1. Nuclear Physics
Lecture 1& 2
© 2015, 2016

2. Objectives: After completing this module, you should be able to:
Define and apply the concepts of mass number, atomic number, and isotopes.
Calculate the mass defect and the binding energy per nucleon for a particular isotope.

3. Composition of Matter
All of matter is composed of at least three fundamental particles (approximations):
Particle
Fig.
Sym
Mass
Charge
Size
Electron e x kg -1.6 x C 
Proton p x kg x C 3 fm
Neutron n x kg fm
The mass of the proton and neutron are close, but they are about 1840 times the mass of an electron.

4. The Atomic Nucleus Compacted nucleus: 4 protons 5 neutrons
Beryllium Atom
Since atom is electri-cally neutral, there must be 4 electrons.
4 electrons

5. Definitions
A nucleon is a general term to denote a nuclear particle - that is, either a proton or a neutron.
The atomic number Z of an element is equal to the number of protons in the nucleus of that element.
The mass number A of an element is equal to the total number of nucleons (protons + neutrons).
The mass number A of any element is equal to the sum of the atomic number Z and the number of neutrons N :
A = N + Z

6. For example, consider beryllium (Be):
Symbol Notation
A convenient way of describing an element is by giving its mass number and its atomic number, along with the chemical symbol for that element.
For example, consider beryllium (Be):

7. Example 1: Describe the nucleus of a lithium atom which has a mass number of 7 and an atomic number of 3.
Lithium Atom
A = 7; Z = 3; N = ?
N = A – Z =
neutrons: N = 4
Protons: Z = 3
Electrons: Same as Z

8. Isotopes of Elements
Isotopes are atoms that have the same number of protons (Z1= Z2), but a different number of neutrons (N). (A1  A2)
Helium - 4
Helium - 3
Isotopes of helium

10. The following are best described as nuclides:
Because of the existence of so many isotopes, the term element is sometimes confusing. The term nuclide is better.
A nuclide is an atom that has a definite mass number A and Z-number. A list of nuclides will include isotopes.
The following are best described as nuclides:

11. Atomic mass unit: 1 u = 1.6606 x 10-27 kg
One atomic mass unit (1 u) is equal to one-twelfth of the mass of the most abundant form of the carbon atom--carbon-12.
Atomic mass unit: 1 u = x kg
Common atomic masses:
Proton: u
Neutron: u
Electron: u
Hydrogen: u

12. Exampe 2: The average atomic mass of Boron-11 is 11. 009305 u
Exampe 2: The average atomic mass of Boron-11 is u. What is the mass of the nucleus of one boron atom in kg? =
Electron: u
The mass of the nucleus is the atomic mass less the mass of Z = 5 electrons:
Mass = u – 5( u)
1 boron nucleus = u
m = 1.83 x kg

13. Mass and Energy Recall Einstein’s equivalency formula for m and E:
The energy of a mass of 1 u can be found:
E = (1 u)c2 = (1.66 x kg)(3 x 108 m/s)2
E = 1.49 x J Or
E = MeV
When converting amu to energy:

14. Similar conversions show other rest mass energies:
Example 3: What is the rest mass energy of a proton ( u)?
E = mc2 = ( u)(931.5 MeV/u)
Proton: E = MeV
Similar conversions show other rest mass energies:
Neutron: E = MeV
Electron: E = MeV

15. Nuclear Size and Mass The Size of Nuclei
The size and structure of nuclei were first investigated in the scattering experiments of Rutherford, Using the principle of conservation of energy, Rutherford found an expression for how close an alpha particle moving directly toward the nucleus can come to the nucleus before being turned around by Coulomb repulsion. In such a head-on collision If we equate the initial kinetic energy of the alpha particle to the maximum electrical potential energy of the system (alpha particle plus target nucleus), we have:

16. R = R0A1/3 Nuclear Size and Mass
The radii of most nuclei is given by the equation:
R = R0A1/3
R: radius of the nucleus, A: mass number, R0= 1.2x10-15m
The mass of a nucleus is given by:
m = A u
u = 1.66…x10-27kg (unified mass unit)

17. Exercise :
Estimate the radius of a uranium-235 nucleus.
Answer: 7.41 x m

18. 5626Fe is the most abundant isotope of iron (91.7%)
Nuclear Density
5626Fe is the most abundant isotope of iron (91.7%)
R = 1.2x10-15m (56)1/3= 4.6x10-15m
ρ= m / V = 2.3x1017kg/m3

19. The Nuclear Force
The force that binds together protons and neutrons inside the nucleus is called the Nuclear Force
Some characteristics of the nuclear force are:
Attractive forces at distance between nucleons equal 0.4fm (in spite of coulomb's force).
•It is very short range ≈10-15m
•It does not depend on charge, Fn-p = Fn-n = Fp-p (Isobars or Mirror nuclei).
•It is much stronger than the electric force
•It is saturated (nucleons interact only with near neighbors)
•It favors formation of pairs of nucleons with opposite spins.
We do not have a simple equation to describe the nuclear force

23. The Mass Defect
The mass defect is the difference between the rest mass of a nucleus and the sum of the rest masses of its constituent nucleons.
The whole is less than the sum of the parts! Consider the carbon-12 atom ( u):
Nuclear mass = Mass of atom – Electron masses = u – 6( u) = u
The nucleus of the carbon-12 atom has this mass.
(Continued )

24. Mass Defect (Continued)
Mass of carbon-12 nucleus:
Proton: u
Neutron: u
The nucleus contains 6 protons and 6 neutrons:
6 p = 6( u) = u
6 n = 6( u) = u
Total mass of parts: = u
Mass defect mD = u – u
mD = u

25. The binding energy for the carbon-12 example is:
The binding energy EB of a nucleus is the energy required to separate a nucleus into its constituent parts.
EB = mDc2 where c2 = MeV/u
The binding energy for the carbon-12 example is:
EB = ( u)(931.5 MeV/u)
EB = 92.2 MeV
Binding EB for C-12:

26. Binding Energy per Nucleon
An important way of comparing the nuclei of atoms is finding their binding energy per nucleon:
Binding energy per nucleon
For our C-12 example A = 12 and:

27. Formula for Mass Defect
The following formula is useful for mass defect:
Mass defect mD
mH = u; mn = u
Z is atomic number; N is neutron number; M is mass of atom (including electrons).
By using the mass of the hydrogen atom, you avoid the necessity of subtracting electron masses.

28. M = 4.002603 u (From nuclide tables)
Mass defect mD
ZmH = (2)( u) = u
Nmn = (2)( u) = u
M = u (From nuclide tables)
mD = ( u u) u
mD = u

29. Since there are four nucleons, we find that
Example 4 (Cont.) Find the binding energy per nucleon for helium-4. (mD = u)
EB = mDc2 where c2 = MeV/u
EB = ( u)(931.5 MeV/u) = 28.3 MeV
A total of 28.3 MeV is required To tear apart the nucleons from the He-4 atom.
Since there are four nucleons, we find that

30. Binding energy per nucleon
- determines stability
- for 4He, BE = 28 MeV so BE/nucleon = 7 MeV
Fusion
Fission
increase in nearest neighbors
increase in Coulomb repulsion dominates

31. Binding Energy Vs. Mass Number
The figure below shows a plot of binding energy per nucleon as a function of nucleon number.
Curve shows that EB increases with A and peaks at A = 60 – about 8 Mev per nucleon. Heavier nuclei are less stable.
Green region is for most stable atoms.
For heavier nuclei, energy is released when they break up (fission). For lighter nuclei, energy is released when they fuse together (fusion).

32. Stable Nuclei
We can begin to understand why certain nuclei are stable and others unstable by realizing that nucleons have spins of 1/2 and obey the Pauli exclusion principle.
Nucleons, like electrons and their electronic energy levels, occupy discrete nuclear energy levels.
Minimum energy configurations (i.e., nucleons in the lowest possible energy levels) give the most stable nuclei.

33. Stability Curve
A plot of N versus Z for the stable nuclides looks like this:
Neutrons produce attractive forces within nuclei, and help hold the protons together.
For small numbers of protons, about an equal number of
neutrons is enough to provide stability, hence NZ for small Z.
As the number of protons gets larger, an excess of neutrons is needed to overcome the proton-proton repulsion.

34. The stability of nuclei follows a definite pattern.
• The majority of stable nuclei have both even Z and even N ("even-even" nuclides).
• Most of the rest have either even Z and odd N ("even-odd") or odd Z and even N ("odd-even").
• Very few stable nuclei have both Z and N odd.
The reasons for this pattern are the Pauli exclusion principle and the existence of nuclear energy levels.
Each nuclear energy level can contain two nucleons of opposite spin.

35. The neutrons and protons occupy separate sets of energy levels.
When both Z and N are even, the energy levels can be filled. The nucleus doesn't "want" to gain or lose nucleons by participating in nuclear reactions. The nucleus is stable.
protons
neutrons
When both Z and N are odd, the nucleus is much more likely to "want" to participate in nuclear
reactions or nuclear decay, because it has unfilled nuclear energy levels.
protons
neutrons

36. All of its neutrons and protons in filled energy levels—very stable.
Example:
All of its neutrons and protons in filled energy levels—very stable.
Energy level diagrams illustrative only; not quantitatively accurate!
protons
neutrons
Example:
“Extra” neutron in higher energy level; therefore unstable.
Decays via  decay into .

37. 2. Isotopes are atoms having
( u).
2. Isotopes are atoms having
(a) Same number of protons but different number of neutrons
(b) Same number of neutrons but different number of protons
(c) Same number of protons and neutrons
(d) None of the above