# 11.1 Nuclear structure and stability

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By the end of this section, you will be able to:
• Describe nuclear structure in terms of protons, neutrons, and electrons
• Calculate mass defect and binding energy for nuclei
• Explain trends in the relative stability of nuclei

Nuclear chemistry is the study of reactions that involve changes in nuclear structure. The chapter on atoms, molecules, and ions introduced the basic idea of nuclear structure, that the nucleus of an atom is composed of protons and, with the exception of ${}_{1}^{1}\text{H},$ neutrons. Recall that the number of protons in the nucleus is called the atomic number (Z) of the element, and the sum of the number of protons and the number of neutrons is the mass number (A). Atoms with the same atomic number but different mass numbers are isotopes of the same element. When referring to a single type of nucleus, we often use the term nuclide    and identify it by the notation ${}_{\text{Z}}^{\text{A}}\text{X},$ where X is the symbol for the element, A is the mass number, and Z is the atomic number (for example, ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}\right).$ Often a nuclide is referenced by the name of the element followed by a hyphen and the mass number. For example, ${}_{\phantom{\rule{0.5em}{0ex}}6}^{14}\text{C}$ is called “carbon-14.”

Protons and neutrons, collectively called nucleons , are packed together tightly in a nucleus. With a radius of about 10 −15 meters, a nucleus is quite small compared to the radius of the entire atom, which is about 10 −10 meters. Nuclei are extremely dense compared to bulk matter, averaging 1.8 $×$ 10 14 grams per cubic centimeter. For example, water has a density of 1 gram per cubic centimeter, and iridium, one of the densest elements known, has a density of 22.6 g/cm 3 . If the earth’s density were equal to the average nuclear density, the earth’s radius would be only about 200 meters (earth’s actual radius is approximately 6.4 $×$ 10 6 meters, 30,000 times larger). [link] demonstrates just how great nuclear densities can be in the natural world.

## Density of a neutron star

Neutron stars form when the core of a very massive star undergoes gravitational collapse, causing the star’s outer layers to explode in a supernova. Composed almost completely of neutrons, they are the densest-known stars in the universe, with densities comparable to the average density of an atomic nucleus. A neutron star in a faraway galaxy has a mass equal to 2.4 solar masses (1 solar mass = ${\text{M}}_{\text{☉}}$ = mass of the sun = 1.99 $×$ 10 30 kg) and a diameter of 26 km.

(a) What is the density of this neutron star?

(b) How does this neutron star’s density compare to the density of a uranium nucleus, which has a diameter of about 15 fm (1 fm = 10 –15 m)?

## Solution

We can treat both the neutron star and the U-235 nucleus as spheres. Then the density for both is given by:

$d=\phantom{\rule{0.2em}{0ex}}\frac{m}{V}\phantom{\rule{5em}{0ex}}\text{with}\phantom{\rule{5em}{0ex}}V=\phantom{\rule{0.2em}{0ex}}\frac{4}{3}\pi {r}^{3}$

(a) The radius of the neutron star is $\frac{1}{2}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.2em}{0ex}}\text{26 km}=\phantom{\rule{0.2em}{0ex}}\frac{1}{2}\phantom{\rule{0.3em}{0ex}}×\phantom{\rule{0.2em}{0ex}}2.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m}=1.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m},$ so the density of the neutron star is:

$d=\phantom{\rule{0.2em}{0ex}}\frac{m}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{m}{\frac{4}{3}\pi {r}^{3}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{2.4\left(1.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)}{\frac{4}{3}\pi {\left(1.3\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{3}}\phantom{\rule{0.2em}{0ex}}=5.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{17}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$

(b) The radius of the U-235 nucleus is $\frac{1}{2}\phantom{\rule{0.3em}{0ex}}×15\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}\phantom{\rule{0.2em}{0ex}}\text{m}=7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}\phantom{\rule{0.2em}{0ex}}\text{m},$ so the density of the U-235 nucleus is:

$d=\phantom{\rule{0.2em}{0ex}}\frac{m}{V}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{m}{\frac{4}{3}\pi {r}^{3}}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{\text{235 amu}\left(\frac{1.66\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-27}\text{kg}}{\text{1 amu}}\right)}{\frac{4}{3}\pi {\left(7.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}\text{m}\right)}^{3}}\phantom{\rule{0.2em}{0ex}}=2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{17}\phantom{\rule{0.2em}{0ex}}{\text{kg/m}}^{3}$

These values are fairly similar (same order of magnitude), but the nucleus is more than twice as dense as the neutron star.

## Check your learning

Find the density of a neutron star with a mass of 1.97 solar masses and a diameter of 13 km, and compare it to the density of a hydrogen nucleus, which has a diameter of 1.75 fm (1 fm = 1 $×$ 10 –15 m).

## Answer:

The density of the neutron star is 3.4 $×$ 10 18 kg/m 3 . The density of a hydrogen nucleus is 6.0 $×$ 10 17 kg/m 3 . The neutron star is 5.7 times denser than the hydrogen nucleus.

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