which originate outside the solar system. Most of these high energy particles, which can damage microelectronic chips and DNA, are fortunately deflected by the earth's magnetic field, so that only a minority of them reach the surface.
Free neutrons, unlike protons, are inherently unstable, undergoing beta decay
n → p+ + e– with a half-life of about 10 minutes.
Although neutrons are without electric charge, their mass enables them to carry sufficient kinetic energy to be considered "ionizing radiation".
Neutrons in Chemistry There are two important chemical applications of neutrons:
Owing to their short lifetimes, neutrons must be generated at their sites of use. Small neutron generators, often of table-top size, are used for routine laboratory work. These consist of a unit that ionizes deuterium and tritium, followed by an an accelator that directs these ions to a target containing D or T. The most widely-employed arrangement carries out the reaction
D + T → n + He4.
Neutrons in industry The primary industrial use of neutrons is the manufacture of radioactive isotopes for use as radiotracers in chemistry, and as medical imaging and cancer treatment applications. The neutrons for these purposes are usually produced by a neutron-producing fission process in a nuclear reactor.
Neutrons in the environment Cosmic particles interact with the atmosphere to generate highly-penetraring particles that in turn produce a small but constant flux of neutrons on (and within) the earth, and even deep into the oceans.
We commonly use the term nuclide to refer to a particular kind of nucleus — that is, one containing specific numbers of protons and neutrons. The number of protons it contains is denoted by Z, which of course corresonds to the total nuclear charge and thus to the atomic number of the element it belongs to. Thus the mass number A of the nucleus is given by
A = Z + N
in which Z is the atomic number and N is the neutron number.
A nuclide "X" has has a mass number of 32 and a neutron number of 16.
Write its symbol in standard nuclide notation.
Solution: The atomic number Z = A – N = 32 – 16 = 16; reference to a table of the elements (or to a periodic table) tells you that this is a nuclide of sulfur. it can be represented as 32S, S-32, or, in nuclide notation, as 16S32.
Note that the "16" is redundant, as that value of I is implied by the symbol S. Nevertheless, it is often convenient to show it anyway, especially when writing balanced equations for nuclear reactions.
Most of the elements found in nature consist of mixtures of isotopes. For example, S32 is the most common isotope of sulfur, accounting for 94.3 percent of the natural element. The other four natural isotopes of 16S are
S33 (.8%), S34 (4.3%) and S35 (0.02%).
These values are readily determined by mass spectrometry.
The number of stable (non-radioactive) isotopes of the elements varies from zero (in 22Tc and all elements above 82Pb), to a high of 10 in 50Sn.
This periodic table shows the number of stable isotopes of the elements, with different colors representing the numbers 0 to 7-or-more according to the legend near the top center. (The larger original Wikipedia image can be viewed here.) Note that
Among the elements having more then one stable isotope, there is a clear pattern in which elements of even atomic number have more stable isotopes than do those of odd atomic number.
Most of the elements are mixtures of two or more isotopes, but 27 of them (shown here in orange) exist as a single nuclide; these elements are said to be monoisotopic.
The proton and neutron are commonly regarded as distinctly separate particles having nearly identical masses and differing only in their electric charges. But shortly after neutron was discovered in 1932, the German physicist Werner Heisenberg showed, on theoretical grounds, that these two particles can be considered as two different states of a single "nucleon" particle.
Protons and neutrons were long thought to be "fundamental" particles in the sense that they are not composed of yet-smaller parts. But in the 1960s, a what became known as the "quark model" was proposed by the theoretical physicists Murray Gell-Mann and George Zweig. A few years later, experiments conducted at Stanford University revealed the evidence of some tiny point-like objects (now known as quarks) within these nucleons.
Present-day theory envisages a variety of quarks which possess quantum properties ("flavors") to which physicists have given whimsical names such as "color", "charm", and "strangeness". Quarks are spin-½ particles, and thus subject to the Pauli exclusion principle. This means that no two quarks within the same particle can have the same set of quantum numbers. (You will recall that electrons in atoms also face the same restrictions, which is essentially the basis of the periodic table.)
The two kinds of quarks present in nucleons are up-quarks and down-quarks; these carry fractional electric charges that of +2⁄3 e or –1⁄3 e respectively, where e is the electron charge. According to the rather exotic rules of quantum chromodynamics, the three quarks in nucleons must differ in the "color" quantum number, as indicated by the arbitrarily-chosen colors in the illustration below. Notice how the fractional electric charges combine to yield the charges +1 and 0 for the two kinds of nucleons.
The quarks are held together by the strong force (represented by the squiggly lines), the strongest of the four fundamental forces of nature described below.
But no matter how strongly particles are bound together, they can dissociate if the temperature is high enough; for the configurations of quarks that we know as nucleons, this temperature is around 1012 K. Temperatures of this magnitude are believed to have existed during the first microsecond following the birth of the universe — a period known as the quark epoch.
It is interesting that the masses of the quarks themselves are very small compared to those of a proton or neutron. The major source of a nucleon's mass is the mass-energy of the strong force that binds the quarks.
There are only four of these. The gravitational force is familiar to us all as the agent that prevents us from floating off the planet, or the Earth from drifting away from the Sun. But in spite of the massive scale at which it operates, gravity is by far the weakest of all the fundamental forces, and plays no role in the microscopic world of individual molecules or atoms.
The electromagnetic force is the source of electrostatic attraction and repulsion, and serves as the principal actor at the atomic and molecular levels. As such, this is the force we experience most intimately. It gives shape to our bodies and to the world around us; by governing the movement of electric charges, it forms the basis of all life processes and the working of our brains. And in the form of electromagnetic radiation, it provides the light that illuminates it all.
The gravitational and electromagnetic forces share the common property of exerting their effects over virtually unlimited distances, although their strenghs drop of with distance according to the inverse square law.
The remaining two fundamental forces, by contrast, operate only within the confines of the nucleus, so as important as they are in allowing matter to exist, we experience them only indirectly.
Older textbooks use the term strong force to describe what is now more properly known as the nuclear force (discussed below) that binds the neutrons and protons within the nucleus, and thus holds the nucleus together. Although this is technically correct, it does not quite convey the whole picture. The nuclear force is described separately in the next section.
What is now properly known as the strong force binds the three quarks of nucleons into the units that manifest themselves as neutrons and protons. Being by far the strongest of the four fundamental forces of nature, it is aptly named. But its full strengh is exerted only in the tiny distances between the point-like quarks.
Unlike the other three forces, the weak force does not attract or bind particles together, its strength being only about 10–7 that of the strong force. A more appropriate name for it might be the weak interaction.
From the standpoint of nuclear chemistry, the significance of the weak force is its ability to change the "flavor" of the quarks in nucleons, leading to the types of radioactive decay we know as beta-decay and electron capture.
Β–-decay is a very common type of radioactivity in nuclei having an excess of neutrons. Free neutrons, which have half-lifes of about 15 minutes, are also subject to this fate. Notice how electric charge is conserved in either process.
As everyone knows, like electric charges repel. So how can a collection of Z protons confined within the tiny volume of an atomic nucleus be prevented from flying apart? The obvious answer is that there must exist a previously unrecognized stronger attractive force that opposes and overcomes the proton-proton repulsion. This is known, appropriately enough, as the nuclear force, or more precisely as the residual strong force.
Prior to the discovery of quarks in the 1970s, the force that holds nuclei together was known as the "strong force". We now use this term to describe the force that binds quarks within nucleons. The binding of nucleons themselves within the nucleus is now called the nuclear force, or more precisely, the residual nuclear force.
The existence of some kind of a "strong force" capable of opposing the electrostatic repulsion between protons was postulated soon after the proton was discovered. This force, unlike the two described above, manifests itself only over the very short distances within the nucleus itself, and drops off completely beyond about 2.5 fm. Its strongest point (the bottom of the potential energy curve) occurs at about 1 fm, where the attractive force is around 25,000 N. At extremely small distances (below about 0.7 fm,) the force becomes repulsive, maintaining a minimum distance between nucleons and thus affecting the size of the nucleus itself.
[insert plot]
The nuclear force is a side-effect of the strong force, which acts mainly over the very short distances required to bind quarks together, and is thus unable to reach out beyond individual nucleons. However, random fluctuations in the distributions of quarks within a nucleon give rise to a net nucleon-nucleon attraction which is the direct source of the nuclear force.
This is similar in concept to the London or dispersion forces that derive from momentary unbalances in the electron distribution within electrically neutral molecules, and which act to loosely bind them together in solids and liquids.
The strength of the nuclear force is only a tiny fraction of that of the strong force. It is also known by the more descriptive name residual nuclear force.
In conventional chemistry, the "binding energy" of a molecule is defined by the quantity of energy released when the molecule is formed from its elements. Thus when hydrogen and oxygen combine to form one mole of liquid water at 298 K, this exothermic process releases 268 kJ of thermal energy into the surroundings:
H2 + ½ O2 → H2O ΔH° = –286 kJ
This means that in order to break H2O up into its elements, which would be the reverse of this reaction, 286 kJ of energy must be supplied to the system, so we can define this as "the binding energy of a mole of H2O":
H2O → H2 + ½ O2 ΔH° = +286 kJ
In the same way, we can define the binding energy of a nucleus in terms of the amount of energy required to break it up into its component nucleons. Taking deuterium (1H2) as an example, we could write
1H2 → 1H1 + 1n0 ΔE = 212,000,000 kJ
Aside from the obvious difference in the magnitudes of these two binding energies, there is an important difference in the way we treat these two kinds of decomposition reactions. In chemical processes involving atoms and molecules, we always consider mass to be conserved. But in nuclear reactions, the huge energy changes require that we take into account the mass-equivalents of these energy changes.
According to the special theory of relativity, energy has mass. To find out how much, we can rearrange the famous equation E = mc2 into the form E/m = c2, in which c is the velocity of light, approximately 3.0 × 108 m sec–1. If E is in joules and m in kilograms, then
E/m = 9.0 × 1016 m2 sec–2 = 9.0 × 1016 J kg–1 = 9.0 × 1015 kJ kg–1
Calculate the change in mass of the system when
a) one mole of water at 298 K is formed from its elements. ΔH = –286 kJ
b) one mole of deuterium nuclei is formed from its component nucleons.
Solution:
a) In this exothermic reaction, the energy lost to the surroundings reduces the mass of the system by the mass-equivalent 286 kJ.
Using the E/m ratio of 9.0E15 kJ kg–1, we have (286 kJ) / (9.0E15 kJ kg–1) =3.5E-30 kg ... a loss in mass that would be too small to observe by many orders of magnitude.
b) The formation 1H2 is exothermic by 2.12E8 kJ mol–1. Proceding as above,
(2.12E8 kJ mol–1) / (9.0E15 kJ kg–1)
= 2.4E–7 kg or an easily observable 0.24 g.
It is common practice in many areas of physics to express energies in electron-volts (eV) instead of the SI unit kJ. The electron volt is defined as the energy required to move a unit [electron] charge through a potential difference of 1 volt. Recalling that 1 volt = 1 joule per coulomb and the electron charge is 1.602 × 10–19 C, then
The equivalance 1 eV = 1.602 × 10–19 J refers to the kinetic energy acquired by a single electron that has been accelerated through a potential difference of one volt. The corresonding value for one mole of electrons is just Avogadro's number times this, or 96.4 kJ.
Thus on a molar basis, 1 eV is comparable to the energy of a typical chemical bond.
1 eV = (1.602 × 10–19 C)(1 J C–1) = 1.602 × 10–19 J
The energy 2.12E8 kJ lost when 1 mole of 1H2 is formed from its nucleons corresonds to (2.12E8 kJ) / (1.60E–21 kJ eV–1) = 1.3 × 1029 eV — which is an awful lot of eV's!
For this reason, physicists find it more convenient to focus on individual particles, rather than on the chemists' moles of particles, so the convention in nuclear science is to divide these massive energy changes by Avogadro's number:
(1.3E29 eV mol–1) ÷ (6.02E23 mol–1) = 2.2E6 eV = 2.2 MeV
... so the MeV (millions of electron volts) is the common energy currency of nuclear science.
MeV's and mass But because mass and energy are essentially the same thing, the eV and MeV are also favored by particle physicists to express mass.
If you are not familiar with this "dimensions" stuff, have a look at this overview of Dimensions and dimensional analysis.
Equating energy and mass can be rather confusing to beginners, because energy has the dimensions of (mass)(length)2(time)–2 or ML2T–2, whereas the dimensions of mass are just plain M.
The way around this is to note that solving E = mc2 for m yields m = E/c2. So if we divide the dimensions of energy by c2, whose dimensions are L2T–2, we get rid of that pesky L2T–2 bit: ML2T–2 ÷ L–2T–2 = M and all is well with the world. But this means that particle masses should really be expressed in units of MeV/c2. And this is exactly what we do, but nobody wants to say the mass of the electron is 0.511 "mev-over-c-squared", so physicists have adopted the convention of dropping the "over-c-squared" part. But bear in mind that it's still there, but just in hiding.
Given that 1 eV = 1.609 × 10–19 J, calculate the mass-equivalent of a 1 GeV particle.
Solution: First, convert from eV to GEV: 1 GeV = 109 eV = 1.609 × 10–10 J
Important: As explained in the fine print above, the energy of the particle is more properly expressed as 1 GeV/c2, rather than "1GeV".
Expressed in mass, this becomes (1.609 × 10–10 J) /c2 =
(1.609 × 10–10 J) ÷ (9.0 × 1016 m2 s–2) = 1.79 × 10–27 kg
In nuclear chemistry, we are dealing mostly with atomic nuclei, so instead of expressing their masses in kilograms, we generally find it more convenient to employ a much smaller measure known as the unified atomic mass unit, denoted by the abbreviation u (Some older texts leave off the "unified" part and refer to it as the amu.)
1 atomic mass unit (1 u)
= 1.66053886 × 10–27 kg
= 931 Mev
= 1.49 x 10–30 J
The u is defined as 1/12 of the mass of one atom of C12; as such, its value of 1.66053886 x 10–27 kg is the mass-equivalent of each unit on the atomic weight scale. Fortunately, there is no need to memorize this number, because, when expressed in grams, it is just the reciprocal of Avogadro's number NA :
1 u = 1/NA gram = 1 ÷ (1000 NA) Kg
Calculate the energy equivalent of the unified atomic mass unit in MeV.
Solution:
From the previous problem example, a mass of 1.79E-27 kg has an energy of equivalent of 1.00 Gev. So 1 u will be 1.66/1.79 of 1 GeV:
(1.00 GeV)(1.66E–27 kg) / 1.79E–27 kg)
= 0.93 GeV
Atomic mass data for the elements have been measured to high precision by mass spectrometry. An especially convenient source, giving both atomic masses and abundances, can be found at this site.
Those nuclei that are not radioactive are thermodynamically stable; this means that a reaction in which their nucleons (protons and neutrons) combine to form the nucleus will be exothermic. Moreover, the magnitude of the energy that is released in such a process will be sufficiently large that the mass of the nucleus will be measurably smaller than that of the nucleons from which it is formed. This mass loss is commonly known as the mass defect.
Consider, for example, the formation of a deuterium nucleus from its components:
1H1 + 0n1 → 1H2
particle |
mass, u |
proton | 1.007276470 |
neutron | 1.008664904 |
1H2 | 2.014102 |
The experimentally-measured masses of the proton and neutron add up to
1.00726 u + 1.008665 u = 2.015941 u. Comparing this with the mass of a deuterium nucleus, it is apparent that the mass lost in this process, and thus the mass defect Δm = 2.015941 u – 2.014102 u = 0.001839 u
But of course, this "missing" mass has been released to the surroundings in the form of thermal energy, so in terms of mass-energy, the equation is balanced:
1H1 + 0n1 → 1H2 + energy
2.015941 u = 2.015941 u
You may have noticed that we have not included electrons in the above reckoning. If we are considering nuclei alone, electrons are not involved at all. But even if the equation represents conversion of a protium (1H1) atom into a deuterium atom, there are equal numbers of electrons on both sides, so they would cancel out. Since the actual materials employed in nuclear chemistry are basically atomic, most tables list atomic masses rather than nuclear masses, and for most purposes either can be used.
The mass defect was discovered by F.W. Aston in connection with his pioneering work with the mass spectrograph in the 1920's. Its theoretical basis was explained by George Gamow a few years later.
It is important to understand that this mass loss applies only to the system defined by the particles themselves; mass is not conserved in this system, but mass-energy is conserved if one takes into account the mass-equivalent of the energy that is released to the surroundings in the form of heat.
It should also be apparent that this same quantity of energy must be added to the deuterium nucleus in order to break it down into its components:
1H2 + energy → 1H1 + 0n1
As such, it corresponds to the binding energy of the 1H2 nucleus. Expressed in MeV, this works out to (.001839 u) (931 MeV/u) = 1.71 MeV.
Calculate the binding energy of the 25Mn55nucleus, whose mass is 54.9380 u.
Solution: This nucleus contains 25 protons and (55–25 = 30) neutrons, having a total mass of (25 × 1.007276 u) + (30 × 1.008665 u) = 55.4560 u. The mass defect of Mn55 is Δm = (55.4560 u – 54.9380 u) = 0.5180 u. The energy equivalent of this mass is (0.5180 u)(931 MeV/u) = 482 MeV, which is also the total binding energy of Mn55.
The total binding energy of 25Mn55 is considerably greater than that of 1H2 because the larger number of nucleons in the heavier nucleus affords more opportunity for the residual strong force which is the ultimate source of the nuclear binding energy.
A nuclide that has not been shown to undergo radioactive decay is said to be "stable". The formation of any nucleus from its component protons and neutrons is a highly exothermic process, and becomes moreso as the mass of the nucleus increases. Thus in this sense, all nuclei (even those that are radioactive) are thermodynamically stable.
But in nuclear science, the term "stable" is commonly used to describe a nuclide that has not been shown to undergo radioactive decay; all other nuclides are said to be "unstable."
And finally, we have "relative stability":
If we divide the total binding energy of a nuclide by the number of protons and neutrons it contains, we obtain what is known as the average binding energy per nucleon. This quantity represents the contribution that a [somewhat fictional] "average" nucleon in a particular nuclide makes to that nuclide's total binding energy. It enables us to compare the stabilities of different nuclei which would otherwise be masked by the general increase in the total binding energy with increasing mass. As such, it provides a clue some of the rules that govern the underlying structure of nuclei.
The overall shape of the average binding energy per nucleon plot is a direct consequence of the opposing effects of the extremely short range nuclear force and the longer-range (but far more weak) coulombic repulsion due to the protons.
It is worth taking some time to understand this chart; take particular note of the following points:
Despite the wealth of experimental results relating to the atomic nucleus that have accumulated since the discovery of the neutron in 1932, we still lack a thorough understanding of what is inside it and of how its component parts interact. Similar kinds of problems arise in many areas of physical science, and are best approached by constructing scientific models that can correlate and "explain" multiple observations. Most imporantly, such models must be testable.
Much of what we call "science" is based on models of some sort. It often happens that no single model is able to encompass all the aspects of a given system. This is certainly the case for the nucleus; each model can give us some insight into the nature of the nucleus, and can yield useful predictions, even while a complete understanding eludes us.
It is an experimental fact that a nucleus can be made by combining protons and neutrons; but the conventional picture that these particles retain their identities inside a nucleus is questionable.
If the nucleus is regarded as a roughly spherical collection of protons and neutrons (or of the corresponding kinds of quarks) in approximately equal numbers, the shape of the relative stability plot can be rationalized as follows:
Because this picture fails to explain anything beyond the general shape of the relative stability plot, it hardly qualifies as a generally useful model of the nucleus. It is, of course, useful for naming and classifying nuclides.
An elaboration of the "nucleus of nucleons" view, known as the liquid drop model, treats the nucleus as a drop of an incompressible fluid subject to surface tension and other effects.
Recall that in an ordinary liquid, the molecules near the surface are stabilized by van der Waals forces originating only from the interior of the liquid rather from all drections around them. The smaller the surface area of a liquid, the more energetically stable it is, so this effect, known as surface tension, causes a liquid not subject to other forces to minimize its surface area by assuming the shape of a spherical drop.
In the same way, a "liquid" made of nucleons (with the residual strong force substituting for van der Waals interactions) will form a drop-like shape.
In its mathematical form, the liquid drop model is expressed as a series of empirical terms relating the relative binding energy Eb to functions of the mass number A (and in some cases, the atomic number Z) to not only the surface area (proportional to A2/3) but also to the proton-proton coulombic repulsion. Additional terms take into account the Pauli exclusion principle (no two nucleons can have the same quantum numbers) and "pairing energy" that takes into account the greater stability of nuclides having even numbers of protons and neutrons.
This model yields results in good agreement with the observed plot for the heavier nuclei, but not for the very lightest ones. Its major weakness is that it is a purely phenomenological model that takes little account of the underlying quantum nature of nucleons.
A more detailed look at the plot of Eb as a function of mass number reveals a number of features that may remind one of the variations that are seen in many atomic properties such as atomic radius and ionization energy when these values are plotted as functions of atomic number.
At the smallest scale are the zigzags in the relative stabilities of nuclides having even and odd mass numbers, with the even ones generally having higher values of Eb.
This trend continues over the entire range of elements, and is reflected in numerous other criteria relating to nuclear stability. It could suggest that nucleons of like type tend to pair up, very much as electrons do in atoms.
This should not be all that surprising, given that nucleons, like electrons, are fermions having half-integral spins, which tend to pair up in accord with the Pauli exclusion principle.
Maria Goeppert Meyer (1906-1972) was born and educated in Germany, but spent most of her career in the U.S. For many of her early years in that country, she worked without pay, owing to strict university policies that prevented married couples from both working at the same universities (where her American-born husband was also a physicist.) Geoppert Meyer was the second woman to become a Nobel laureate, after Marie Curie.
On a larger scale, nuclides having proton or neutron numbers of 2, 8, 20, 28, 50, or 82, and also a neutron number of 128, are especially stable; these are the so-called magic numbers.
This fact was independently discovered by a group led by Hans Jensen, and later by Maria Goeppert Meyer. Goeppert Meyer, while investigating the origin of the elements, observed that nuclides having this sequence of numbers are especially abundant in the universe. She also found that other nuclear properties relating to stability followed the same trends.
Discovering the magic numbers and observing other regularities in nuclear properties was one thing, but working out the shape of a potential well that would yield a shell structure that corresponds to the observed periodicity of nuclear porperties was a far more difficult task which was not accomplished until Goeppert Meyer published her work in 1946. Goeppert Meyer and Jensen (who later became collaborators) were awarded the Nobel Prize in Physics in 1963.
Even more favored are those nuclides in which both the proton and neutron numbers are magic. These nuclei, which include 2He4, 8O16, 20Ca40 and
82Pb208 are said to be doubly magic.
The details of the shell model as they relate to the actual possible nuclear energy levels are rather complicated, and well beyond the level we can go into here. You can get a fairly understandable glimpse of the major ideas at this HyperPhysics page; this Florida State U. page offers a more comprehensive view without a lot of mathematics.
Make sure you thoroughly understand the following essential concepts that have been presented above.