Nucleosynthesis Essay, Research Paper

The “big bang” which created the universe, only created the elements Hydrogen (H) and Helium (He) and possibly a very small amount of Lithium (Li). However, a glance at the periodic table of the elements shows that today (some 15 billion years after the big bang) there are at least 108 known elements. Every atom of every element heavier than Li has been produced since the big bang! The “factories” which make these elements are stars. “Nucleosynthesis” or the synthesis of nuclei, is the process by which stars (which start out consisting mostly of H and He) produce all other elements.

The key is nuclear fusion, in which small nuclei are joined together to form a larger nucleus. (This contrasts with nuclear fission, in which a large nucleus breaks apart to form two smaller nuclei). Fusion requires an extremely large amount of energy (see fig. 1), and can typically only take place in the centers of stars.

FIGURE 1

a) Low energy proton is strongly repelled by the 7Be nucleus.b) High energy proton moves so fast that it can strike the 7Be nucleus. Once the proton touches the nucleus, it has a chance to stick. If the proton sticks, the 7Be becomes a 8B nucleus.c) 8B is radioactive and changes into 8Be plus a positron (b+) and a neutrino (n). 8Be is itself radioactive, and almost immediately breaks into two 4He nuclei.

Protons repel each other. This repulsion becomes stronger as the protons get closer together (just like when you try to stick two magnets together north to north, or south to south. Try this! As you push the magnets closer together, it becomes harder to do). However, if the protons can actually touch each other, they have a chance to stick together! This is because of the “strong nuclear force” which attracts nucleons (protons or neutrons) together, and is much stronger (at close range) than the “electromagnetic force” repulsion that makes protons repel other protons. (Magnets do not do this: two like poles will never stick together).

In order to get a proton to strike another proton (or a nucleus that contains several protons) they must be traveling at high relative speeds; if their “closing velocity” is not great enough, they will never get close enough to stick together, because they strongly repel each other. But, just as you can make two of the same magnetic poles touch each other by providing sufficient force, so too can protons “touch” when they have sufficient relative speed. This can take place in the center of the sun, where the temperature is extremely high. Temperature is related to atomic motion: the hotter something is, the faster its atoms are moving [] see demo “food coloring in water”[].

Table 1 shows the nuclear reactions that are taking place in our sun, as well as nuclear reactions that take place in stars that are either older than our sun, or hotter than our sun. The reactions in columns 2 and 3 occur after a star has entered the “red giant phase”. How fast a star evolves to this point depends on its mass: stars heavier than the sun can reach this phase in less than 5 billion years (the age of the sun) whereas stars with about our sun’s mass take about 10 billion years to get there. The particles you may be unfamiliar with are: n the neutrino, g a gamma ray (high energy light wave), and b+ the positron (the antimatter version of the electron).

TABLE 1. NUCLEAR REACTIONS IN STARS

OUR SUN NOWOLDER, OR HOTTER STARS

p + p ? 2H + b+ + n4He + 4He ? 8Be + g12C + p ? 13N + g

2H + p ? 3He + g8Be + 4He ? 12C + g 13N ? 13C + b+ + n

3He + 3He ? 4He + p + p12C + 4He ? 16O + gT1/2 = 10 min

16O + 4He ? 20Ne + g13C + p ? 14N + g

3He + 4He ? 7Be + g20Ne + 4He ? 24Mg + g14N + p ? 15O + g

7Be + p ? 8B + g 15O ? 15N + b+ + n

8B ? 8Be + b+ + nT1/2 = 120 ms

8Be ? 4He + 4He15N + p ? 12C + 4He

He burning (core)H burning shell

In our sun, the first three nuclear reactions {shaded} are the major source of energy. The second group (of four) reactions also occur in the sun, but much less frequently than the first group (which is called the p-p chain). In both cases, the fuel (hydrogen) is converted into the product (helium), and energy (in the form of heat and light) is produced. The third column of reactions is called the CNO cycle, because carbon (C), nitrogen (N) and oxygen (O) are produced and C is recycled. The CNO reaction cycle is now occurring in the sun (the energy required for these reactions is roughly the same as that required for the p-p chain) because the sun had carbon to begin with (it hasn’t made any C yet!). Since the sun had carbon present when it formed, it is referred to as a ‘later generation’ star. The ‘first generation’ of stars contained only H, He and Li from the big bang. Later generation stars contain material that has been processed in other stars.

As a star like the sun evolves, a vast amount of H is consumed, and a vast amount of 4He is produced. This 4He can not combine with other 4He because this reaction requires more energy than is available, i.e. the 4He nuclei are not moving fast enough. However, as the 4He accumulates in the star’s core, the pressure rises (which causes the temperature to rise since the core consists of gas). With increased temperature, more energy becomes available, and the star eventually reaches a point where the second column of above reactions can occur; this is called He burning. As we proceed down the second column, the reactions become increasingly less likely since they require increasing amounts of energy. This is because all nuclei have a positive electric charge, and like charges repel each other. The bigger the charge, the stronger the repulsion. This set of reactions can not produce any nucleus heavier than Fe (iron, atomic number 26). Again, a brief glance at the periodic table reveals that there are many elements heavier than Fe; these are also produced in stars, but not by any kind of fusion reaction. What takes us beyond Fe are two nucleosynthetic processes, called the ’s-process’ and the ‘r-process’.

The S-Process

Since a star’s H is usually not fully used up when He burning begins, the star’s energy comes from two distinct zones: the He burning core and the H burning “shell”. At the boundary between the two zones, material from both regions is free to mix together. Thus we can get the following reactions:

12C + p ? 13N + g14N + 4He ? 18F + g

13N ? 13C + e+ + n18F ? 18O + e+ + n

[13C + 4He ? 16O + n]18O + 4He ? 22Ne + g

[22Ne + 4He ? 25Mg + n]

The bracketed equations are important because in these reactions, a free neutron is liberated. Neutrons are not repell through a beta decay. This reaction is in the form: This process restores the balance between protons and neutrons in the nucleus. However, it also changes the chemical element (because the number of nuclear protons increases). This new element will also absorb neutrons until it reaches an unstable isotope, and then it will beta decay to yet a different element (see fig. 2). In this way, a star can produce elements heavier than Fe.

figure 2

The solid line shows the progression of the s-process starting from the seed nucleus 56Fe. We can get a rough idea of the neutron flux (the number of neutrons hitting a given location each second) by comparing the half-lives of ‘branching isotopes’ with non-branching isotopes. For instance, 69Zn has a half-life of 13.8 hours, and 75Ge has a half-life of 82.8 minutes, whereas 63Ni has a half-life of 100 years and 85Kr of 10.7 years. Thus neutrons are only absorbed very infrequently (probably on the order of weeks between absorptions).

An important parameter regarding an unstable nucleus is its half-life. This is the time during which the nucleus has a 50/50 chance of decaying. Another way of thinking of this is that if we have a large collection of a certain unstable atom, after a length of time equal to one half-life, half of these atoms will have decayed. For some of the unstable isotopes along the s-process path, the half-life is sufficiently long that some will absorb another neutron before they decay, the rest will simply decay. The s-process path is said to branch at these isotopes (see fig. 2).

Thus we see that in the s-process, neutron absorptions and beta decays cause an 56Fe nucleus to ‘march up the chart of the nuclides’ along the so-called ‘valley of beta stability’ which is simply the location of the stable isotopes in the chart of the nuclides. The reason it is called this may be clear from figure 2 in the “Radioactive Decay” module. The stable member(s) of each isobar are those that have minimum nuclear energy. The nuclei on either side of the stable nuclide have a higher energy. This makes it look like the stable nuclides lie in the bottom of a valley). Of course, not all of the 56Fe nuclei, or any other nuclei for that matter, are used up in this process, and new 56Fe (the so-called ’seed nucleus’) is continually made in He burning. Thus the star eventually has a distribution of nuclides between 56Fe and 209Bi (above 209Bi, a decays happen rapidly. These have the effect of regenerating Pb or Bi. Thus the s-process can not produce elements heavier than Bi).

Again, a brief glance at the periodic table reveals that there are elements heavier than Bi in nature. Also, we see from figure 2 that not all of the stable isotopes are produced in the s-process (for instance 70Zn, 76Ge, 82Se, etc., the nuclides to the right side of the s-process path) though these isotopes exist in our solar system. These isotopes, and the elements above Bi are produced in the ‘r-process’. (Other isotopes, like 58Ni, 74Se, 78Kr, etc., the nuclides to the left side of the s-process path, are produced in a third nucleosynthetic process, called the ‘p-process’. Because this process did not contribute much matter to the solar system (note the low abundances of these isotopes in the chart of the nuclides) it will not be discussed here).

The R-Process

Why can’t the s-process make isotopes like 70Zn? Basically it’s because there just aren’t that many free neutrons available, so they don’t ‘come around’ very often. When a 68Zn nucleus absorbs a neutron to become 69Zn, it has a chance to absorb another neutron to become 70Zn. However, with a half-life of 13.8 hours, by the time another neutron is absorbed, the 69Zn has already decayed to 69Ga. If 70Zn is to be produced (and we know it must be produced) we must increase the rate of neutron production. With more free neutrons available, the 69Zn would have a better chance of absorbing one. This is the r-process, the rapid addition of neutrons.

The word “rapid” is actually an understatement; it could be called “explosive”; the r-process occurs in supernova explosions! Here’s how it works: Before a supernova, a star has produced an excessive amount of 56Fe. This accumulates in the core (recall that we can’t go beyond 56Fe with fusion). As always, there is a battle between gravity (which tries to compact the core) and heat (which tries to expand the core). Eventually, after enough 56Fe is produced, gravity wins. When the Fe core collapses, it does so dramatically, and generates pressures which are truly incredible. The pressure is so great that the orbital electrons are pushed into their nucleus! Thus in one incredible electron capture reaction, all of the Fe in the core is converted to neutrons (1.4 solar masses worth!). An implosion shock wave reaches the core’s center and rebounds. As it does so, it sweeps vast numbers of neutrons out with it, and they smash into the matter above them. Now the neutron absorptions occur rapidly enough to bridge the gap out to nuclides like 70Zn. In fact, they happen rapidly enough to bridge the gap from Bi to the heavier elements (we know that 244Pu existed in the early solar system. This means that a 209Bi nucleus would have to absorb at least 35 neutrons before any a decay could occur!). Thus in one brief event, lasting at most only a few seconds, we produce all known elements heavier than Fe.

Appendix (applications)

There is an interesting application of s-process branching, through which we can roughly calculate the temperature that existed inside a red giant star even though the star exploded almost five billion years ago!

The key is that dust is produced in the atmospheres of red giant stars, and the unique isotopic distributions of the elements made in the star are “frozen in” as solids. These dust grains survive to the present day, preserved in primitive meteorites []see “Interstellar Grains” module[]. Let’s look at the specific example of 85Kr to see how this ‘remote thermometer’ works.

Our bodies, at a temperature of about 40 ?C (~100 ?F) give off infrared radiation which can be seen with special cameras. A log in a fire, at a temperature of about 600 ?C (~1100 ?F) glows red. Molten metal in a furnace, at a temperature of about 1500 ?C (~2700 ?F) shines with intense white light. Thus as temperature increases, the radiation (light) emitted becomes more energetic (changes color to shorter wavelengths) as well as more intense (more photons emitted per second). This is basically a result of the increased energy of the atomic collisions in the hot material []see “Blackbody Radiation” module[]. For temperatures characteristic of star cores (hundreds of millions of ?C) the collisions produce nuclear reactions as well as an abundant supply of high energy gamma rays. When these gammas are absorbed by a nucleus, they can make the nucleus transition to an excited energy state (just as visible or ultraviolet light can make an atomic electron transition to a higher orbital. This is the first step in making a laser beam).

As we saw in figure 1 of the “Radioactive Decay” module, 85Kr has a ‘metastable excited state’ which is only 0.305 MeV above the ground state (a fairly small energy when considering nuclear transitions). The temperature in the star will dictate how much of the 85Kr present will be in its excited state (the temperature determines the number of photons and their energy distribution. This together with the amount of 85Kr nuclei present in the star (which can be roughly calculated) gives the amount of 85Kr nuclei which should be in the first excited state.) But from figure 2 of the “Radioactive Decay” module, we see that this excited state has a much shorter half-life than the ground state (4.48 hours vs. 10.7 years) and that this excited state can decay directly to 85Rb. Thus the more 85Kr that can reach this excited state, the shorter its effective half-life will be. Finally figure 2 of this module shows that 85Kr is a branch point on the s-process path. The 10.7 year half-life of 85Kr is sufficiently long that many nuclei will absorb a neutron to become 86Kr. However, a half-life of 4.48 hours is not long enough to absorb another neutron before beta decay (which happens 79% of the time from this excited state) and will not produce 86Kr. Therefore, the amount of 86Kr present in a dust grain tells us what percentage of the 85Kr absorbed a neutron, which in turn tells us the temperature that existed inside the star.

This is a difficult idea. If you understood it, then you really understood this module