.. rtant 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.