The Periodic Table: A nuclear view
Purpose: Students will learn how interactions among the protons and neutrons in the atomic nucleus affect the properties and stability of chemical elements, and how these properties could inform the creation of future elements.
Procedural overview: After reviewing the semi-empirical mass formula as a class, students will apply the knowledge to calculate nuclear binding energies and to understand the formula’s implications for nuclear reactions and nuclear decays.
Approximate class time: One class period.
Supplies:
Activity Guide for Students: The Periodic Table: A Nuclear View
Calculators (graphing calculators work best)
Optional: 3-D graphing program, such as Mathematica or Maple, and a computer projector
Directions for teachers:
Have your students read the article “Prospecting the periodic table,” in the March, 2019 issue of Science News, and ask them to think about what a future periodic table might look like. Explain that the class period will be used to explore the properties of atomic nuclei, including equations that explain how the makeup of the nucleus and interactions among protons and neutrons affect its stability.
Plan to introduce the full semi-empirical mass formula on the board and use the class discussion questions below to review the meaning of the different terms in the formula with your students. Finally, have students explore patterns in nuclear stability by applying the semi-empirical mass formula to the student questions provided below.
If you have access to a 3-D graphing program such as Mathematica or Maple, show the students 3-D plots of the semi-empirical mass formula (plot -EB/A on the vertical axis vs. Z and N on the other axes), or even have the students make 3-D plots themselves.
Background information:
The electrons orbiting the atomic nucleus participate in chemical reactions and govern the properties of elements. The protons and neutrons within the atomic nucleus participate in nuclear reactions and govern the nuclear stability and other nuclear properties of elements. The protons and neutrons within the nucleus follow rules that are similar to those that electrons follow.
Protons and neutrons are collectively termed nucleons. The number of protons is Z, the atomic number. The number of neutrons is N. The atomic mass A is Z + N, not counting tiny fractions removed for the binding energy (see explanation below).
Nucleons within a nucleus have a lower total energy than free nucleons. The amount by which the energy of nucleons in the nucleus is lower is the binding energy EB, which is defined here as a positive value (electron binding energies are measured in eV and nuclear binding energies are measured in MeV).
Since E = mc2, if the energy of the nucleus decreases by EB, the mass of the nucleus decreases by EB/c2. Nuclear reactions that change the binding energy can convert small fractions of the mass of the nucleus into relatively large amounts of energy.
The semi-empirical mass formula treats the nucleus as a collection of a large number of nucleons, so it is important to recognize that the resulting number is not very accurate for small nuclei, basically nuclei of the first six elements or so.
A simple version of the semi-empirical mass formula for the nuclear binding energy is:
EB (in MeV) = 16(A) – 18(A)(2/3) – 0.71(Z)(Z – 1)/(A)(1/3) – 24(N – Z)2/(A) + pairing term + shell correction
The pairing term:
if both N and Z are even numbers = 34(A)(-3/4)
if N is odd and Z is even, or vice versa (meaning A is odd) = 0
if both N and Z are odd numbers = -34(A)(-3/4)
The shell correction is approximately A/20 if N or Z is a stable number (2, 8, 20, 28, 50, 82 or 126) for a completely filled shell.
The binding energy per nucleon is EB/A. The larger that number is, the more stable the average nucleon is in a nucleus.
Class discussion questions:
1. The first term (16(A)) in the formula is the average binding energy of a nucleon with its nearest neighbors, due to the strong nuclear force. The strong force attracts protons and protons, neutrons and neutrons, and protons and neutrons, but it only acts at very short distances, basically for nearest neighbors. Often this first term is described as being proportional to the volume of the nucleus. Why does the nuclear volume increase in direct proportion to A?
Neglecting space between nucleons, if there are twice as many nucleons, the volume of the nucleus would be twice as large. So the volume of the nucleus is proportional to A. This relationship assumes that the number of nucleons is large enough that the nucleus can be treated as a large sphere of a given volume, not a lumpy shape of a handful of nucleons.
2. The second term (-18(A)(2/3)) in the formula corrects for those nucleons at the surface of the nucleus that do not have nearest neighbors on one side. The term can also be thought of as a surface tension term, which pulls the nucleus into a spherical shape just as surface tension pulls water droplets into a spherical shape. Why is this term negative, and why is it proportional to A(2/3)?
If the volume of the nucleus is proportional to the radius3 or A, the surface area of the nucleus is proportional to radius2 or A(2/3). The first term assumes that all nucleons have nearest neighbors on all sides, so it overestimates the binding energy. The second term is negative to subtract off that excess energy, in order to account for the fraction of nucleons that do not have nearest neighbors on one side.
3. The third term in the formula (0.71(Z)(Z – 1)/(A)(1/3)), often called the Coulomb term, is due to the electric repulsion among the positively charged protons in the nucleus. Why is it negative? Why is it proportional to (Z)(Z – 1) and inversely proportional to A(1/3)?
The strong force attraction among the nucleons makes the first term positive, but the electric repulsion among the protons opposes that, so the corresponding third term is negative. As the number Z of protons increases, the number of pairs of protons that repel each other increases like (Z)(Z – 1). (The 1 is subtracted because a proton cannot repel itself.) The repulsive energy between a pair of protons is inversely proportional to the distance between them, and the average distance increases like the nuclear radius (volume1/3), A(1/3), so the third term is inversely proportional to A(1/3) .
4. The fourth term in the formula (-24(N – Z)2/(A)), often called the asymmetry term, acts to try to make the nucleus have equal numbers of protons and neutrons. Review how the Pauli exclusion principle and the Aufbau principle apply to electrons. The principles also apply to protons and neutrons. Why does the nucleus want to have approximately equal numbers of protons and neutrons and why is the fourth term negative?
From the Pauli exclusion and Aufbau principles, protons cannot occupy the same state as each other, so they occupy higher and higher energy levels. The same thing happens with the neutrons, except they are in their own set of energy levels. If there are equal numbers of protons and neutrons, they fill their respective set of energy levels up to the same level. If there are more neutrons than protons, they fill the neutron energy levels to a higher level than the proton energy levels. In this case, the neutrons at the highest filled levels are inclined to become protons (via beta decay) and thereby qualify to move to a lower energy level. (The same reasoning applies if there are more protons than neutrons.) The asymmetry term is negative because as |N – Z| increases, the excess neutrons (or excess protons) are increasingly unhappy to be stuck at higher and higher energies, so they are less tightly bound. Thus, the binding energy decreases.
5. Just considering the fourth term (-24(N – Z)2/(A)) for a modest-sized atom with A = 24, how much energy would be released if the difference between neutrons and protons went from |N – Z| = 3, 2 or 1 to |N – Z| = 0? What does that suggest about the instability of certain nuclei and their likelihood to undergo beta decay?
For A = 24, going from |N – Z| = 3, 2 or 1 to |N – Z| = 0 would release 32 = 9, 22 = 4 or 12 = 1 MeV, respectively. Deviating from the preferred balance between protons and neutrons rapidly decreases the binding energy, making those nuclei quite unstable and thus likely to undergo beta decay. That instability helps to explain why elements are predominantly made up of an isotope that is at or near balanced numbers of protons and neutrons, with few or no isotopes that deviate significantly from that balance.
6. The fifth or pairing term is positive if both N and Z are even, negative if both N and Z are odd and 0 otherwise. What does this mean physically?
The pairing term:
if both N and Z are even numbers = 34(A)(-3/4)
if N is odd and Z is even, or vice versa (meaning A is odd) = 0
if both N and Z are odd numbers = -34(A)(-3/4)
The protons like to double up in pairs, with one spin-up and one spin-down. Likewise the neutrons like to double up in pairs, with one spin-up and one spin-down. The nucleus is happier (the binding energy is greater) if all the protons are in pairs and all the neutrons are in pairs. The nucleus is less stable (the binding energy is lower) if one proton is left unpaired and one neutron is left unpaired. The binding energy is intermediate if all the protons are paired but there is an unpaired neutron, or vice versa.
Student questions:
1. From examining the periodic table, what numbers of total electrons result in completely filled energy shells and therefore what are the most chemically stable elements?
From the last column of the period table, the numbers of total electrons that completely fill energy shells (and the corresponding elements) are 2 (He), 10 (Ne), 18 (Ar), 36 (Kr), 54 (Xe), 86 (Rn), etc.
2. There are also total numbers of protons that completely fill proton energy shells within the nucleus, and neutron totals that completely fill neutron energy shells. The stable numbers for protons or neutrons are different than those for electrons, since the forces within the nucleus are different. The numbers that result in stability, and high binding energy, for protons or neutrons are Z or N = 2, 8, 20, 28, 50, 82 and 126. Why do these number of protons and/or neutrons lead to stability? What elements and isotopes do the first three stable numbers (2, 8, 20) correspond to?
These numbers of protons and/or neutrons lead to stability because they completely fill shells within the nucleus. The first three stable numbers indicate that especially stable nuclei are helium-4 (Z = 2, N = 2), oxygen-16 (Z = 8, N = 8), and calcium-40 (Z = 20, N = 20).
3. The most stable nuclei (highest value of EB/A) are those located around iron and nickel on the periodic table. What is EB/A for iron-56?
EB (in MeV) = 16(56) – 18(56)(2/3) – 0.71(26)(26 – 1)/(56)(1/3) – 24(30 – 26)2/(56) + 34(56)(-3/4) + 0
EB/(56) = 510 MeV/56 = 9.1 MeV/nucleon
4. The semi-empirical mass formula is not designed to work well for very small nuclei, but for a rough estimate we can apply it to helium-4 and see what happens. What is EB/A for helium-4?
EB (in MeV) = 16(4) – 18(4)(2/3) – 0.71(2)(2 – 1)/(4)(1/3) – 24(2 – 2)2/(4) + 34(4)(-3/4)+ (4)/20
EB/(4) = 30 MeV/4 = 7.5 MeV/nucleon
5. The official value of EB/A for helium-4 is approximately 7 MeV per nucleon. How close was your answer? Approximately how much energy would be produced by fusing four protons together to form helium-4, as occurs in stars? (Two of the protons become neutrons.)
For helium-4, the official value of 7 MeV/nucleon is close to the calculated value of 7.5 MeV. Protons are free nucleons so they have zero binding energy. Fusing them to produce helium-4 would yield 7 MeV/nucleon, or 28 MeV per helium-4 nucleus.
6. What is EB/A for uranium-235? If the nucleons from a uranium-235 nucleus could split to form iron-56 nuclei, what would be the average energy released per nucleon, and the total energy released by the uranium nucleus? (In practice, 235 nucleons might divide up into nuclei that are somewhat larger or smaller than iron-56, but those new nuclei would have binding energies close to what you have already calculated for iron-56, so we will just use that.)
EB (in MeV) = 16(235) – 18(235) (2/3) – 0.71(92)(92 – 1)/ (235)(1/3) – 24(143 – 92)2/(235) + 0 + 0
EB/A = 1850 MeV/235 = 7.9 MeV
Going from uranium to iron would release 9.1 – 7.9 = 1.2 MeV/nucleon, or 1.2 MeV/nucleon x 235 nucleons/uranium = 280 MeV per uranium atom.
Note to the teacher: Uranium fission typically results in elements that are somewhat heavier and somewhat less tightly bound than iron, so it produces around 200 MeV per uranium atom. Without getting into those more detailed calculations, the simple calculation for students gives the right ballpark answer of 200-ish MeV per uranium atom.
7. What is EB/A for a hypothetical new element with Z = 120 and N = 180? If the nucleons from an element 120 nucleus could split to form iron-56 nuclei, what would be the average energy release per nucleon, or the total energy released by element 120 nucleus? (In practice, 300 nucleons might divide up into nuclei that are somewhat larger or smaller than iron-56, but those new nuclei would have binding energies close to what you have already calculated for iron-56, so we will just use that.)
EB (in MeV) = 16(300) – 18(300)(2/3) – 0.71(120)(120 – 1)/(300)(1/3) – 24(180 – 120)2/(300) + 34(300)(-3/4) + 0
EB/A = 2190 MeV/300 = 7.3 MeV
Going from element 120 to iron would release 9.1 – 7.3 = 1.8 MeV/nucleon, or
1.8 MeV/nucleon x 300 nucleons/atom = 540 MeV per atom.
Note to the teacher: As with uranium, the actual amount of energy released would be somewhat less, but this is still a very large amount of energy released and shows why such heavy elements are so unstable. If a heavy element can release around half a billion electron volts per atom by decaying, it will likely take the opportunity to get to a lower energy state.
8. Based on your calculations for uranium and element 120, what term in the semi-empirical mass formula makes nuclei increasingly unstable as they get very large? What is the physical explanation for that effect?
The third term, the Coulomb term, becomes a very large negative number, significantly decreasing the binding energy for very large nuclei. Physically, there are so many protons that the electric repulsion among them begins to overcome the nuclear strong force attraction among the protons and neutrons.
9. Maximizing EB for a given value of A gives the optimal fraction of nucleons that should be protons in order to achieve the greatest stability with respect to beta decay. The following equation, which gives the optimal fraction, can be derived from the semi-empirical mass formula:
Z/A = 0.5/(1 + 0.0074(A)(2/3))
For A = 16, what optimal fraction and number of protons does the equation predict? (Round your answer to a whole number.) What element is that?
Z/A = 0.48 or Z = 8, oxygen
10. For A = 235, what optimal fraction and number of protons does the equation predict? (Round your answer to a whole number.) What element is that?
Z/A = 0.39 or Z = 92, uranium
11. For A = 300, what optimal fraction and number of protons does the equation predict? (Round your answer to a whole number.) What element is that? From the “Long life” illustration in “Prospecting the periodic table,” for the optimal number of protons calculated, what N value might be the most stable? How does that compare with the N value when A = 300 and Z is the value calculated?
Z/A = 0.375 or Z = 113
According to the predicted island of stability, shown in the “Long life” illustration, if Z = 113, the most stable values of N would be from about 183 to 186. The estimate landed in the right ballpark as an N value of 187 and right outside the information provided in the illustration. The complex nuclear forces in larger, heavier element also make it difficult to predict exactly where the hoped-for “island of stability” would be, or just how stable the elements on it would be.
12. In general, what does this equation predict about protons and neutrons in light elements? What is the physical reason for that?
The numbers of protons and neutrons should be approximately equal, due to the asymmetry term.
13. In general, what does this equation predict about protons and neutrons in very heavy elements? What is the physical reason for that?
Neutrons should make up a larger and larger fraction of the nucleus. Extra neutrons are needed to provide extra strong force “glue” to overcome the increasing repulsion among the large numbers of protons.
14. What implications does this equation have for the composition and radioactivity of nuclei produced by fission reactions?
Stable forms of heavy nuclei such as uranium are very neutron-rich. When they split through fission, the resulting product nuclei will also be very neutron-rich. However, smaller nuclei do not like to be so neutron-rich, so they will be highly radioactive, undergoing lots of beta decay to convert some of their neutrons into protons.
15. What implications does this equation have for smashing smaller nuclei together as building blocks to create new very heavy elements?
New very heavy elements would need to be more neutron-rich than smaller nuclei would prefer to be. Thus it is difficult to find smaller nuclei that are sufficiently neutron-rich, or if you can find them, they rapidly undergo beta decay to be less neutron-rich, so you have to smash them together quickly before they decay. The article mentions that some of the experiments use calcium-48, a rare calcium isotope that has 20 protons but 28 neutrons. That makes it much more neutron-rich than normal calcium-40 (with just 20 neutrons). Whereas most smaller nuclei that are that neutron-rich would rapidly undergo beta decay, calcium-48 is unusually stable because it has stable numbers of both protons (Z = 20) and neutrons (N = 28).