1. What is the structure of an atom?
The nuclear atom
More than 2000 years ago, the Greeks suggested that matter was made up of very tiny (small) particles which they called atoms.
What is inside the atom?
What is the structure of an atom?
19.1 The atomic model

2. Models of an atom
Plum pudding model
19.1 The atomic model

3. The model was disproved by the 1909 gold foil experiment,
The plum pudding model of the atom by J.J. Thomson, who discovered the electron in 1897, was proposed in 1904.
The atom is composed of electrons, surrounded by a soup of positive charge to balance the electron's negative charge, like negatively-charged "plums" surrounded by positively-charged "pudding".
The model was disproved by the 1909 gold foil experiment,
Plum pudding model
19.1 The atomic model

4. Geiger-Marsden scattering experiment
a particles are made to hit the thin (super thin) gold foil.
Flashes of light will be observed when a particles hit the zinc sulphide screen.
19.1 The atomic model

5. Geiger-Marsden scattering experiment
We expect all the a particles can pass through the gold foil.
Results:
Nearly all a particles pass straight through the gold foil.
Some α-particles (about 1/8000) were scattered by angles greater than 90∘and very few even rebounded back along original paths.
19.1 The atomic model

6. Rutherford’s Remark
It was quite the most incredible event that has ever happened to me in my life.
It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.
On consideration, I realized that this scattering backward must be the result of a single collision, and when I made calculations I saw that it was impossible to get anything of that order of magnitude unless you took a system in which the greater part of the mass of the atom was concentrated in a minute nucleus.
It was then that I had the idea of an atom with a minute massive centre, carrying a charge.
19.1 The atomic model

7. How can the heavy a particles bounce back after hitting the thin gold foil?
a particles – He2+
Explanation:
All positive charge of the atom and most of the mass were concentrated in a tiny core called nucleus.
The rest of the atom was largely empty space.
19.1 The atomic model

8. How can the heavy a particle bounce back after hitting the thin gold foil?
Explanation:
Most of a particles passed straight through the empty space of the gold atoms.
Some come close to the nucleus were repelled by a strong electrostatic force, so they were deflected or bounced back.
Bounced back
deflected
19.1 The atomic model

9. -particle lose their K. E
-particle lose their K.E. on approaching the +ve charged nucleus, being repelled by an electrostatic force.
At P, distance of closest approach K.E. lost = P.E. due to -particle location in electric field of nucleus.
-particle is then ‘reflected’ away from nucleus and finally acquires the same K.E. as it had initially. Collision is elastic.
19.1 The atomic model

10. Estimated Upper Limit of the Size of a Nucleus
PE = 2Ze2/40r
At P, distance of nearest approach K.E. of ’s, ½mv2 = 2Ze2/40r (P.E.)
hence an estimate of r which is upper limit to size of nucleus.
19.1 The atomic model

11. 1 Nuclear fission When a heavy nucleus splits up,
 huge amount of energy is released.
Two typical nuclear fission reactions are:
+ energy released
+ energy released
The total mass of the nuclear products is slightly less than the total mass of the original particles.
Energy is released and can be calculated by using Einstein’s mass-energy relation (E = mc2)
19.1 The atomic model

12. 3 Nuclear fusion
A huge amount of energy is also released when two light nuclei join together to form a heavy nucleus.
This process is called nuclear fusion.
19.1 The atomic model

13. Example of Nuclear Fusion
Deuterium-Tritium Fusion Reaction.
The total mass of the nuclear products is slightly less than the total mass of the original particles.
19.1 The atomic model

14. 3 Nuclear fusion H 2 1 + H 3 1 He 4 2 + n 1 + energy+
energy
Since nuclei carry +ve charges, they repel each other.
For fusion to occur, the 2 hydrogen nuclei must approach each other with very high speed.
 hydrogen gas of high temperature
(108 C)!
19.1 The atomic model

15. 3 Nuclear fusion
Nuclear fusion occurs in the core of the Sun, giving out heat and light. The reaction takes place continuously for billions of years.
(Photo credit: US NASA)
19.1 The atomic model

16. Nuclear Power – controlled fission
The schematic diagram of a nuclear reactor is shown below:
19.1 The atomic model

17. Fuel Enriched uranium is used as the fuel (uranium dioxide).
Natural uranium contains only 0.7% uranium-235, which does not undergo fission in these circumstances.
Treatment is required to increase the concentration to 3%.
19.1 The atomic model

18. Moderator + energy released
The probability that fission caused by a high energy neutron (fast moving neutron) is quite low.
Use materials to slow down neutrons so that the probability of causing a fission is significantly higher.
These neutron slowing down materials are the so called moderators.
19.1 The atomic model

19. Moderator At rest u V v m M Before collision After collision
These neutron slowing down materials are the so called moderators.
At rest u v V m M
Before collision
After collision
By momentum conservation: mu = mv + MV
By energy conservation: ½mu2 = ½mv2 + ½MV2
19.1 The atomic model

20. Moderator V v After collision The choice of the moderator
The atoms of an ideal moderator should have the same mass as a neutron (M = m). So a neutron colliding elastically with a moderator atom would lose almost all its KE to the moderator atom.
The moderator atoms should not absorb neutrons but should scatter them instead.
In practice, heavy water (D2O i.e. 2H2O) is chosen as the moderator. (Heavy water is different from hard water)
19.1 The atomic model

21. Control rods
The control rods are made of boron-steel, which absorbs neutrons.
They are raised and lowered to vary the number of neutrons to control the rate of fission.
19.1 The atomic model

22. 2 Chain reaction
When a uranium nucleus splits, 2 or 3 neutrons are emitted.
neutron
U-235 nucleus splits
If these neutrons carry on splitting other uranium nucleus...
19.1 The atomic model

23. 2 Chain reaction self-sustaining  chain reaction neutron
escapes
neutron
U-235 nucleus splits
19.1 The atomic model

24. Chain reaction
The fission neutrons enter the moderator and collide with moderator atoms, transferring KE to these atoms.
So the neutrons slow down until the average KE of a neutron is about the same as that of a moderator atom.
19.1 The atomic model

25. The fission neutrons could be absorbed by the U-238 nuclei without producing further fission.
The fission neutron could escape from the isolated block of uranium block without causing further fission.
The critical mass of fuel is the minimum mass capable of producing a self-sustaining chain reaction.
19.1 The atomic model

26. Two typical nuclear fission reactions are:
+ energy released
+ energy released
At the fission of U-235 on the average 2.5 neutrons are released but not all of these cause fission.
multiplication factor (k)
19.1 The atomic model

27. escapes
neutron
U-235 nucleus splits
19.1 The atomic model

28. critical reactor
The number of neutrons in the system is constant, i.e. they cause the same number of fissions in every second.
19.1 The atomic model

29. k > 1, the system is supercritical
k < 1, the system is subcritical
k can be varied by lowering or raising the control rods
19.1 The atomic model

30. 1 Nuclear power b Potential hazards
The nuclear waste remains radioactive for thousands of years.
 serious handling and storage problems
Nuclear accidents lead to the leakage of radiation.
 widespread and long-lasting disasters
(Photo credits: BREDL; IAEA)
19.1 The atomic model

31. 1 Nuclear power c Controlled nuclear fusion Fuel = H-2, H-3
(plentiful in sea water)
Waste product = He-4
(inert and non-radioactive)
(Photo credit: Princeton Plasma Physics Lab)
Cheaper, abundant and safer, but not yet in practice
19.1 The atomic model

32. contentious 1 Nuclear power d Benefits and disadvantages
Solve energy shortage crisis 
Create serious social and environmental problems 
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33. 1 Nuclear power d Benefits and disadvantages Benefits
 Solve energy shortage crisis
 No fuel transportation problem
 Cheaper than coal/oil for generating power in most cases
 Little environment pollution
19.1 The atomic model

34. 1 Nuclear power d Benefits and disadvantages Disadvantages
 Accident  serious consequence
 Expensive in maintaining safety standards
 Unnecessary as alternative energy sources exist
 Lead to widespread of nuclear weapons
19.1 The atomic model

35. END
19.1 The atomic model

36. Pre-lesson assignment Einstein's mass-energy relation (E = mc2)
According to Einstein’s mass-energy equation E = mc2, what is the amount of energy produced if 1 kg of a certain element completely changes into energy?
Solution:
Energy released
= (1)(3 x 108)2
= 9 x 1016 J
19.1 The atomic model

37. Q2 Mass of U-235: 3.9030 x 10-25 kg Ba-144: 2.3899 x 10-25 kg
Kr-90: x kg Neutron: x kg
Calculate the nuclear energy released in the nuclear fission.
Solution:
Mass difference
= x – ( x x x 10-27)
= x kg
Energy released = (3.251 x )(3 x 108)2
= 2.93 x J
19.1 The atomic model

38. Nuclear Energy unified atomic mass unit unit of energy: eV
Binding energy
19.1 The atomic model

39. Atomic mass 1 u = unified atomic mass unit Particle or element
Atomic mass / kg
Proton (p)
1.673 x 10-27
Neutron (n)
1.675 x 10-27
Electron (e)
9.14 x 10-31
Hydrogen (H)
1.674 x 10-27
Helium (He)
6.646 x 10-27
Lithium (Li)
1.165 x 10-26 u u u u u u
1 u = unified atomic mass unit
19.1 The atomic model

40. The unified atomic mass unit (u) is defined as one twelfth of the mass of the carbon-12 atom.
History
Question:
How to express the unified atomic mass unit (1u) in kg?
Solution:
Given that:
mass of a C-12 atom = x kg
∴1u = x kg /12
= 1.66 x kg
19.1 The atomic model

41. Example 1
Given that the atomic mass of a hydrogen atom is u. Find the mass of a hydrogen atom.
Solution
Mass of a hydrogen atom
= x 1.66 x 10-27
= 1.67 x kg
19.1 The atomic model

42. Units of Energy Energy = charge Q x voltage V
Very large unit of energy: kWh
(electric bill)
S.I. unit: J
Very small unit of energy: eV
(Atomic physics)
Energy = charge Q x voltage V
19.1 The atomic model

43. Energy in eV – + 1.6 x 10-19 C – By W = QV 1 eV = (1.6 x 10-19)(1)
∴ 1 eV = 1.6 x J
1 V
1 eV is the energy (K.E.) gained by an electron when it is accelerated through 1 V.
19.1 The atomic model

44. Energy in eV +
1 V
1.6 x C + –
1 eV is the work done in moving a charge of 1.6 x C through 1 V.
19.1 The atomic model

45. Example 1 page 25 1 u = 1.66 x 10-27 kg 1 eV = 1.602 x 10-19 J
Show that 1 u of mass is equivalent to 931 MeV by the mass-energy relation.
Solution:
The energy equivalent to 1 u of mass
= mc2
= (1.66 x 10-27)(2.998 x 108)2
= x J
= (1.492 x / x 10-19) eV
= 931 x 106 eV
= 931 MeV
1 u = 1.66 x kg
1 eV = x J
1 u = 931 MeV
19.1 The atomic model

46. Example 2 Consider the following nuclear fission.
Given that: 1u = 931 MeV
Mass of neutron: u
U-235: u
Ba-144: u
Kr-90: u
(a) Find the amount of nuclear energy released in nuclear reaction.
Solution:
Mass difference = ( – – – )u
= u
Energy released = x 931 MeV
= 180 MeV
19.1 The atomic model

47. Energy released = (2.563 x 1024) x (180 MeV)
(b) Hence, show that the energy released from 1 kg of U-235 is about 7.4 x 1013 J which is the energy released by burning about 3 x 106 tonnes of coal.
Solution:
No of U-235 atom in 1 kg fuel
= 1/( x 1.66 x 10-27)
= x 1024
Energy released = (2.563 x 1024) x (180 MeV)
= (2.563 x 1024) x (180 x 106) x (1.6 x 10-19)
= 7.4 x 1013 J
19.1 The atomic model

48. Example 3 Consider the following nuclear fusion.
Given that: 1u = 1.66 x kg = 931 MeV
Mass of neutron: u
H-2: u
H-3: u
He-4: u
(a) Find the amount of nuclear energy released in the reaction.
Solution:
Difference in mass
= ( – – )u
= u
Energy released
= x 931 MeV
= 17.1 MeV
19.1 The atomic model

49. Hence, show that the energy released from 1 kg of fuel is about 3
Hence, show that the energy released from 1 kg of fuel is about 3.3 x 1014 J.
Solution:
No of H-2 and H-3 in 1 kg fuel
= 1/[( ) x 1.66 x 10-27] = x 1026
Total energy released
= ( x 1026) x (2.74 x 10-12)
= 3.3 x 1014 J
19.1 The atomic model

50. Nuclear Energy – nucleus
Protons & neutrons are collectively called nucleons.
A = mass number / Nucleon number
19.1 The atomic model

51. Nuclear Energy – nucleus
Consider a helium nucleus.
Mass of a helium atom (He – 4)
Mass of components
2 protons: 2 x u
2 neutrons: 2 x u
2 electrons: 2 x u u
Total mass: u +
19.1 The atomic model

52. Mass of a helium atom (He – 4)
The difference between the mass of an atom and the total mass of the particles in the atom taken separately is known as mass defect Dm. + –
Dm =
Mass of a helium atom (He – 4)
Mass of components u
2 protons: 2 x u
2 neutrons: 2 x u
2 electrons: 2 x u
Total mass = u
Mass defect of He = u – u = u
19.1 The atomic model

53. Binding Energy mass of a nucleus < total mass of separated nucleons
The energy is required to separate the nucleons (to overcome nuclear force binding the nucleons)
This energy is called the binding energy Eb.
19.1 The atomic model

54. Binding Energy binding energy = Δm c2
where Δm is the mass defect of the nucleus.
19.1 The atomic model

55. Find the binding energy per nucleon in a He-4 atom.
m = u – u
= u
(1u = 931 MeV)
Eb = x 931 MeV = 28.3 MeV
Binding energy per nucleon = 7.08 MeV
19.1 The atomic model

56. Binding Energy
The greater the binding energy per nucleon, (Eb/A), the more / less stable the nuclei.
19.1 The atomic model

57. Binding energy Curve
Fe – 56 has the largest binding energy per nucleon. Therefore, it is very stable.
Either side of maximum binding energy per nucleon are less stable.
19.1 The atomic model

58. Binding energy Curve Fusion
3. When light nuclei are joined together, the binding energy per nucleon is also increased and become more stable.
large binding energy per nucleon ⇔ nucleons are at low energy state. (energy is released.)
19.1 The atomic model

59. Binding energy Curve Fission
4. When a big nucleus disintegrates, the binding energy per nucleon increases and become more stable (energy is released)
19.1 The atomic model

60. Example 4 Find the binding energy per nucleon for a Pb-206 nucleus.
Solution:
Mass defect
= (82 x x u x u) – u
= u
Binding energy = x 931 MeV
= 1690 MeV
Binding energy per nucleon Eb/A
= 1690 MeV / 206
= 8.20 MeV
19.1 The atomic model

61. END
19.1 The atomic model

62. Conditions for a Fusion Reaction (2)
Confinement
The hot plasma must be well isolated away from material surfaces in order to avoid cooling the plasma and releasing impurities that would contaminate and further cool the plasma.
In the Tokamak system, the plasma is isolated by magnetic fields.
19.1 The atomic model

63. History of unified atomic mass unit
The chemist John Dalton was the first to suggest the mass of one atom of hydrogen as the atomic mass unit.
Francis Aston, inventor of the mass spectrometer, later used 1/16 of the mass of one atom of oxygen-16 as his unit.
Oxygen was chosen because it forms chemical compounds with many other elements.
19.1 The atomic model

64. The discovery of isotopes
Oxygen-17 and Oxygen-18 was discovered.
Chemists: one-sixteenth of the average mass of the oxygen atoms.
Physicists: one-sixteenth of the mass of an atom of oxygen-16. 
Difference in atomic weights: about 275 parts per million
19.1 The atomic model

65. Since 1961, by definition the unified atomic mass unit is equal to 1/12 of the mass of a carbon-12 atom.
Physicists: carbon-12 was already used as a standard in mass spectroscopy.
Chemists: difference in atomic mass: 42 parts per million which was acceptable
19.1 The atomic model