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# Substructure of the Nucleus

## Substructure of the Nucleus

What is inside the nucleus? Why are some nuclei stable while others decay? (See this figure.) Why are there different types of decay ($$\alpha$$, $$\beta$$ and $$\gamma$$)? Why are nuclear decay energies so large? Pursuing natural questions like these has led to far more fundamental discoveries than you might imagine.

We have already identified protons as the particles that carry positive charge in the nuclei. However, there are actually two types of particles in the nuclei—the proton and the neutron, referred to collectively as nucleons, the constituents of nuclei. As its name implies, the neutron is a neutral particle ($$q=0$$) that has nearly the same mass and intrinsic spin as the proton. This table compares the masses of protons, neutrons, and electrons. Note how close the proton and neutron masses are, but the neutron is slightly more massive once you look past the third digit. Both nucleons are much more massive than an electron. In fact, $${m}_{p}=\text{1836}{m}_{e}$$ (as noted in Medical Applications of Nuclear Physics and $${m}_{n}=\text{1839}{m}_{e}$$.

This table also gives masses in terms of mass units that are more convenient than kilograms on the atomic and nuclear scale. The first of these is the unified atomic mass unit (u), defined as

$$\text{1 u}=1\text{.}\text{6605}×{\text{10}}^{-\text{27}}\phantom{\rule{0.25em}{0ex}}\text{kg.}$$

This unit is defined so that a neutral carbon $${}^{\text{12}}\text{C}$$ atom has a mass of exactly 12 u. Masses are also expressed in units of $$\text{MeV/}{c}^{2}$$. These units are very convenient when considering the conversion of mass into energy (and vice versa), as is so prominent in nuclear processes. Using $$E={\text{mc}}^{2}$$ and units of $$m$$ in $$\text{MeV/}{c}^{2}$$, we find that $${c}^{2}$$ cancels and $$E$$ comes out conveniently in MeV. For example, if the rest mass of a proton is converted entirely into energy, then

$$E={\text{mc}}^{2}=(\text{938.27 MeV/}{c}^{2}){c}^{2}=\text{938.27 MeV.}$$

It is useful to note that 1 u of mass converted to energy produces 931.5 MeV, or

$$\text{1 u}=\text{931.5 MeV/}{c}^{2}.$$

All properties of a nucleus are determined by the number of protons and neutrons it has. A specific combination of protons and neutrons is called a nuclide and is a unique nucleus. The following notation is used to represent a particular nuclide:

$${}_{Z}^{A}{\text{X}}_{N},$$

where the symbols $$A$$, $$\text{X}$$, $$Z$$ , and $$N$$ are defined as follows: The number of protons in a nucleus is the atomic number $$Z$$, as defined in Medical Applications of Nuclear Physics. X is the symbol for the element, such as Ca for calcium. However, once $$Z$$ is known, the element is known; hence, $$Z$$ and $$\text{X}$$ are redundant. For example, $$Z=\text{20}$$ is always calcium, and calcium always has $$Z=\text{20}$$. $$N$$ is the number of neutrons in a nucleus. In the notation for a nuclide, the subscript $$N$$ is usually omitted. The symbol $$A$$ is defined as the number of nucleons or the total number of protons and neutrons,

$$A=N+Z,$$

where $$A$$ is also called the mass number. This name for $$A$$ is logical; the mass of an atom is nearly equal to the mass of its nucleus, since electrons have so little mass. The mass of the nucleus turns out to be nearly equal to the sum of the masses of the protons and neutrons in it, which is proportional to $$A$$. In this context, it is particularly convenient to express masses in units of u. Both protons and neutrons have masses close to 1 u, and so the mass of an atom is close to $$A$$ u. For example, in an oxygen nucleus with eight protons and eight neutrons, $$A=\text{16}$$, and its mass is 16 u. As noticed, the unified atomic mass unit is defined so that a neutral carbon atom (actually a $${}^{\text{12}}\text{C}$$ atom) has a mass of exactly 12 $$\text{u}$$. Carbon was chosen as the standard, partly because of its importance in organic chemistry (see this appendix).

### Masses of the Proton, Neutron, and Electron

Let us look at a few examples of nuclides expressed in the $${}_{Z}^{A}{\text{X}}_{N}$$ notation. The nucleus of the simplest atom, hydrogen, is a single proton, or $${}_{1}^{1}\text{H}$$ (the zero for no neutrons is often omitted). To check this symbol, refer to the periodic table—you see that the atomic number $$Z$$ of hydrogen is 1. Since you are given that there are no neutrons, the mass number $$A$$ is also 1. Suppose you are told that the helium nucleus or $$\alpha$$ particle has two protons and two neutrons. You can then see that it is written $${}_{2}^{4}{\text{He}}_{2}$$.

There is a scarce form of hydrogen found in nature called deuterium; its nucleus has one proton and one neutron and, hence, twice the mass of common hydrogen. The symbol for deuterium is, thus, $${}_{1}^{2}{\text{H}}_{1}$$ (sometimes $$\text{D}$$ is used, as for deuterated water $${\text{D}}_{2}\text{O}$$). An even rarer—and radioactive—form of hydrogen is called tritium, since it has a single proton and two neutrons, and it is written $${}_{1}^{3}{\text{H}}_{2}$$. These three varieties of hydrogen have nearly identical chemistries, but the nuclei differ greatly in mass, stability, and other characteristics. Nuclei (such as those of hydrogen) having the same $$Z$$ and different $$N$$ s are defined to be isotopes of the same element.

There is some redundancy in the symbols $$A$$, $$\text{X}$$, $$Z$$, and $$N$$ . If the element $$\text{X}$$ is known, then $$Z$$ can be found in a periodic table and is always the same for a given element. If both $$A$$ and $$\text{X}$$ are known, then $$N$$ can also be determined (first find $$Z$$; then, $$N=A-Z$$). Thus the simpler notation for nuclides is

$${}^{A}\text{X},$$

which is sufficient and is most commonly used. For example, in this simpler notation, the three isotopes of hydrogen are $${}^{1}\text{H,}\phantom{\rule{0.25em}{0ex}}{}^{2}\text{H,}$$ and $${}^{3}\text{H,}$$ while the $$\alpha$$ particle is $${}^{4}\text{He}$$. We read this backward, saying helium-4 for $${}^{4}\text{He}$$, or uranium-238 for $${}^{\text{238}}\text{U}$$. So for $${}^{\text{238}}\text{U}$$, should we need to know, we can determine that $$Z=\text{92}$$ for uranium from the periodic table, and, thus, $$N=\text{238}-\text{92}=\text{146}$$.

A variety of experiments indicate that a nucleus behaves something like a tightly packed ball of nucleons, as illustrated in this figure. These nucleons have large kinetic energies and, thus, move rapidly in very close contact. Nucleons can be separated by a large force, such as in a collision with another nucleus, but resist strongly being pushed closer together. The most compelling evidence that nucleons are closely packed in a nucleus is that the radius of a nucleus, $$r$$, is found to be given approximately by

$$r={r}_{0}{A}^{1/3},$$

where $${r}_{0}=\text{1.2 fm}$$ and $$A$$ is the mass number of the nucleus. Note that $${r}^{3}\propto A$$. Since many nuclei are spherical, and the volume of a sphere is $$V=(4/3){\mathrm{\pi r}}^{3}$$, we see that $$V\propto A$$ —that is, the volume of a nucleus is proportional to the number of nucleons in it. This is what would happen if you pack nucleons so closely that there is no empty space between them.

Nucleons are held together by nuclear forces and resist both being pulled apart and pushed inside one another. The volume of the nucleus is the sum of the volumes of the nucleons in it, here shown in different colors to represent protons and neutrons.

### Example: How Small and Dense Is a Nucleus?

(a) Find the radius of an iron-56 nucleus. (b) Find its approximate density in $$kg/{m}^{3}$$, approximating the mass of $${}^{56}\text{Fe}$$ to be 56 u.

Strategy and Concept

(a) Finding the radius of $${}^{56}\text{Fe}$$ is a straightforward application of $$r={r}_{0}{A}^{1/3},$$ given $$A=56$$. (b) To find the approximate density, we assume the nucleus is spherical (this one actually is), calculate its volume using the radius found in part (a), and then find its density from $$\rho =\mathrm{m/V}$$. Finally, we will need to convert density from units of $$u/{fm}^{3}$$ to $$kg/{m}^{3}$$.

Solution

(a) The radius of a nucleus is given by

$$r={r}_{0}{A}^{1/3}.$$

Substituting the values for $${r}_{0}$$ and $$A$$ yields

$$\begin{array}{lll}r& =& {(1.2 fm)(56)}^{1/3}=(1.2 fm)(3.83)\\ & =& 4.6 fm.\end{array}$$

(b) Density is defined to be $$\rho =\mathrm{m/V}$$, which for a sphere of radius $$r$$ is

$$\rho =\cfrac{m}{V}=\cfrac{m}{(4/3){\mathrm{\pi r}}^{3}}.$$

Substituting known values gives

$$\begin{array}{lll}\rho & =& \cfrac{56 u}{(1.33)(3.14){(4.6 fm)}^{3}}\\ & =& 0.138 u/{fm}^{3}.\end{array}$$

Converting to units of $$kg/{m}^{3}$$, we find

$$\begin{array}{lll}\rho & =& (0.138 u/{fm}^{3})(1.66×{10}^{–27}\phantom{\rule{0.25em}{0ex}}\text{kg/u})\left(\cfrac{1 fm}{{10}^{–15}\phantom{\rule{0.25em}{0ex}}\text{m}}\right)\\ & =& 2.3×{10}^{17}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}.\end{array}$$

Discussion

(a) The radius of this medium-sized nucleus is found to be approximately 4.6 fm, and so its diameter is about 10 fm, or $${10}^{–14}\phantom{\rule{0.25em}{0ex}}\text{m}$$. In our discussion of Rutherford’s discovery of the nucleus, we noticed that it is about $${10}^{–15}\phantom{\rule{0.25em}{0ex}}\text{m}$$ in diameter (which is for lighter nuclei), consistent with this result to an order of magnitude. The nucleus is much smaller in diameter than the typical atom, which has a diameter of the order of $${10}^{–10}\phantom{\rule{0.25em}{0ex}}\text{m}$$.

(b) The density found here is so large as to cause disbelief. It is consistent with earlier discussions we have had about the nucleus being very small and containing nearly all of the mass of the atom. Nuclear densities, such as found here, are about $$2×{10}^{14}$$ times greater than that of water, which has a density of “only” $${10}^{3}\phantom{\rule{0.25em}{0ex}}{\text{kg/m}}^{3}$$. One cubic meter of nuclear matter, such as found in a neutron star, has the same mass as a cube of water 61 km on a side.
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