1. Nuclear Chemistry

2. The alpha particle (a)
The beta particle (b)
Gamma radiation (γ)

3. a particle ---the nucleus of a He atom
a particle ---the nucleus of a He atom. b Particle --- the same as an electron. gamma rays --- a type of electromagnetic radiation made up of photons (packets of energy).

4. Electron Capture (K-Capture)
Addition of an electron to a proton in the nucleus
As a result, a proton is transformed into a neutron. p 1 + e −1
 n

5. Stable Nuclei
There are no stable nuclei with an atomic number greater than 83.
These nuclei tend to decay by alpha emission.

6. Radiation and Table O
The Regents Reference Tables provides us with a summary of the different types of radiation:

7. Penetrating power of radiation
The ability of radioactive particles to pass through air and other materials is inversely related to their mass.
Alpha particles – the least penetrating, they travel only a few centimeters through air. They can be stopped by a single sheet of paper.
Beta particles – more penetrating, they travel several meters through air. They can be stopped by a sheet of Al or plastic.
Gamma Rays – most penetrating, thick sheets of lead or concrete are required to stop gamma rays.

8. Diagram showing penetrating ability

10. Where does the radiation come from?
Rutherford suggested that the radiation resulted from the breakdown of the nucleus of an atom, resulting in radiation being given off, and the nucleus of the atom changing into a new element.
For instance, the fact that U-238 undergoes alpha decay (emits an a particle) can be shown by this reaction:

11. Why does the atom break up?
Remember that the nucleus of the atom is held together by the strong nuclear force. This force is normally strong enough to hold the protons and neutrons together.
However, sometimes the force of repulsion due to the protons having the same charge overcomes the strong nuclear force and the atom breaks apart.

12. How does beta decay occur?
Sometimes an atom will emit a b particle when it breaks up.
In beta decay a neutron apparently “spits out” an electron (the b particle) and becomes a proton.

13. Balanced nuclear equations
Nuclear reactions can be represented by equations. These reactions are governed by two “laws”:
The law of conservation of mass number – the sum of the mass numbers on the reactant sides must be equal to the sum of the mass numbers on the product side.
This law applies to all nuclear equations!

14. Balanced nuclear equations
The second law is:
The law of conservation of charge (atomic #) – the sum of the atomic numbers on the reactant sides must be equal to the sum of the atomic numbers on the product side.
This law applies to all nuclear equations!

15. Predicting products in alpha decay
The Law of Conservation of Mass Number and the Law of Conservation of Charge allows us to predict products in a nuclear reaction.
For instance, suppose we wanted to predict the atom produced when Radon-222 undergoes alpha decay.

16. Predicting products Based on the two laws we can predict:
The mass number of particle X must be 218.
The charge (atomic #) of particle X must be 84.
The symbol of the element can then be determined from the Periodic Table.

17. How about beta decay?
It works the same way. Let’s look at the beta decay of Strontium-90.
Remember the sum of the mass numbers and atomic number on both sides MUST be the same.
So atom X must be:

18. Using balanced nuclear equations to identify the type of radioactivity.
Suppose we know that a particular atom undergoes radioactive decay and we are able to identify the atom that is produced.
For instance, Iodine-131 is known to form Xenon-131 when it decays. What radioactive particle must it emit?
Using the Laws of Conservation of Mass # and Charge, we can identify the type of radiation given off.
Particle X must be a b particle:

19. Another type of radioactive decay
Some atoms undergo a decay process that produces a positron. A positron has the same mass as an electron, but is positively charged.
Symbols for the positron include:
Positrons are a form of anti-matter. Antimatter is made up of particles with the same properties as normal matter, but are opposite in charge.

20. Positrons are also listed in Table O

21. Positron Emission
We can use our ability to balance nuclear equations to predict what will be given off when Potassium-37 undergoes positron emission.
There’s only one atom that will work, and that’s Argon-37.

22. Your turn!
Using the Laws of Conservation of Mass # and Charge, write balanced nuclear equations for the following nuclear reactions:
Beta decay of Phosphorus-32
Alpha decay of U-238
Positron decay of Iron-53
Decay of Oxygen-17 into Nitrogen-17
Decay of Potassium-42 into Calcium-42
Decay of Plutonium-239 into Uranium-235

23. Balancing Nuclear Equations
Conserve mass number (A).
The sum of protons plus neutrons in the products must equal the sum of protons plus neutrons in the reactants. 1n U
235 92 + Cs
138 55 Rb 96 37
+ 2
= x1
Conserve atomic number (Z) or nuclear charge.
The sum of nuclear charges in the products must equal the sum of nuclear charges in the reactants. 1n U
235 92 + Cs
138 55 Rb 96 37
+ 2
= x0

24. Neutron-Proton Ratios
Like charges repel - Any element with more than one proton (i.e., anything but hydrogen) will have electrostatic repulsions between the protons in the nucleus.
A strong nuclear force helps keep the nucleus from flying apart.
Neutrons play a key role stabilizing the nucleus.
Therefore, the ratio of neutrons to protons (n/Z) is an important factor for stability.

25. Stable Nuclei Belt of stability
The shaded region in the figure shows what nuclides would be stable, the so-called ….
Belt of stability

26. Neutron-Proton Ratios
For smaller nuclei (Z  20) stable nuclei have a neutron-to-proton ratio close to 1:1.

27. Neutron-Proton Ratios
As nuclei get larger, it takes a greater number of neutrons to stabilize the nucleus. Notice the n/Z ratio has increased in the “Belt of stability”

28. Beta Emitters Nuclei above this belt have too many neutrons.
They tend to decay by emitting beta particles.
Because a n0 has been converted to a p+, n/Z has decreased into the “belt of stability”

29. Nuclear Stability and Radioactive Decay
Beta decay
14C N + 0b 6 7 -1
Decrease # of neutrons by 1
40K Ca + 0b 19 20 -1
Increase # of protons by 1
1n p + 0b 1 -1

30. Positron or Electron capture
Nuclei below the belt have too many protons.
They tend to become more stable by positron emission or electron capture.
Because the number of protons has been reduced, the n/Z has increased.

31. Nuclear Stability and Radioactive Decay
Positron decay
11C B + 0b 6 5 +1
Increase # of neutrons by 1
38K Ar + 0b 19 18 +1
Decrease # of protons by 1
1p n + 0b 1 +1

32. Nuclear Stability and Radioactive Decay
Electron capture decay
37Ar + 0e Cl 18 17 -1
Increase number of neutrons by 1
55Fe + 0e Mn 26 25 -1
Decrease number of protons by 1
1p + 0e n 1 -1

33. Neutron-Proton Ratio and Nuclear Stability

34. Neutron-Proton Ratio and Nuclear Stability
Special stability is associated with certain proton and neutron numbers
due to shell effects in nuclei similar to the s, p, d, and f shells in atoms
These proton and neutron numbers are called “Magic Numbers.”
Magic numbers are:

35. Neutron-Proton Ratio and Nuclear Stability
Example nuclides with magic numbers of nucleons includes:

36. Mass Defect Some of the mass can be converted into energy
Shown by a very famous equation!
E=mc2
Energy
Mass
Speed of light

37. Nuclear Stability and Radioactive Decay
Beta decay
14C N + 0b 6 7 -1
Decrease # of neutrons by 1
40K Ca + 0b 19 20 -1
Increase # of protons by 1
1n p + 0b 1 -1
Positron decay
11C B + 0b 6 5 +1
Increase # of neutrons by 1
38K Ar + 0b 19 18 +1
Decrease # of protons by 1
1p n + 0b 1 +1

38. Nuclear Stability and Radioactive Decay
Electron capture decay
37Ar + 0e Cl 18 17 -1
Increase number of neutrons by 1
55Fe + 0e Mn 26 25 -1
Decrease number of protons by 1
1p + 0e n 1 -1
Alpha decay
Decrease number of neutrons by 2
212Po He + 208Pb 84 2 82
Decrease number of protons by 2
Spontaneous fission
252Cf In + 21n 98 49

39. Nuclear binding energy + 19F 91p + 101n
Nuclear binding energy is the energy required to break up a nucleus into its component protons and neutrons.
Nuclear binding energy + 19F p + 101n 9 1
DE = (Dm)c2
9 x (p mass) + 10 x (n mass) = amu
Dm= amu – amu
Dm = amu
DE = amu x (3.00 x 108 m/s)2
= x 1016 amu m2/s2
Using conversion factors:
1 kg = x 1026 amu
1 J = kg m2/s2
DE = x 10-11J

40. Nuclear binding energy = 1.43 x 1010kJ/mol
DE = (-2.37 x 10-11J) x (6.022 x 1023/mol)
DE = x 1013J/mol
DE = x 1010kJ/mol
Nuclear binding energy = 1.43 x 1010kJ/mol
binding energy per nucleon =
binding energy
number of nucleons =
2.37 x J
19 nucleons
= 1.25 x J/nucleon

41. 21.1
The atomic mass of is amu. Calculate the nuclear binding energy of this nucleus and the corresponding nuclear binding energy per nucleon.

42. 21.2
Strategy
To calculate the nuclear binding energy, we first determine the
difference between the mass of the nucleus and the mass of all the protons and neutrons, which gives us the mass defect. Next, we apply Equation (19.2) [ΔE = (Δm)c2].
Solution
There are 53 protons and 74 neutrons in the iodine nucleus.
The mass of atom is
53 x amu = amu
and the mass of 74 neutrons is
74 x amu = amu

43. 21.3
Therefore, the predicted mass for is = amu, and the mass defect is
Δm = amu amu
= amu
The energy released is
ΔE = (Δm)c2
= ( amu) (3.00 x 108 m/s)2
= x 1017 amu · m2/s2

44. 21.4
Let’s convert to a more familiar energy unit of joules. Recall that 1 J = 1 kg · m2/s2. Therefore, we need to convert amu to kg:
Thus, the nuclear binding energy is 1.73 x J . The nuclear binding energy per nucleon is obtained as follows:

45. Half-lifes
The rate at which a particular radioisotope decays is described by its half-life.
The half-life is defined as the time that it takes for one half of a sample of a radioactive element to decay into another element.
The half-life of a radioisotope is dependent only on what the radioisotope is.

46. Radioactive Decay Rates
Relative stability of nuclei can be expressed in terms of the time required for half of the sample to decay
Examples: time for 1 g to decay to .5 g
Co yr
Cu h
U x 109 yr
U x 108 yr

47. The time required for half of a sample to decay
Half-Life
The time required for half of a sample to decay

49. Half-Life

50. Kinetics of Radioactive Decay
A wooden object from an archeological site is subjected to radiocarbon dating. The activity of the sample that is due to 14C is measured to be 11.6 disintegrations per second. The activity of a carbon sample of equal mass from fresh wood is 15.2 disintegrations per second. The half-life of 14C is 5715 yr. What is the age of the archeological sample?

51. Kinetics of Radioactive Decay
First we need to determine the rate constant, k, for the process.
= t1/2
0.693 k
= 5715 yr
0.693 k
= k
0.693
5715 yr
= k
1.21  10−4 yr−1

52. Kinetics of Radioactive Decay
Now we can determine t:
= kt Nt N0 ln
= (1.21  10−4 yr−1) t
11.6
15.2 ln
= (1.21  10−4 yr−1) t
ln 0.763
= t
6310 yr

53. Table N provides us with a list of various nuclides, their decay modes, and their half-lifes.
Using Table N, what is the decay mode and half-life for Radium-226?

54. Using Table N Table N indicates that Radium-226 undergoes alpha decay.
Based on this we can write a balanced nuclear equation to represent this reaction:
This tells us that for every atom of Radium that decays an atom of Radon is produced.

56. Using Half-life
Table N also tells us that Radium-226 has a half-life of 1600 years.
Starting with a 100g sample, after 1 half-life (or 1600 years), 50g remain.
After another 1600 years, half of the 50g will remain (25g).

57. Carbon-14 Dating
The age of objects that were once alive can be determined by using the C-14 dating test. In this test, scientists determine how much C-14 is left in a sample and from this determine the age of the object.
From Table N we can determine that C-14 undergoes b decay:

58. Where does the Carbon-14 come from?
C-14 is created in the atmosphere by cosmic rays.
It becomes part of living things through photosynthesis and the food chain.
When the plant or animal dies, the C-14 begins to decay.

59. Using C-14 to Age Objects
By comparing the amount of C-14 left in a sample to the amount that was present when it was alive, and using the half-life of 5700 years (Table N), one can determine the age of a sample.

60. Uranium-238 Series
The Uranium-238 Decay Series is used to determine the age of rocks.
In this series, the ratio of the U-238 to the Pb-206 is used to determine the age of the rock.

61. Sample Half-life Problem 1
A 10 gram of sample of Iodine-131undergoes b decay, what will be the mass of iodine remaining after 24 days?
From Table N, the ½ life of iodine is determined to be approximately 8 days.
That means that 24 days is equivalent to 3 half-lifes.
The decay of 10 grams of I-131 would produce:
1.25 grams of I-131 would remain after 24 days.

62. Sample Half-life Problem 2
A sample of a piece of wood is analyzed by C-14 dating. The percent of C-14 is found to be 25% of what the original C-14 concentration was. What is the age of the sample?
First, let’s analyze how many half-lives have taken place.
Two half-lives have gone by while the sample decayed from the original C-14 concentration to 25% of that concentration.
Based on Table N, the half-life of C-14 is 5730 years, so…

64. Continued…
Nuclear Chemistry:
Fission and Fusion

65. nuclear fission is either a nuclear reaction or a radioactive decay process in which the nucleus of an atom splits into smaller parts (lighter nuclei). The fission process often produces free neutrons and photons (in the form of gamma rays), and releases a very large amount of energy even by the energetic standards of radioactive decay.

66. Nuclear Fission How does one tap all that energy?
Nuclear fission is the type of reaction carried out in nuclear reactors.

68. Nuclear Fission
This process continues in what we call a nuclear chain reaction.

69. Nuclear Fission
If there are not enough radioactive nuclides in the path of the ejected neutrons, the chain reaction will die out.

70. Energy in Nuclear Reactions
There is a tremendous amount of energy stored in nuclei.
Einstein’s famous equation, E = mc2, relates directly to the calculation of this energy.

71. Energy in Nuclear Reactions
For example, the mass change for the decay of 1 mol of uranium-238 is − g.
The change in energy, E, is then
E = (m) c2
E = (−4.6  10−6 kg)(3.00  108 m/s)2
E = −4.1  1011 J

72. Nuclear Fission
Bombardment of the radioactive nuclide with a neutron starts the process.
Neutrons released in the transmutation strike other nuclei, causing their decay and the production of more neutrons.

73. nuclear fusion is a nuclear reaction in which two or more atomic nuclei collide at very high speed and join to form a new type of atomic nucleus

75. Nuclear Fusion Fusion would be a superior method of generating power.
The good news is that the products of the reaction are not radioactive.
The bad news is that in order to achieve fusion, the material must be in the plasma state at several million kelvins.