1. The Nucleus and Radioactivity
Chapter 20 MHR
Chapter 30 Giancoli 5th

2. Canadian Connections
Cobalt Bomb Therapy for cancer treatment developed by Dr. Harold John’s at the University of Saskatchewan
Chalk River Nuclear Plant, Chalk River Ontario supplies medical radioactive isotopes globally
Atomic Energy of Canada Limited (AECL) sells Canadian Deuterium (CANDU) nuclear reactors around the world
Uranium mines in Western Canada

3. Nuclear Mass and Energy
After Rutherford and Bohr, questions persisted about atomic theory and the nucleus-what made up nucleus and what keeps it together
Rutherford discovered the proton when experimenting with α particles fired at nitrogen gas
In 1932 James Chadwick discovered the neutron.
Physicists determined that neutrons had almost same mass as protons and that nucleus had protons and neutrons. These two are called nucleons

4. Representing the Atom ZAX
X is the chemical symbol for the element e.g. U (uranium) H (hydrogen)
Z is the atomic number i.e number of protonz  (same as # electrons for neutral atom)
A is the atomic mass number aka the nucleon number A = Z + N where N = # neutrons

5. Mass and Charge of Atomic particles
Mass (kg)
Mass (u)
Charge (C)
Proton
x 10-27
x 10-19
Neutron
x 10-27
Electron
x 10-31
x 10-19

6. The Strong Nuclear Force
Around the end of 1930’s physicists discovered that there were strong interactions between any two protons, any two neutrons and a proton and a neutron-called the strong nuclear force i.e. strongest force known
p↔p n↔n p↔n

7. The Strong Nuclear Force
If protons r about 2 fm (femtometres ) apart, the strong nuclear force is 100X bigger than the Coulomb repulsive force, but at 3 fm, the nuclear force is almost zero-very short range force. It only acts between adjacent nucleons. If the separation distance decreases to about 0.5 fm, the strong nuclear force becomes repulsive. This may occur because nucleons can’t overlap.

8. Stability of Nucleus

9. Stability of the Nucleus
Nuclei with large numbers of nucleons tend to be unstable. Why?
In nuclei with more than 20 nucleons, the nucleons on one side of the nucleus are so far from the other side that they no longer attract each other. The Coulomb repulsive force between protons is still quite strong.
Calcium 40 has 20 protons and 20 neutrons. Beyond 40 nucleons, the ratio of neutrons to protons gradually increases up to the largest element e.g. Fe 56 has 26 protons and 30 neutrons so the neutron to proton ratio is 30/26 = Au 197 has 79 protons and 118 neutrons See table in next slide

10. Stability of the Nucleus
Element
Neutron #
Atomic #
Ratio
Fe 56 30 26
1.15
Au 197
118 79
1.49
Lr 262
159
103
1.54

11. Stability of the Nucleus
The increasing ratio results in more nucleons experiencing the attractive strong nuclear force than each pair of protons experiencing the repulsive Coulomb force. This appears to stabilize larger nuclei. If a nucleus does not lie in the range of stability, it will disintegrate.

12. Nuclides and Isotopes
Refer back to the graph of N vs Z. The dots represent stable combinations of protons and neutrons. These unique combinations are called nuclides.
Vertical columns of dots indicate different forms of the same element called isotopes i.e. same # protons but different numbers of neutrons e.g. N 14 and N 15 both have 7 protons

13. Nuclides and Isotopes
Hydrogen has three isotopes H 1 H 2 and H 3. H 2 is called deuterium and H 3 is called tritium. Most of the time, isotopes are named by the element and the atomic mass number e.g. C 14.

14. Nuclear Binding Energy and Mass Defect
If the strong nuclear force is the strongest known force, then in order to remove a nucleon from the nucleus a vast emount of energy would be required. For hydrogen, 13.6 eV is required to ionize the atom i.e. to remove an electron. For helium (He), a stable, noble gas, about 20 MeV is required to remove a neutron

15. Nuclear Binding Energy and Mass Defect
The mass of a nucleus is less than the mass of its constituents. Where has this difference in mass gone? This mass difference or defect has gone into energy that keeps the nucleus together. It can be determined by E = mc2
The total energy required to break a nucleus apart into its constituent protons and neutrons is called the total binding energy of the nucleus. If the total binding energy (BE) is divided by the number of nucleons, we discover the average binding energy per nucleon.

16. Nuclear Binding Energy and Mass Defect
e.g. 24He has a total BE of 28.3 MeV. It also has 4 nucleons so
Avg BE/nucleon is 28.3 MeV/4 = 7.1 MeV

17. Nuclear Binding Energy and Mass Defect

18. Calculating the Binding energy for Fe (iron)
Fe 56 has 56 nucleons and 26 protons

19. Calculating the Binding energy for Fe (iron)
Nuclear masses are given in unified atomic mass units or ‘u’
1 u = x kg
Since E = mc2 = x10-27kg x c2
and x J = 1 eV, then
1 u = MeV/ c2

20. Calculating the Binding energy for Fe (iron)
mproton = u
mneutron = u
Fe has 26 protons and (56-26) = 30 neutrons and a nuclear mass (i.e w/out electrons) of u

21. Calculating the Binding energy for Fe (iron)
Mass of protons = 26( )= u
Mass of neutrons =30( )= u
Total mass nucleons = u
Less nuclear mass = u
Mass defect = u
Change to energy x 931.5
Total BE = MeV
Avg BE/nucleon = 492.3/56 = MeV