Subsections


14.2 The Simplest Nuclei

This subsection introduces the simplest nuclei and their properties.


14.2.1 The proton

The simplest nucleus is the hydrogen one, just a single proton. It is trivial. Or at least it is if you ignore the fact that that proton really consists of a conglomerate of three quarks held together by gluons. A proton has an electric charge $e$ that is the same as that of an electron but opposite in sign (positive). It has the same spin $s$ as an electron, $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Spin is the quantum number of inherent angular momentum, chapter 5.4. Also like an electron, a proton has a magnetic dipole moment $\mu$. In other words, it acts as a little electromagnet.

However, the proton is roughly 2,000 times heavier than the electron. On the other hand the magnetic dipole moment of a proton is roughly 700 times smaller than that of an electron. The differences in mass and magnetic dipole moment are related, chapter 13.4. In terms of classical physics, a lighter particle circles around a lot faster for given angular momentum.

Actually, the proton has quite a large magnetic moment for its mass. The proton has the same spin and charge as the electron but is roughly 2,000 times heavier. So logically speaking the proton magnetic moment should be roughly 2,000 times smaller than the one of the electron, not 700 times. The explanation is that the electron is an elementary particle, but the proton is not. The proton consists of two up quarks, each with charge $\frac23e$, and one down quark, with charge $-\frac13e$. All three quarks have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. Since the quarks have significantly lower effective mass than the proton, they have correspondingly higher magnetic moments. Even though the spins of the quarks are not all aligned in the same direction, the resulting net magnetic moment is still unusually large for the proton net charge, mass, and spin.


Key Points
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The proton is the nucleus of a normal hydrogen atom.

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It really consists of three quarks, but ignore that.

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It has the opposite charge of an electron, positive.

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It has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

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It is roughly 2,000 times heavier than an electron.

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It has a magnetic dipole moment. But this moment is roughly 700 times smaller than that of an electron.


14.2.2 The neutron

It is hard to call a lone neutron a nucleus, as it has no net charge to hold onto any electrons. In any case, it is somewhat academic, since a lone neutron disintegrates in on average about 10 minutes. The neutron emits an electron and an antineutrino and turns into a proton. That is an example of what is called “beta decay.” Neutrons in nuclei can be stable.

A neutron is slightly heavier than a proton. It too has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$. And despite the zero net charge, it has a magnetic dipole moment. The magnetic dipole moment of a neutron is about two thirds of that of a proton. It is in the direction opposite to the spin rather than parallel to it like for the proton.

The reason that the neutron has a dipole moment is that the three quarks that make up a neutron do have charge. A neutron contains one up quark with a charge of $\frac23e$ and two down quarks with a charge of $-\frac13e$ each. That makes the net charge zero, but the magnetic dipole moment can be and is nonzero.


Key Points
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The neutron is slightly heavier than the proton.

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It too has spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 1}\kern-.2em
/\kern-.2em\lower.4ex\hbox{\the\scriptfont0 2}\kern.05em$.

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It has no charge.

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Despite that, it does have a comparable magnetic dipole moment.

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Lone neutrons are unstable. They suffer beta decay.


14.2.3 The deuteron

The smallest nontrivial nucleus consists of one proton and one neutron. This nucleus is called the deuteron. (An atom with such a nucleus is called deuterium). Just like the proton-electron hydrogen atom has been critical for deducing the structure of atoms, so the proton-neutron deuteron has been very important in deducing knowledge about the internal structure of nuclei.

However, the deuteron is not by far as simple a two-particle system as the hydrogen atom. It is also much harder to analyze. For the hydrogen atom, spectroscopic analysis of the excited states provided a gold mine of information. Unfortunately, it turns out that the deuteron is so weakly bound that it has no excited states. If you try to excite it, it falls apart.

The experimental binding energy of the deuteron is only about 2.22 MeV. Here a MeV is the energy that an electron would pick up in a one-million voltage difference. For an electron, that would be a gigantic energy. But for a nucleus it is ho-hum indeed. A typical stable nucleus has a binding energy on the order of 8 MeV per nucleon.

In any case, it is lucky that that 2.22 MeV of binding energy is there at all. If the deuteron would not bind, life as we know it would not exist. The formation of nuclei heavier than hydrogen, including the carbon of life, begins with the deuteron.

The lack of excited states makes it hard to understand the deuteron. In addition, spin has a major effect on the force between the proton and neutron. In the hydrogen atom, that effect exists but it is extremely small. In particular, in the hydrogen atom ground state the electron and proton align their spins in opposite directions. That produces the so-called singlet state of zero net spin, chapter 5.5.6. However, the electron and proton can also align their spins in the same direction, at least as far as angular momentum uncertainty allows. That produces the so-called triplet state of unit net spin. For the hydrogen atom, it turns out that the triplet state has very slightly higher energy than the singlet state, {A.38}.

In case of the deuteron, however, the triplet state has the lowest energy. And the singlet state has so much more energy that it is not even bound. Almost bound maybe, but definitely not bound. For the proton and neutron to bind together at all, they must align their spins into the triplet state.

As a result, a nucleus consisting of two protons (the diproton) or of two neutrons (the dineutron) does not exist. That is despite the fact that two protons or two neutrons attract each other almost the same as the proton and neutron in the deuteron. The problem is the antisymmetrization requirement that two identical nucleons must satisfy, chapter 5.6. A spatial ground state should be symmetric. (See addendum {A.39} for more on that.) To satisfy the antisymmetrization requirement, the spin state of a diproton or dineutron must then be the antisymmetric singlet state. But only the triplet state is bound.

(You might guess that the diproton would also not exist because of the Coulomb repulsion between the two protons. But if you ballpark the Coulomb repulsion using the models of {A.40}, it is less than a third of the already small 2.22 MeV binding energy. In general, the Coulomb force is quite small for light nuclei.)

There is another qualitative difference between the hydrogen atom and the deuteron. The hydrogen atom has zero orbital angular momentum in its ground state. In particular, the quantum number of orbital angular momentum $l$ equals zero. That makes the spatial structure of the atom spherically symmetric.

But orbital angular momentum is not conserved in the deuteron. In terms of classical physics, the forces between the proton and neutron are not exactly along the line connecting them. They deviate from the line based on the directions of the nucleon spins.

In terms of quantum mechanics, this gets phrased a bit differently. The potential does not commute with the orbital angular momentum operators. Therefore the ground state is not a state of definite orbital angular momentum. The angular momentum is still limited by the experimental observations that the deuteron has spin 1 and even parity. That restricts the orbital angular momentum quantum number $l$ to the possible values 0 or 2, {A.39}. Various evidence shows that there is a quantum probability of about 95% that $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 0 and 5% that $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2.

One consequence of the nonzero orbital angular momentum is that the magnetic dipole strength of the deuteron is not exactly what would be expected based on the dipole strengths of proton and neutron. Since the charged proton has orbital angular momentum, its acts like a little electromagnet not just because of its spin, but also because of its orbital motion.

Another consequence of the nonzero orbital angular momentum is that the charge distribution of the deuteron is not exactly spherically symmetric. The probability of finding the proton is not independent of the direction from the center of gravity. It is said that the deuteron has a nonzero electric quadrupole moment. The asymmetric charge distribution affects how the deuteron interacts with electromagnetic fields.

Roughly speaking, you may think of the charge distribution of the deuteron as elongated in the direction of its spin. That is not quite right since angular momentum has uncertainty in direction. Therefore, instead consider the quantum state in which the deuteron spin has its maximum component, $\hbar$, in the chosen $z$-​direction. In that state, the charge distribution is elongated in the $z$-​direction.

The nonzero orbital momentum also shows up in experiments where various particles are scattered off deuterons.

To be sure, the precise probability of the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state has never been established. However, assume that the deuteron is modeled as composed of a proton and a neutron. (Although in reality it is a system of 6 quarks.) And assume that the proton and neutron have the same properties as they have in free space. (That is almost certainly not a good assumption; compare the next section.) For such a model the $l$ $\vphantom0\raisebox{1.5pt}{$=$}$ 2 state needs to have about 4% probability to get the magnetic moment right. Similar values can be deduced from the quadrupole moment and scattering experiments, [30].


Key Points
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The deuteron consists of a proton and a neutron.

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It is the simplest nontrivial nucleus. The diproton and the dineutron do not exist.

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It has spin 1 and even parity. The binding energy is 2.225 MeV.

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There are no excited states. The ground state is all there is.

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The deuteron has a nonzero magnetic dipole moment.

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It also has a nonzero electric quadrupole moment.


14.2.4 Property summary

Table 14.1 gives a summary of the properties of the three simplest nuclei. The electron is also included for comparison.


Table 14.1: Properties of the electron and of the simplest nuclei.
\begin{table}\begin{displaymath}
\setlength{\arraycolsep}{10pt}
\begin{array}...
...}{2m_{\rm p}} = 5.051\;10^{-27}\mbox{ J/T}
\end{displaymath}
\end{table}


The first data column gives the mass. Note that nuclei are thousands of times heavier than electrons. As far as the units are concerned, what is really listed is the energy equivalent of the masses. That means that the mass is multiplied by the square speed of light following the Einstein mass-energy relation. The resulting energies in Joules are then converted to MeV. An MeV is the energy that an electron picks up in a 1 million voltage difference. Yes it is crazy, but that is how you will almost always find masses listed in nuclear references. So you may as well get used to it.

It can be verified from the given numbers that the deuteron mass is indeed smaller than the sum of proton and neutron masses by the 2.225 MeV of binding energy. It is a tenth of a percent, but it is very accurately measurable.

The second column gives the charge. Note that all these charges are whole multiples of the proton charge $e$. However, that is not a fundamental requirement of physics. In particular, up quarks have charge ${\textstyle\frac{2}{3}}e$ while down quarks have charge $-{\textstyle\frac{1}{3}}e$. The proton contains two up quarks and a down one, producing net charge $e$. The neutron contains one up quark and two down ones, producing zero net charge.

The third column gives the charge radius. That is a measure of the spatial extent of the charge distribution. The electron is, as far as is known, a point particle with no internal structure. For the neutron, with no net charge, it is not really clear what to define as charge radius.

The fourth column shows the quantum number of net angular momentum. For the first three particles, that is simply their spin. For the deuteron, it is the nuclear spin. That includes both the spins and the orbital angular momenta of the proton and neutron that make up the deuteron.

The fifth column is parity. It is even in all cases. More complicated nuclei can have negative parity.

The sixth column is the magnetic dipole moment. It is expressed in terms of the so-called nuclear magneton $\mu_{\rm N}$. A proton circling around with one quantum unit of orbital angular momentum has one nuclear magneton of magnetic moment due to its motion. (Which would be in addition to the intrinsic magnetic moment listed in the table. Note that magnetic moments are vectors like angular momenta; they may cancel each other when summed.)

As the table shows, nuclei have much smaller magnetic moments than electrons. That is due to their much larger masses. However, using a magnetic field of just the right frequency, nuclear magnetic moments can be observed. That is how nuclear magnetic resonance works, chapter 13.6. Therefore nuclear magnetic moments are important for many applications, including medical ones like MRI.

The last column lists the electric quadrupole strength. That is a measure for the deviation of the nuclear charge distribution from a spherically symmetric shape. It is a complicating factor in nuclear magnetic resonance. Or an additional source of information, depending on your view point. Nuclei with spin less than 1 do not have electric quadrupole moments. (That is an implied consequence of the relation between angular momentum and symmetry in quantum mechanics.)

Note that the SI length unit of femtometer works very nicely for nuclei. So, since physicists hate perfection, they define a new non-SI unit called the barn b. A barn is 100 fm$\POW9,{2}$. So you will likely find the quadrupole moment of the deuteron listed as 0.00286 $e$b. Note the additional leading zeros. Some physicists do not like them, for good reason, and then use millibarn, giving 2.86 $e$mb. However, the quadrupole moments for many heavy nuclei are quite large in terms of millibarn. For example, einsteinium-253 has around 6,700 $e$mb. Anytime now, physicists are bound to figure out that centibarn works even better than millibarn. The charge $e$ is commonly left away from the definition of the quadrupole moment.


Key Points
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The properties of the simplest nuclei are summarized in table 14.1.