Physics Bootcamp

Subsubsection 57.1.3.1 \(N/Z\) Ratio as a Measure of Stability

There are 254 known stable nuclides and hundreds of unstable nuclides, some occurring naturally while others made in the laboratory. The unstable nuclides are also called radionuclides. To get a better understanding of the type of nucleus that would be stable, in Figure 57.1.2 we plot all the stable nuclides found in nature in a diagram with the neutron number on one axis and the proton number on the other axis. The figure shows that small stable nuclides with \(A \leq 21\) have a balance between neutrons and protons with \(N=Z\text{.}\) As \(Z\) of the nuclei gets larger, the repulsive electric force between the protons increases and it takes additional neutrons to keep them sufficiently separated so as to form stable nuclei. Therefore, stable larger \(Z\) nuclides have \(N>Z\text{.}\)

There is no stable nuclei for neuclides that have \(Z>83\text{.}\) Atoms of all the elements beyond Bismuth, \(Z=83\) are unstable and decay through one or other mechanisms of radioactivity to be discussed later in the chapter. Even the Bi isotope \(^{209}_{\ 83}\)Bi, which was long thought to be stable, was found to decay with a half life of \(1.9\times 10^{19}\) yr, which is much longer than the age of the universe.

Subsubsection 57.1.3.2 Evenness of \(Z\text{,}\) \(N\text{,}\) and \(A\) as measures of stability

Among the 254 or so stable nuclides observed in nature so far, 153 have even \(A\) and 101 have odd \(A\text{.}\) Of the 153 even \(A\) stable nuclides 148 have both \(Z\) and \(N\) even and only 5 have both both \(Z\) and \(N\) odd. Among the 110 odd \(A\) stable nuclides, 53 are even \(Z\) and odd \(N\) and 48 are odd \(Z\) and even \(N\text{.}\) These abundances suggest that evenness of \(Z\text{,}\) \(N\text{,}\) and \(A\) contributes to the stability of a nucleus.

Subsubsection 57.1.3.3 Binding energy as a measure of stability

It turns out that the mass of nuclei (except that of the hydrogen atom) is always less than the sum of the masses of the protons and neutrons contained in the nucleus. The difference in mass is called the missing mass. From the Einstein's relation,

we interpret the difference in mass to be the energy that must have been taken out of the system when putting them together to form the nucleus. In other words, this energy must be supplied if we were to separate the nucleons of the nucleus far apart from each other. This energy is called the binding energy \(E_b\) of the nucleus. For a nucleus of mass \(M_{\textrm{nuc}}\) with \(Z\) protons and \(N\) neutrons the binding energy will be

We can include the electrons with protons and same number of electrons with the atomic nucleus and write this in terms of mass \(m_H\) of H atom and the nuclide, \(M_{\textrm{atom}}\text{.}\)

From this binding energy we define binding energy per unit nucleon \(\mathcal{E}_b\) by dividing it by the number of nucleons.

Figure 57.1.3 shows a plot of binding energy per nucleon versus the mass number \(A\text{.}\) The plot shows that the highest binding energy per nucleon is near \(A = 60\text{.}\) In general, a nucleus that has higher binding energy per nucleon will be more stable than the nucleus with less.

Due to the higher binding energy per nucleon for intermediate size nuclei, if a larger nucleus, say \(A=120\text{,}\) splits into two nuclei of \(A=60\text{,}\) energy will be released. Similarly, if two smaller nuclei, say two nuclei of \(A = 30\) were to combine to form one nucleus with \(A = 60\text{,}\) energy will also be released. The process of breaking of nuclei is called fission while the process of combination of nuclei is called fusion.

Subsubsection 57.1.4.1 Nuclear Spin

The protons and neutrons also have spin angular momentum. The nuclear spin is denoted by \(I\) rather than \(S\text{.}\) The magnitude of the spin angular momentum to the spin quantum number is similar here as was the case for electronic spin. Both proton and neutron are spin \(\frac{1}{2}\) particles, meaning that the spin quantum number is \(\frac{1}{2}\) and the magnitude of their spin angular momentum is

Nuclear spin quantum number is also space-quantized. That is, the projection of the spin vector along any axis will take only distinct values. For spin 1/2, the quantum numbers for \(z\)-projections are \(+ 1/2\) and \(-1/2\) and corresponding angular momenta components are

The spins of the protons and neutrons in a nuclei add vectorially to give the total nuclear spin \(I\) of a nucleus. %When you have a proton and a neutron in a nucleus, you can have spin quantum number 0 or 1. If spin quantum number is 0, you would have the projection \(I_z\) just take one value 0. If the spin quantum number is \(1\text{,}\) the \(z\)-projection quantum number of the spin will have three values, viz., \(1, 0, -1\text{.}\) A state of total nuclear spin \(I\) has the spin angular momentum equal to

The space-quantization of the nuclear spin is similar to the space quantization of electron spin. Therefore, the projection of the nuclear spin vector along any axis has values from \(-I\hbar\) to \(+I\hbar\) in steps of \(\hbar\text{.}\) For instance, the projection on the \(z\)-axis of the nuclear spin \(I\) state will be

The value \(I\) is called the spin quantum number and the values \(-I, -I+1, \cdots, I-1, +I\) are called the spin quantum numbers of the \(z\)-projection of the spin.

For instance, the spin quantum number of the nucleus of deuterium (D) is 1 since the spins of the proton and neutron in a deuterium nucleus are pointed in the same direction. The nuclear spin angular momentum of the deuterium nucleus is

which can have three projections along any axis, say on \(z\)-axis, giving three possible values of the components of the nuclear spin angular momentum.

Some of the larger nuclei, e.g. \(^{13}_{\ 6}\)C, has 12 of the spins paired up leaving only one nucleon with unpaired spin. The spin of \(^{14}_{\ 7}\)N is 1 and that of \(^{23}_{11}\)Na is actually \(\frac{3}{2}\text{.}\)

Subsubsection 57.1.4.2 Nuclear Magnetic Moment

If a charged particle has a spin angular momentum \(\vec I\text{,}\) then it will have a magnetic dipole moment \(\vec \mu\) also.

where \(\gamma\) is the gyromagnetic ratio of the particle. For an electron, the gyromagnetic ratio is given by

where \(g\) is called the Land\'e \(g\) factor, which has a value close to 2. Similarly, the gyromagnetic ratios of a nucleus is defined similarly,

where \(m_p\) is the mass of a proton and \(g_n\) is the \(g\)-factor of the particular nucleus. The constants in this equation are combined to include \(\hbar\) and a new quantity called nuclear magneton is introduced for convenience.

In terms of \(\mu_N\text{,}\) the gyromagnetic ratios of a nucleus becomes

Putting this in Eq. (57.1.3) we get the relation between the magnetic dipole moment of a nucleus and its spin angular momentum vector .

The components of the magnetic dipole moment are related to the components of the spin angular momentum quantum number as

where \(I_z/\hbar\) are equal to the quantum numbers since we have divided out \(\hbar\text{.}\) The gyromagnetic ratio and g-factors of proton and neutron are

Due to the spin angular momentum and charge of the proton, proton has a magnetic dipole moment. The neutron also has a magnetic dipole moment; even though, neutron is overall neutral in charge, it is made up of particles, called quarks, which we will discuss later in the book, that are charged and have spin angular momentum as well. The magnetic dipole moments of the the proton and neutron are

where \(\mu_N\) is nuclear magneton given by

where \(m_p\) is the mass of a proton. Recall a similar unit of magnetic dipole moment used for the magnetic dipole moment of an electron. There we have the Bohr magneton, \(\mu_B\text{,}\)

where \(m_e\) is the mass of an electron. The negative value for the magnetic dipole moment of neutron is indicative of the fact that its spin and magnetic dipole moment are in opposite direction.

There can be three total spin states if a nucleus has three nucleons, eg. tritium. We can label them with their spin quantum numbers, which would be

You get a nucleus of total spin half if the spins of two of the nucleons are paired up and the spin of the other nucleon is unpaired, and you get total spin three-half when spins of all the nucleons are pointed in the same direction.

The protons and neutrons also have magnetic dipole moments. The magnetic dipole moment of proton is \(2.7928\mu_N\text{,}\) where \(\mu_N\) is nuclear magneton with a value \(5.05\times 10^{-27}\) J/T. The magnetic dipole moment of a neutron has a lower value, \(1.9135\mu_N\text{,}\) and points in the opposite direction than the magnetic dipole moment of the proton in an external magnetic field.

Subsubsection 57.1.4.3 Energy in a magnetic field

The magnetic dipole moments of the proton and neutron give rise to a net magnetic dipole moment of a nucleus. As is generally the case, a magnetic dipole \(\vec \mu\) experiences a torques when placed in a magnetic field \(\vec B\text{.}\)

You can also show that the energy of a magnetic dipole in an external magnetic field is given by

For concreteness, let the magnetic field be pointed towards the positive \(z\)-axis. Then, there will be no torque if the magnetic dipole \(\vec \mu\) points either towards the magnetic field or opposite to it. The energy of the dipole of proton is lowest when it is aligned with the field and that of the neutron is lowest when it is in the opposite direction to the field.

The energy of a proton \(U_p\) when it is aligned with the external field will be

when it is in the opposite direction to the field, it will be

Therefore, there will be two energy states for the proton when it is placed in an external magnetic field. The difference in the energy of the two states is \(\Delta E\text{,}\)

A proton can absorb a photon to make quantum transition between these states. The frequency \(f\) of the photon will be given by

where \(h\) is the Planck constant. Often we write this using this using angular frequency \(\omega\) rather than \(f\text{.}\)

where \(\hbar = h/2\pi\text{.}\) Classically, if the magnetic dipole vector is not pointed towards or opposite to \(\vec B\text{,}\) then it will experience a torque which will make the dipole moment vector rotate about the magnetic field. The frequency of precession is called the Larmor frequency and often denoted as \(\omega_L\text{.}\)

The precession of the magnetic dipole of the proton can be used to put energy to the proton nucleus and cause the spin to flip by exciting the precessing proton with a photon of the same frequency as the precession frequency. When the frequencies match, the proton absorbs energy from the photon field in its transition from the lower energy state to the higher energy state. This is the basis of the technique called Nuclear Magnetic Resonance (NMR).

The NMR techniques can be applied to any nucleus that has a net nuclear magnetic dipole moment. For instance, NMR of \(^{13}_{\ 6}\)C and \(^{31}_{15}\)P are commonly used in chemical analysis. Their Larmor frequencies are different from that of the proton given above. The Larmor frequencies of nuclear dipole precession also depend upon the chemical environment of the atom which has made NMR an invaluable tool for chemists.