Nuclear Physics
In the early part of the 1900’s Ernest Rutherford demonstrated that the atom consisted of a small, positive nucleus that contained most of the mass surrounded by electrons. In the previous chapter we discussed the configurations and energies of the electrons. In this chapter we discuss the makeup of the nucleus.
The nucleus of an atom consists of protons and neutrons. The proton has a charge e = 1.602 x 10-19 C and the neutron has no net charge. The nucleus is characterized by atomic number (Z), neutron number (N), and mass number (A = Z + N):
- Atomic number Z: number of protons in nucleus
- Neutron number N: number of neutrons in nucleus
- Mass number A: number of nucleons (neutrons and protons) in nucleus
Symbolically, we describe a nucleus as , where X is the element (e.g., , etc.)
The Z number determines the element. Elements with different N and A numbers are isotopes. For example, the isotopes of carbon are .
The mass number, A, which is given for the elements in the periodic table is generally not an integer. The reason is that it represents an average of the naturally occurring isotopes and it includes the mass equivalent of the binding energy of the nucleons.
Atomic mass unit (u)
The atomic mass unit (u) is defined as 1/12th the mass of . Since 12 grams of is a mole, or Avogadro’s number of atoms, the atomic mass unit is
The rest mass energy of 1u is
From the equation E = mc2, we can write
The masses of the proton and neutron are slightly larger than 1 u. This is because of the binding energy of the nucleus, as will be discuss later. The proton, neutron and electron masses are
Proton: m = 1.6726 x 10-27 kg = 1.007276 u = 938.28 Mev/c2
Neutron:m = 1.6750 x 10-27 kg = 1.008665 u = 938.57 Mev/c2
Electron:m = 9.109 x 10-31 kg = 5.486 x 10-4 u = 0.511 Mev/c2
Size of Nucleus
The size of the nuclei can be determined by experiments in which charged particles such as alpha particles (helium nuclei) are scattered from a material. The angular distribution of the scattering depends on the distance of closest approach of the alpha particles to the nuclei. The radii of the nuclei has been found to follow the relationship
,
where r0 = 1.2 x 10-15 m and A is the atomic number. This equation suggests that the densities of the nuclei of the different elements are all approximately the same.
The density of the nuclei can be estimated as follows:
Nuclear Stability
For small atomic numbers, the number of protons and neutrons in the nucleus is about the same. For example, . But for Z greater than about 20, the number of neutrons exceeds the number of protons (e.g., ). The figure below is a plot of the number of neutrons (N) versus the number of protons (Z). The solid points represent stable nuclei while the shaded area represents radioactive nuclei. The stability of the nucleus depends on competition between electrostatic repulsion between the protons and the attractive nuclear force between nucleons. Without the nuclear force the nucleus would fly apart due to the electrostatic repulsion. The nuclear force is a short-range force that exists mainly between nearest-neighbor nucleons; whereas, the electrostatic force is comparatively long range. If there were no neutrons in the nucleus, then the repulsive electrostatic force would be too great for stability. In a sense, neutrons help to spread the protons apart, reducing their repulsion. If there were no neutrons, then the electrostatic force on the outer protons would increase in proportion to the nuclear radius. If Q is the total nuclear charge and the nuclear charge density, then the force on an outer proton (e) would be
So, as the number of protons increases, there must be an increasing excess of neutrons to reduce the nuclear charge density.
Binding Energy of Nuclei
The mass of a nucleus is less than the sum of the masses of the individual protons and neutrons because of the binding energy of the nucleus. An amount of energy equal to the binding energy must be supplied to the nucleus to break it apart into individual protons and neutrons. The mass difference is related to the binding energy by the equation
Example:
Calculate the binding energy of deuterium.
Solution:
The deuterium nucleus consists of a proton and a neutron. So we find the difference between the sum of the masses of the protons and neutron and the mass of the deuterium nucleus (deuteron),
If we use the atomic mass for md, then this includes the mass of the nucleus and the mass of the electron. If we also use the atomic mass of hydrogen in place of the mass of the proton, then the mass of the electron will cancel out in the calculation. Then
Per nucleon, the binding energy is
Example:
Calculate the binding energy of .
Solution:
Per nucleon,
The binding energy per nucleon generally increases rapidly with increasing A up to about A ~ 60 and then more slowly decreases with further increase in A. This means that the nuclei with intermediate masses are the most stable compared with the light and heavy nuclei. The binding energy per nucleon is shown below as a function of A. This figure suggests that we can convert mass to energy by combining lighter nuclei to make nuclei of intermediate size (fusion) or breaking apart heavy nuclei into nuclei of intermediate size (fission).
Radioactive Decay
An unstable nucleus can transform to a more stable nucleus by emitting an alpha particle or a beta particle (electron or positron). A nucleus in an excited state can also emit gamma rays. The alpha particle is the nucleus of the helium-4 atom, . The positron is the antiparticle of the electron. Its mass is the same as that of the electron but it has a positive charge +e. A gamma ray is a high energy photon and has no mass or charge.
Alpha Decay
Alpha decay can be expresses symbolically as
Note that both the total mass number A and the atomic number Z are the same before and after the decay. An example of alpha-decay is the decay of radium-226 into radon-222:
Example:
Calculate the energy released in the alpha decay of radium.
Solution:
Since the alpha particle is much lighter than the radon nucleus, most of the released energy appears as the kinetic energy of the alpha particle.
Beta Decay
Beta decay can be expressed as
(negative beta decay)
or
(positive beta decay)
Note that the mass number doesn’t change, but the nuclear charge either increases (for electron emission) or decreases (for positron emission).
The emission of an electron is equivalent to the decay inside the nucleus of a neutron into a proton and an electron –
Similarly, the emission of a positron is equivalent to the conversion of a proton in the nucleus into a neutron and a positron.
A process related to beta decay is K-capture. In this process a nucleus ‘captures’ an orbital electron in the K-shell which combines with a proton to form a neutron. This process is expressed as
(K-capture)
Example:
Calculate the energy released in the beta decay of C-14 into N-14.
Solution:
The process is
We can neglect the mass of the electron on the right side of the expression if we use the atomic masses, since atomic N has one more electron than atomic C. Then
If the electron were the only particle emitted in this decay process, then essentially all of this energy would be imparted to the electron since its mass is small compared to the mass of the nitrogen nucleus. However, it is observed that most of the time the electron’s kinetic energy is significantly less than this amount. The observed distribution of kinetic energies is as shown in the figure on the next page.
This discrepancy led to the prediction by Pauli that a third particle must be involved to carry away the unaccounted for energy. In addition, the third particle was required to have a spin angular quantum number of ½ to account for conservation of angular momentum. This particle was expected to have no charge and a very small mass and was called the ‘neutrino’. It was eventually discovered in 1956. The complete decay process would be given as
is an anti-neutrino. In + beta decay processes, a neutrino () is emitted. It has recently been shown that there are several types of neutrinos and that they have a finite, but small, mass. Because of their small mass they travel very near the speed of light. Neutrinos are extremely abundant (the sun produces a large flux of neutrinos); however, they are extremely difficult to detect.
Gamma Decay
Quite often in alpha or beta decay or in nuclear reactions a nucleus is left in an excited stated, somewhat like an excited electronic state in an atom. The nucleus then decays to its ground state by emitting one or more high energy photons (gamma rays). This process is described as
Below is an example of gamma decay:
Penetrating power of radiation
The penetrating power of radiation depends on the type and energy. Gamma rays are generally the most penetrating since they have no charge to interact with the electrons and nuclear charges in matter. Several centimeters of lead would be required to provide safe shielding from a high energy gamma emitter source. Alpha particles are the least penetrating both because of their charge and their mass. They easily lose energy in air and can be shielded by thick paper. Beta particles have intermediate penetrating capacity.
Radioactive Decay Rates
The number of nuclei N that will decay in a time interval t is proportional to the number of radioactive nuclei N and the time interval,
By using calculus it can be shown that this leads to an exponential dependence of N and the decay rate N/t on time –
is the decay constant and has units of inverse time. The unit for decay rate, R, is the curie.
The half-lifeT1/2 is the time for one-half of the nuclei to decay. The relationship between the half-life and the decay constant can be obtained as follows.
Or,
The exponential decay formula is somewhat like that for the discharge of a capacitor through a resistor, where is analogous to the RC time constant.
Example:
Tritium (hydrogen-3) decays into helium-3 by beta decay with a half-life of 12.3 yrs:
What is the activity (decay rate) of 1 mg of tritium?
Solution:
Example:
A curie is a very large anddangerous amount of radioactivity. How long would one have to wait for the tritium activity to reduce to 1 mCi?
Solution:
Carbon Dating
CO2 in the earth’s atmosphere has a small amount of radioactive C-14. The C-14 is produced from N-14 in the upper atmosphere by neutrons which are generated in cosmic ray interactions :
C-14 is radioactive and decays back into N-14 through beta decay:
The lifetime of C-14 is 5730 yr, and the fraction of C-14 in the atmosphere is about 1 part in 1012. Living organisms ingest C-14 from the atmosphere and from ingesting other organisms and contain this same fraction of C-14. The C-14 in a living organism produces a decay rate of 15 decays/min·g.
When a living organism dies, it no longer ingests C-14 and its fraction of C-14 decreases with time. By comparing the decay rate of a dead organism with that of a live organism, the time of death can be determined.
Example:
20 g of carbon is extracted from a fossil and the C-14 activity is measured to be100 decays/min. What is the age of the fossil?
Solution:
The activity of 20-g of ‘live’ carbon would be
So,
Radioactive Series
All elements for which Z > 83 are radioactive. These elements decay through alpha or beta decay into lower Z elements which may also be radioactive. Eventually, they decay into an element which is stable. The various radioactive elements that are produced in this process can be classified into four series depending on the starting and ending nuclei. These four radioactive series are given listed as follows:
Series / Starting isotope / Half-life (yrs) / Stable end productUranium / / 4.47x109 /
Actinium / / 7.04x108 /
Thorium / / 1.41x1010 /
Neptunium / / 2.14x106 /
The figure below shows the Thorium Series.
Nuclear Reactions
In a nuclear reaction a target nucleus is struck by an energetic particle such as a neutron, alpha particle or another nucleus, and the particles are transformed into one or more other particles.
Examples:
Nuclear reactions can be either exothermic or endothermic, depending on whether the final kinetic energy is greater or smaller than the initial kinetic energy. The ‘Q-value’ of the reaction is the energy released (or absorbed). For exothermic reactions Q > 0; for endothermic Q < 0. The Q-value can be obtained from the difference in mass of the initial and final products.
A reaction with a negative Q requires that the incident particle have a minimum amount of kinetic energy in order for the reaction to occur. If the target particle is at rest, then the threshold kinetic energy for the reaction must exceed |Q| in order to conserve momentum. For example, if KEi = |Q|, then KEf = 0 and the final momentum is zero, so momentum is not conserved. By requiring that both energy and momentum be conserved, it can be shown that the threshold kinetic energy of the incident particle for an endothermic reaction to occur is
where m is the mass of the incident particle and M is the mass of the target particle.
Example:
Calculate the Q for the following reaction, and calculate the minimum energy of the incident neutron for the reaction to occur.
Solution:
1