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Chapter 12 The Atomic Nucleus Enrico Fermi (1901 – 1954) Nobel 1938

12.1 Discovery of the Neutron James Chadwick (1891 – 1974) Nobel 1935 discovered the neutron in 1932. Thus it was learned that the nucleus was composed of two particles of approximately equal mass, protons and neutrons.

Can Electrons Exist inside the Nucleus? Before neutrons were discovered, it was thought that the nucleus was composed of protons and electrons. Can electrons exist inside the nucleus as independent particles? No. Why not? Nuclear size. Because of the small size of the nucleus, the uncertainty principle puts a lower limit on the kinetic energy that is greater than the energy of any electron emitted from the nucleus.

Can Electrons Exist inside the Nucleus? Nuclear spin. Nuclear magnetic moment.

Example 12.1 What is the minimum kinetic energy of a proton in a medium-sized nucleus having a diameter of 8 x 10-15 m?

12.2 Nuclear Properties The primary constituents of nuclei are the proton and the neutron. Neutrons and protons are each made up of three quarks. The nuclear mass is equal to the sum of the masses of the constituent particles plus the binding energy. The nuclear charge is + e times the number (Z) of protons.

Atomic Notation AXN Z = atomic number (number of protons) N = neutron number (number of neutrons) A = mass number (Z + N) X = chemical element symbol

The Atomic Mass Unit 1 u = 1.66054 x 10-27 Kg 1 u = 931.494 Mev/c2

Sizes and Shapes of Nuclei Nuclear radius Matter radius (physical size) Force radius (range of nuclear (strong) force. Charge radius (electrons are not affected by nuclear force, but by electromagnetic field of the nucleus.) Nuclear force radius Mass radius Charge radius (experimentally determined)

Nuclear Radius R = r0 A1/3 where r0 1.2 x 10 –15 m. The nuclear radius may be approximated from a spherical charge distribution to be R = r0 A1/3 where r0 1.2 x 10 –15 m. 10 –15 m is called a femtometer or in physics a fermi (fm)

Example 12.2 What is the nuclear radius of 40Ca? What energy electrons and protons are required to probe the size of 40Ca if one wants to “see” at least half the radius?

Example 12.3 What is the ratio of the radii of 238U and 4He?

Intrinsic Spin Protons and neutrons are both fermions with spin quantum numbers s = ± ½. They obey the same spin quantum rules as do electrons.

12.3 The Deuteron After the proton, the next simplest nucleus is the deuteron: 2H. The deuteron mass is: 2.013553 u, and the mass of the deuterium atom is 2.014102. 1

Binding Energy md = mp + mn – Eb /c2 The deuteron nucleus is bound by an energy Bd, which represents mass energy. The mass of the deuteron nucleus is: md = mp + mn – Eb /c2

Eb(AX) = [Nmn + ZM(1H) – M(AX)] c2 Binding Energy Eb(AX) = [Nmn + ZM(1H) – M(AX)] c2 Z Z

12.4 Nuclear Forces Most straightforward techniques of studying nuclear forces are scattering experiments. Two common ones are p – n and p – p scattering. Nuclear forces are attractive short range forces. Two nucleons within about 2 fm of each other feel and attractive force. Outside 3 fm the strong force is essentially zero. Outside 3 fm the repulsive (proton – proton) Coulomb force dominates.

r 3 fm

12.5 Nuclear Stability Equation 12.10 presented a method to determine binding energies. If B is positive, the nucleus is stable against dissociating into free neutrons and protons. A more general statement is a nucleus containing A nucleons is stable if its mass is smaller than any other possible combination of A nucleons.

Eb = [M(R) + M(S) – M(A X )] c2 Nuclear Stability The binding energy of a nucleus AX against dissociation into any other possible combination of nucleons, for example, R and S nuclei, is Eb = [M(R) + M(S) – M(A X )] c2 Z Z

Example 12.4 Show that the nuclide 8Be has a positive binding energy but is unstable with respect to decay into two alpha particles.

A 40, Z~N A 40, N > Z

Nuclear Coulomb Repulsion Energy DECoul = 3 Z (Z – 1) e2 5 4pe0 R

DECoul = 0.72 [Z(Z – 1)] A-1/3 MeV (12.19) Example 12.5 Show that Equation (12.18) can be written as DECoul = 0.72 [Z(Z – 1)] A-1/3 MeV (12.19) and use this equation to calculate the total Coulomb energy of 238U. 92

Example 12.6 Calculate the binding energy per nucleon for 20Ne, 56Fe, and 238U. 10 26 92

Nuclear Models Physicists do not yet fully understand the nuclear force or how nucleons interact inside the nucleus. Current research stresses the constituent quarks that make up the nucleus. Because of this lack of knowledge, physicists have used models to explain nuclear behavior. Nuclear models fall into two categories.

Nuclear Models Independent Particle Models: Nucleons move nearly independently in a common nuclear potential. The shell model has been the most successful of these. Strong Interaction Models: Nucleons are strongly coupled together. The liquid drop model has been quite successful in explaining nuclear masses as well as nuclear fission. Accounts for nuclear binding energy.

Shell Model The shell model is based on the fact that each nucleon moves in a well-defined orbital state within the nucleus in an averaged field produced by the other nucleons.

Neutron energy levels slightly lower than the proton levels because of the additional Coulomb repulsion.

Liquid Drop Model The semi-empirical binding energy formula (Carl F. von Weizsäcker – 1935) is an example of the liquid drop model. This model treats the nucleons as though they were molecules in a drop of liquid. The nucleons strongly interact with each other and undergo frequent collisions as they jiggle around within the nucleus. The jiggling is analogous to thermal vibrations of the molecules in a liquid drop.

Semi – Empirical Binding Energy Formula Eb = avA – aAA2/3 – - aS + d 3 Z (Z – 1) e2 (N – Z)2 5 4pe0 R A aV = 14 MeV Volume aA = 13 MeV Surface aS = 19 MeV Symmetry = + D even-even nuclei d = 0 odd-A (e-o, o-e) d = - D odd-odd nuclei where D = 33 MeV • A-3/4

Semi – Empirical Binding Energy Formula avA: The volume term indicates that the binding energy is approximately the sum of all the interactions between nucleons. aAA2/3: This is a correction term due to the surface effect. That is, that nucleons on the surface do not feel the effect of being completely surrounded by other nucleons. The third term is the Coulomb energy term. The d term is due to the various ways nucleons can pair. It shows that the nucleus is more stable for even-even nucleus.

Even-Even

12.6 Radioactive Decay Unstable nuclei decay spontaneously to some other combination of A nucleons that has a lower mass. We will cover the three primary forms of radioactive decay: a (alpha) b (beta) g (gamma)

Definitions Activity (R) = decays per unit time. 1 Bq = 1 decay/s N (t) = number of unstable nuclei at time t. The decay constant (l) is the probability per unit time that any given nucleus will decay. N0 = number of nuclei in the sample at t = 0. R0 = activity at t = 0.

Decay Formulas N(t) = N0 e-lt R(t) = R0 e-lt

Half - Life The half-life of a given element t1/2 is the time it takes for half the nuclei in a sample to decay. t1/2 = The mean lifetime t is defined as 1/l. 0.693 l

Example 12.7 A sample of 210Po that a decays with t1/2 = 138 days is observed by a student to have 2000 disintegrations/s (2000 Bq). (a) What is the activity in mCi for this source? (b) What is the mass of 210Po?

Example 12.8 A sample of 18F is used internally as a medical diagnostic tool to look for the effects of the positron decay (t1/2 = 110 min). How long does it take for 99% of the 18F to decay?

Example 12.9 What is the alpha activity of a 10-kg sample of 235U that is used in a nuclear reactor?

12.7 Alpha, Beta, and Gamma Decay All these methods of decay had been observed by the early 20th century. When a nucleus decays, all the conservation laws must be conserved: mass-energy, linear momentum, angular momentum, and electric charge. An additional decay law must be added – the law of conservation of nucleons.

Law of Conservation of Nucleons The total number of nucleons (A, the mass number) must be conserved in a low-energy nuclear reaction (say, less than 100 MeV) or decay. Neutrons may be converted into protons, and vice versa, but the total number of nucleons must remain constant. At higher energies enough rest energy may be available to create nucleons, but the other conservation laws still apply.

Radioactive Decay M (AX ) = MD + My + Q/c2 Radioactive decay may occur for a nucleus when some other combination of the A nucleons has a lower mass. Applying the law of conservation of energy: M (AX ) = MD + My + Q/c2 Daughter Lighter Parent Z Q is the energy released and is equal to the total kinetic energy of the reaction products.

Energy Released (Disintegration Energy) Q = [M (AX ) - MD - My] c2 Z

Example 12.10 Show that 230U does not decay by emitting a neutron or proton. 92

Alpha Decay AX A-4D + a Z Z-2

Disintegration Energy for Alpha Decay Q = [M(AX) – M(A-4D) – M(4He)]c2 2 Z Z-2

Beta (b -) Decay n p + b - +n A X Z+1D + b - +n Special example: General case: A X Z+1D + b - +n A Z n is an antineutrino

Disintegration Energy b - Decay Q = [M(AX ) - M(Z+1D )]c2 A Z

b + (Positron) Decay p n + b + +n AX Z-1D + b + +n Special example: General case: AX Z-1D + b + +n A Z n is an neutrino

Disintegration Energy b + Decay Q = [M(AX ) - M(Z-1D ) – 2me]c2 A Z

Electron Capture p + e- n = n ZX + e- Z-1D + n Special case for a proton: p + e- n = n General case: ZX + e- Z-1D + n A A

Disintegration energy for Electron Capture Q = [M(AX ) – M( Z-1D)]c2 A Z

Example 12.11 Show that the relations expresses for the disintegration energy Q in Equations (12.38), (12.41), and (12.44) are correct.

Example 12.12 Show that 55Fe may undergo electron capture, but not b + decay.

Example 12.13 Find whether alpha decay or any of the beta decays are allowed for 226Ac. 89

Gamma Decay AX* AX + g

Example 12.14 Consider the g decay from the 0.072-MeV excited state to the ground state of 226Th at rest in Figure 12.15. Find the exact expression for the gamma-ray energy by including both the conservation of momentum and energy. Determine the error obtained by using the approximate value in Equation (12.46).

12.8 Radioactive Nuclides “Standard Model” tells us the universe was created in the Big Bang about 13 billion (13 x 109) years ago. Neutrons and protons fused together in the first few minutes to form deuterons and other light nuclei. Unstable nuclei that were formed exhibit natural radioactivity.

Radioactive Nuclides There are many natural radioactive nuclides left on Earth with lifetimes long enough to be observed. Since the Earth (and solar system) is about 4.5 billion years old, only those nuclides with half-lives longer than a few billion years can now exist. Most, but not all, have A > 150.

Decay Modes In addition to nuclear fission, heavy radioactive nuclides can change their mass number (A) only by alpha decay. They can change their charge number (Z) by either alpha or beta decay. As a result, there are only four paths that heavy naturally occurring radioactive nuclides may take to decay to stable end products.

The Four Paths of Decay The four paths have mass numbers expressed by either 4n, 4n +1, 4n +2, 4n +3 (n = integer). This is because only alpha decay can change the mass number.

Dating Using Lead Isotopes 204Pb is not radioactive and no other nuclides decay to it. Thus is abundance is constant. 206Pb and 207Pb are stable, but are the end products of decay of 238U and 235U respectively. 235U has a relatively short half-life (0.70 x 109 years), most 235U has already decayed to 207Pb, the the ratio 207Pb/ 204Pb has been relatively constant for the past 2 billion years.

Dating Using Lead Isotopes Since the half-life of 238U (4.47 x 109 years) is about the same of the age of the Earth, only about half the original abundance has decayed to. Therefore, the ratio 206Pb/ 204Pb is still increasing. A plot of the abundance of 206Pb/ 204Pb versus 207Pb/ 204Pb can be a sensitive indicator of the age of lead ores. Using this technique has been used to show that moon rocks are as old as the solar system.

Example 12.15 Assume that all the 206Pb found in a given sample of uranium ore resulted from the decay of 238U and that the ratio 206Pb/ 238U is 0.60. How old is the ore?

Radioactive Carbon Dating Radioactive 14C is produced in our atmosphere by bombardment of 14N by neutrons produced by cosmic rays. n + 14N 14C + p A natural equilibrium of 14C to 12C exists in molecules of CO2 in the atmosphere that all living organisms take in.

Radioactive Carbon Dating When living organisms die, their intake of 14C ceases, and the ration of 14C/ 12C decreases as 14C decays. The half-life of 14C is 5730 years. Thus the age of objects can be determined up to about 45,000 years.

Example 12.16 A bone suspected to have originated during the period of the Roman emperors was found in Great Britain. Accelerator techniques gave its 14C/ 12C ratio as 1.1 x 10-12. Is the bone old enough to have Roman origins?