# 31.3 Substructure of the nucleus

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## Learning objectives

By the end of this section, you will be able to:

• Define and discuss the nucleus in an atom.
• Define atomic number.
• Define and discuss isotopes.
• Calculate the density of the nucleus.
• Explain nuclear force.

The information presented in this section supports the following AP® learning objectives and science practices:

• 3.G.3.1 The student is able to identify the strong force as the force that is responsible for holding the nucleus together. (S.P. 7.2)

What is inside the nucleus? Why are some nuclei stable while others decay? (See [link] .) 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. [link] 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 ${m}_{n}=\text{1839}{m}_{e}$ .

[link] 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

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

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 ,

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