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13. Nuclear Physics

Links to specific sections in the text:

13.a. Nuclear Composition, Forces, and Size
13.b. Nuclear Binding and Semiempirical Formula
13.c. Nuclear decays
13.d. Radioactivity

13.a. Nuclear Composition, Forces, and Size

Nuclei have been known to be composed of nucleons, neutrons and protons, since 1930. Neutrons and protons have almost equal masses m_p = 938.3 MeV/c², m_p = 939.6 MeV/c². Neglecting the less then 1% binding energies, the mass of nuclei is obtained from the total mass of neutrons and protons. The number of protons in a nucleus is Z, the atomic number characterizing elements. The number of protons, Z, equals to the number of electrons in a neutral atom, thus it determines chemistry. The mass number, A, is the total number of nucleons in the nucleus, A=Z+N, where N is the number of neutrons. The number of neutrons does not affect chemistry, but only nuclear mass, and thus physical density. Nuclei of the same element (Z values are equal) with different values of A, and consequently N, are called isotopes. Nuclei of different isotopes can have vastly different properies.

Numerous experiments (scattering and other) have shown that neutrons and protons interact with each other the same way as protons and protons (save the Coulomb repulsion) and neutrons and neutrons. Nuclear forces are oblivius of the charge of the nucleon. Nuclear forces are of very short distance. Their range is about 1 fermi= 10^-15m, five orders of magnitude smaller then the Bohr radius. At the same time nuclear forces are very strong. A fair comparision is given by the effective nuclear potential proposed by Yukawa and the Coulomb potential. The coulomb potential between two protons is

V_C(r) = ke²/r,

while the Yukawa potential has the form of

V_Y(r) = -g² e^-mr/r,

where m µ 1/1 fermi and g>>e. In other words, at large distances, r >>1/ m the Coulomb potential completely overwhelms the nuclear potential, at short distances the nu clear potential is much stronger and attractive, thus able to create a bound state of positively charged protons.

The very short range of nuclear forces determines the nature of nuclear binding about which we will say much more in the next section. The very short range of nuclear forces implies that if use the classical picture of nucleons that in the nucleus a nucleon only interacts with its neighbors. Inside the nucleus every nucleon sees the same types of potential, created by its neighbors. In other words, the potential energy for a nucleon inside the nucleus can be well represented by a flat potential well. The collection of nucleons inside the nucleus is called nuclear matter. Due to the short range of nuclear forces nuclear matter is fairly homogeneous inside nuclei. It follows then the volume of nuclei is proportional to the total number of nucleons, A. Then the volume is V = 4 p /3 R³ and we obtain that the nuclear radius is proportional to A^1/3, R = R₀ A^1/3. Empirically, R₀ turns out to be 1.2 fermi. This shows that the radii of all nuclei are between 1 and 6 f.

Let us determine the density of nuclear matter. The best guess is that, since the radius of nuclei is five orders of magnitude smaller then those of atoms, and atoms packed together create measured densities, the density of nuclear matter is (10⁵)³ = 10¹⁵ times as large as that of ordinary matter. Since we know the nuclear mass and size the calculation can be made more precise.

d_n = m_p A / (4 p R₀³ A / 3 ) = 3 m_p / 4 p R₀³ = 2.3 x 10¹⁷ kg m^-3 . This is indeed 2.3x10¹⁴ times as large as the density of water.

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13.b. Nuclear binding and the semiempirical formula

One consequence of the homogeneity of nuclear matter is the flat nuclear potential. The depth of this potential is approximately 40 MeV. That of course does not mean that every nucleon has a binding energy of 40 MeV. Both protons are fermions, have s=1/2, so they satisfy the Pauli exclusion principle. If we quantize the nucleons as particles in a box, then on every energy level (unless levels are degenerate) there can only be two nucleons, one with spin up one with spin down. The fermi energy level for neutrons, as we learned in statistical physics is

E_Fⁿ = (h²/2m) (3N/8 pV)^2/3

If we neglect coulomb interactions (as it is true for light nucleit) then the number of neutrons is the same as the number of protons, N=Z=A/2. Then the fermi energy for neutrons and for protons is the same

E_Fⁿ = E_F^p = (h²/2m) (9/64 p²R₀³)^2/3,

independent of the size of the nucleus as well. Then the average kinetic energy is 3/5 E_F is also independent of the size. The average energy is then E=K+V µ -8 MeV is also roughly constant. Stable nuclei have the same number of neutrons and protons because if the numbers are not equal then at the top energy level it is advantageous for a neuton to transform into a proton by beta decay or for a proton to transform into a neutron by a positron decay. Therefore in the most stable nuclei the top energy levels of neutrons and protons are equal, as shown in the figure below.

Coulomb interactions cannot be neglected however, mostly for heavier nuclei, which, due to their long range inrease with size. Coulomb repulsion acts on protons only, so in larger nuclei the depth of the potential well for protons is less then 40 MeV. Consequently one needs less protons to fill up the levels to the same energy level where the highest energy neutrons are. Therefore, in heavier nuclei Z is smaller than N.

The simplest nuclear model is the liquid droplet model, which uses the same notions we have been applying until now. One of the important aims of models to succesfully describe the dependence of the binding energy on N and Z. The binding energy is defined as the the total mass energy of constituent nucleons minus the mass energy of the nucleus. This is the total energy one needs to invest to decompose the nucleus into nucleons.

B(Z,N) = c² ( Z m_p + N m_n - M(Z,N) ),

where M(Z,N) is the mass of the nucleus. The interesting quantity is the binding energy per nucleon. B(Z,N) /A. In the roughest nuclear model it is a constant (8 MeV). Experimentally for light nuclei it is much less than 8 MeV, it reaches its maximum in the range of 50<A<80 and starts to decrease for heavier nuclei again. The semiemprical formula provides an approximate description of B(Z,N) using physical principles to find various types of terms, but using phenomenological coefficients.

B(Z,N) = C₁ A - C₂ A^2/3 - C₃ Z(Z-1)/A^1/3 - C₄ (N-Z)²/ A.

The first term of the semiempirical formula expresses the fact that the depth of the nuclear potential well and the average kinetic energy of nucleons is independent of the size of the nucleus and thus the total contribution of all nucleons is proportional to their numbers. If we divide by A, this term alone would give the (not quite correct) constant prediction for B(Z,N)/A. The second term is the so called surface term. Nucleons near the surface of the nucleus feel a less deep potential well, because they are surrounded by nucleons on one side only. Due to the constant density the number of these nucleons is proportional to the size of the surface of the nucleus, 4 pR² = 4 p R₀² A^2/3. In other words, this term is proportional to A^2/3. This term has the strongest influence on the binding energy of small nuclei, because the ratio of the first two terms is the largest for small A. This is the term that causes the smallness of the average binding energy of small nuclei.

The third term of the semiempirical formula is the Coulomb term. Each pair of protons contributes to the Coulomb term equally. The number of protons is Z, so the number of proton pairs is Z(Z-1)/2. The Coulomb energy is proportional to the inverse distance as well. The average inverse distance grows as the radius of the nucleus, R that is proportional to A^1/3. Thus the total Coulomb energy is proportional to Z(Z-1)/A^1/3. The fourth term of the formula expresses the fact that there is a gain of energy if a neutron is transformed into a proton (or vica versa) if the number of protons is larger than that of the neutons.

If one optimizes B(Z,N) with respect to Z (or N) at fixed A, then one obtains a prediction for the binding energy of most stable isotopes. Then plotting the curve as function of A one can get a very good agreement with experimental date when one fits coefficients C_i.

If one increases Z, the fastest growing term of the semiempirical formula is the Coulomb term. If we roughly assume that A increase linearly with Z then the coulomb term is proportional to Z^5/3. Clearly at some not too large value of Z it will completel overtake the volume term C₁ A that will make the nucleus completely unstable. No nuclei with Z>120 can exist. This is the direct consequence of the long range nature of coulomb forces.

The simple minded liquid droplet model can be replace by more sophisticated models that take into account the very important spin-orbit interaction of nucleons.Maria Goeppert-Meyer has succeeded to explain the so called magic numbers by using a shell model of nuclei. If Z or N takes on any of the magic number values, 2,8,20, 28, 50, 82, 126 then the nuclei are especially stable. Adding one extra nucleon decreases the binding energy substantially.

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13.c. Nuclear decays

Many isotopes are radioactive. Radioactive nuclei are not stable they decay into other nuclei after a certain amount of time. The nature of these decays will be studied later.

The basic tenet of decays of particles atoms, nuclei, molecules and other quantum systems is that they do not at all know how long they lived before. As in the military they only know the name and identification number, quantum systems carry a limited amount of information, a certain number of quantum numbers and that is all. Nothing else distinguishes two nuclei in excited states than their excitation energy, and momentum and orientation of angular momentum. Thus for each and every nucleus in the same excited state has the same chance to decay in every second. Of course this probability depends on the excited state itself; different excited states have different probabilities of decay. Then let us suppose that at a given moment we have N nuclei in a given state. After decay these nuclei transform into a different system, so the number of nuclei in this state will decrease. The change of the number N in time dt is dN. Sice every nucleus has the same chance to decay and the probability of decay is proportional to dt we have

dN = - dt N l,

where l is a constant of proportionality, the decay constant, characterizing the decay process. This equation is separable, giving

dN / N = - dt l,

or after integration

log N = log N₀ - t l,

where log N₀ is an integration constant. Exponentiating gives the exponential decay law

N = N₀ e^-lt,

where the significance of the integration constant N₀ is clear: it is the number of nuclei in the given state at t=0. The significance of l can be understood easily if we calculate the time needed for half of a certain sample of radioactive nuclei to decay. This time T_1/2 is called half-life. At t= T_1/2 we have

N₀/2 = N₀ exp{- l T_1/2 },

or taking the log of both sides

T_1/2 = log 2 / l.

in other words, l = log2 / T_1/2. After k halflives, i.e. t=k T_1/2 the size of the sample decreases to

N = N₀ exp{- l k T_1/2 } = N₀ exp{- ( log2 / T_1/2) k T_1/2 } = N₀ / 2^k.

When one measures radioactive decays then one cannot measure N directly. Using counters one measures the number of decaying nuclei per unit time, dN/dt = R, called activity. Using the exponential decay law one obtains

|dN/dt |= R = N₀ l e^-lt = R₀ e^-lt,

where R₀ is the activity at t=0. The activity is measured in curies or in bequerels.

Radioactive carbon dating is based on the folllowing facts:

1. Carbon has a radioactive isotope, ¹⁴₆C, with a half-life of 5730 years.
2. The proportion of this isotope, created by cosmic radiation in CO₂ in the air is constant, a fraction of a percent (1.3x10^-12). Plants metabolize carbon, animals eat plants and create carbon compounds in their body.
3. Fossils burried under ground are not affected by the types of cosmic radiation that creates ¹⁴₆C, so the proportion of this isotope in fossils is slowly decreasing.
4. By measuring the activity of a fossil fragment, and knowing the total amount of carbon in the fragment, one can calculate the percentage of the isotope ¹⁴₆C in the sample and deduce its age.

   Example: Chemical analysis shows that a bone fragment contains 10 g of C. Its activity is 50 counts/min= 0833 / s. How old is the bone fragment?

   Solution: The total number of C atoms in the fragment is N = (6 x 10²³ mole^-1 / 12 g mole^-1) 10 g = 5 x 10²³ Nuclei.
   The initial number of ¹⁴C nuclei is N¹⁴ = 5 x 10²³ x 1.3 x 10^-12 = 6.5 x 10¹¹
   Initial activity is R₀ = N₀ l = 6.5 x 10¹¹ log2 /(5730 x 365 x 24 x 3600) = 2.53 s^-1
   Since R = R₀ e^-lt we obtain
   l t = log ( 2.53/0.833) = 1.11
   Then we obtain t = 1.11/ l = 1.11 x 5730 / log(2) = 9180 years.

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13.d. Radioactivity

Thre most frequent radioactive decays are a, b, and g decays. These are briefly described as

1. a decay. These are decays of heavier nuclei in which an He nucleus is emitted from the parent nucleus. The process was discussed before when we studied tunneling. The decay has the form
   ^A_ZX --> ⁴₂He + ^A-4_Z-2Y. Here X is called the parent nucleus, while Y is the daughter nucleus. Note that the number of neutrons and the number of protons is conserved in the reaction. This is not a real requirement in nuclear reaction, only the total number of nucleons and the total charge must be conserved. Since no other charged particles are present in the reaction this implies the separate conservation of neutron and proton numbers.
   The decay happens because by this decay the system goes into a lower energy state. The energy of the state is lower because, for nuclei with A > 80 the binding energy/nucleon increases if the nucleus becomes lighter. He nuclei are especially stable (it is at double magic number), they have a large binding energy nucleon. Typical example: ²³⁸₉₂U --> ⁴₂He + ²³⁴₉₀Th. The lifetime of the uranium is 4.47x10⁹y, roughly speaking the age of the earth. Thus no wonder that one can still find U on earth. The alpha decay produces energy. The system goes into a lower enrgy state. The total energy produced can be calculated as
   Q = (m_X - m_a - m_Y ) c².
   As an example ²³⁶₉₂U --> ⁴₂He + ²³²₉₀Th (2.3x10⁷y) The total energy produced is
   Q = (236.045561 - 4.002603 - 232.03805) 931.4943 MeV = 4.57 MeV
   Now the kinetic energy of the a particle can also be easily determined. Since
   p_a = p_Th = p, Q = (p²/2)[ 1/m_a + 1/ M_Th ] Thus
   K_a = [ M_Th /( M_Th + m_a) ] Q = (232/236) Q = 4.49 MeV, while
   K_Th = [ m_a /( M_Th + m_a) ] Q = (4/236) Q = 0.078 MeV
   The lifetime of nuclei decaying through a decay can be estimated using the Tunneling model of Gamow. a particles cannot escape, thoughthe process is energetically favorable, because of a coulomb barrier. They eventually tunnel through the barrier.
2. b decay. These decays are characteristic of isotopes that have an excess number of neutrons. The way the nucleus tries to address this problem and go to a lower energy state is that one of the neutrons of the nucleus decays into a proton. The general form of the decay is
   ^A_ZX --> e^- + n + ^A_Z+1Y. Again, note that the number of nucleons is conserved and charge is conserved, because one of the neutrons is transformed into a proton and an electron.
   These reactions happen after a particle decays because for the lighter a nucleus is the smaller the value of N/Z becomes for the most stable isotopes. At th esame time if N/Z>1, as it is true for heavy nuclei then the ratio N/Z increases a decay. It should decrease, so the dfaughter nucleus becomes unstable and undergoes b decay to decrease N/Z. The presence of the neutrino can be clearly seen experimentally, by the fact that the spectrum of the emitted electrons in the CMS of nucleus X is not monoenergetic, as it should be in a two body decay. The neutrinos cannot be seen in this reaction, because they interact so little with matter that they always escape the measuring apparatus. Their presence is also needed due to angular momentum conservation. The "integerness" of the spin of nuclei X and Y is the same. I.e. they are either both fermions or they are both bosons, having the same number of spin 1/2 nucleons. Electrons are however fermions with s=1/2. Thus total angular momentum can only be conserved if an additional s=1/2 particle, the neutino is produced.
   Properties of neutrinos: massless (or rather have masses that are less then a few eV/c²) Very weakly interact with matter (earth is essentially transparent to them). detection is very difficult, need a very large number of them to see a few. Neutrinos are different from antineutrinos that have opposite polarizaations (rigt handed vs. left handed). Experimentlly shown to be different.
   Sometimes nuclei decay through electron capture. They absorb some of their inner electrons that have wavefunctions with large overlaps and undergo a reaction
   ^A_ZX + e^- -> n + ^A_Z-1Y. Example
   ⁷₄Be + e^- -> n + ⁷₃Li
3. g decay. In these decays neither Z nor N changes. This is the preferred decay mode of excited states of nuclei. g particles are photons, only much more energetic than those emitted in atomic or molecular decays. They have an energy of O(1 MeV). Many times a g decay follows a b decay, which lands the nucleus in an excited state.

There are many isotopes that are naturally radioactive, especially of heavy elements. All isotopes of all elements Z> 83 (heavier then Pb and Bi) are radioactive. All of them follow chains of decays that end up in stable isotopes of either Pb or Bi. Typical example is the most stable isotope of U, ²³⁸₉₂U. Its lifetime is 4.5 billion years, so much of it created at the formation of the solar system is still with us. After decaying into Th, as shown above after a series of decays (8 a decays and 4 b decays) it reaches the final state of ²⁰⁶₈₀Pb. This is called a radioactive series. There are several known series, all combinations of a decays and b decays.

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