2. © J Parkinson 1

3. 2 Mass Defect The difference between the mass of the atom and the sum of the masses of its parts is called the mass defect (  m). Careful measurements show that the mass of a particular atom is always slightly less than the sum of the masses of the individual protons, neutrons and electrons of which the atom consists. e.g. a helium nucleus consists of 2 protons and 2 neutrons 2 protons & 2 neutronsHelium atom

4. 3 m p = mass of a proton (1.007277 amu) m n = mass of a neutron (1.008665 amu) m e = mass of an electron (0.000548597 amu) 1 atomic mass unit ( amu ) = 1.661 x 10 -27 kg  m = [ Z(m p + m e ) + (A-Z)m n ] - m atom 1 amu = 1/12 of the mass of an atom of Carbon-12 Z = PROTON NUMBER (ATOMIC NUMBER) AND ALSO EQUALS THE NUMBER OF ELECTRONS A = NUCLEON NUMBER (MASS NUMBER) AND ALSO EQUALS THE NUMBER OF PROTONS + NEUTRONS THE NUMBER OF NEUTRONS MUST BE A - Z

5. 4 For Helium the mass defect will be  m = [ 2( 1.007277 + 0.000548597 ) + 2 x 1.008665 ] – 4.021435 = 0. 0115462 amu 0. 0115462 amu x 1.661 x 10 -27 = 1. 91782 x 10 -29 kg Using Einstein’s E = mc 2, this is equivalent to a loss of energy given by E = 1.91782 x 10 -29 x (3 x 10 8 ) 2 Joules = 1.726 x 10 -12 J This figure is the BINDING ENERGY of the Helium nucleus. THE BINDING ENERGY of a nucleus is defined as the energy which must be input to separate all of its protons and neutrons.

6. 5 Binding Energies are usually expressed in MeV1 amu = 931.3 MeV The binding energy of the Helium nucleus is therefore 0. 0115462 amu x 931.3 = 10.75 MeV To compare the stabilities of different nuclei, Binding Energies PER NUCLEON in the nucleus are compared. For Helium this will be 10.75 ÷ 4 = 2.69 MeV per nucleon The higher the binding energy per nucleon, the greater the stability of the nucleus

7. 6 NUCLEON NUMBER BINDING ENERGY Per NUCLEON 2H2H 238 U 56 Fe 8.8 MeV Iron is the most stable nucleus

8. 7 BE/A A ALTERNATIVELY BE/A A 0 250 9.0 As energy is lost when nuclei are synthesised

9. 8 FISSION IRON Heavy nuclei may increase their stability by Nuclear Fission Light nuclei may increase their stability by Nuclear Fusion FUSION

10. 9 As atomic mass increases, the neutron to proton ratio for stable nuclei increases because proton- proton repulsion becomes significant, as the number of protons increases. Cohesive nuclear forces arise form neutrons, so the neutron to proton ratio must increase for heavier elements. Belt of Stability Proton number, Z Neutron number, N = A - Z N = Z Belt of Stable Isotopes For helium He- 4 the N:P ratio is 1 : 1 For uranium U- 238 the ratio is 1 : 1.6

11. 10 Nuclear Fission is the fragmentation of heavy nuclei to form lighter, more stable ones. The Fission of U - 235 This is only one of several fissures that are possible. On average 2.5 neutrons are released

12. © J Parkinson 11 Neutrons released in the fission of 235 U can induce three more fissions, then nine, and so on leading to a chain reaction.

13. 12 Critical mass is the mass required for the chain reaction to become self- sustaining. neutron Some neutrons may : Cause more fission Get lost Be absorbed by an atom lost For a chain reaction to be self sustaining, every fission must produce at least one more neutron that will initiate further fissions.

14. 13 Nuclear Reactor Pressure vessel and biological shield Graphite Moderator slows neutrons to thermal energies [1 MeV] Fuel Rods remain in reactor until spent Coolant [CO 2 or H 2 O] to extract energy To heat exchanger where steam is produced Boron control rods absorb neutrons to control the reaction