1. Chapter 21 Nuclear Chemistry

2. The Nucleus
Remember that the nucleus is comprised of the two nucleons, protons and neutrons.
The number of protons is the atomic number.
The number of protons and neutrons together is effectively the mass of the atom. (We will look at this in more detail later.)

3. The Nucleus
Also considered to be the nuclear charge.
Remember that the nucleus is comprised of the two nucleons, protons and neutrons.
The number of protons is the atomic number.
The number of protons and neutrons together is effectively the mass of the atom.

4. Isotopes
Not all atoms of the same element have the same mass due to different numbers of neutrons in those atoms.
There are three naturally occurring isotopes of uranium:
Uranium-234
Uranium-235
Uranium-238

5. Radioactivity There are three types of Radioactivity:
Spontaneous Decay
Nuclear Fission
Nuclear Fusion

6. Spontaneous Decay
It is not uncommon for some nuclides of an element to be unstable, or radioactive.
We refer to these as radionuclides.
There are several ways radionuclides can decay into a different nuclide.

8. He U Th He Alpha Decay: Loss of an -particle (a helium nucleus) + 4 2
238 92
 Th
234 90 He 4 2 +

9. Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation): U
238 92
 Th
234 90 He 4 2 +

10. Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
Total on top = 238 U
238 92
 Th
234 90 He 4 2 +

11. Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
Total on top = 238
Total on top = 238 U
238 92
 Th
234 90 He 4 2 +

12. Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation): U
238 92
 Th
234 90 He 4 2 +
Total on bottom = 92

13. Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation): U
238 92
 Th
234 90 He 4 2 +
Total on bottom = 92
Total on bottom = 92

14. Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
Total on top = 238
Total on top = 238 U
238 92
 Th
234 90 He 4 2 +
Total on bottom = 92
Total on bottom = 92

15. Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation): Ra
226 88 X ? He 4 2
 +

16. Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Total on top = 226 Ra
226 88 X ? He 4 2
 +
Total on bottom = 88

17. Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Total on top = 226
Total on top = 226 Ra
226 88 X ? He 4 2
 +
Total on bottom = 88
Total on bottom = 88

18. Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Total on top = 226
Total on top = 226 Ra
226 88 X
222 86 He 4 2
 +
Total on bottom = 88
Total on bottom = 88

19. Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Total on top = 226
Total on top = 226 Ra
226 88 Rn
222 86 He 4 2
 +
Total on bottom = 88
Total on bottom = 88

20.  e I Xe e Beta Decay: n  p + e +
Loss of a -particle (a high energy electron)  −1 e or
n  p e 1 -1 I
131 53 Xe 54
 + e −1

21.  e I Xe e Beta Decay: + Loss of a -particle (a high energy electron)−1 e or
53 p, 78 n I
131 53 Xe 54
 + e −1

22.  e I Xe e Beta Decay: + Loss of a -particle (a high energy electron)−1 e or
53 p, 78 n
54 p, 77 n I
131 53 Xe 54
 + e −1

23.  e I Xe e Beta Decay: + Loss of a -particle (a high energy electron)−1 e or
53 p, 78 n
54 p, 77 n I
131 53 Xe 54
 + e −1
Good for decreasing the ratio of neutrons to protons!

24. e C B e Positron Emission: Anti-matter!
Loss of a positron (a particle that has the same mass as but opposite charge than an electron)
Anti-matter! e 1 C 11 6
 B 5 + e 1

25. e C B e Positron Emission: p  n + e
Loss of a positron (a particle that has the same mass as but opposite charge than an electron) e 1
p  n e 1 C 11 6
 B 5 + e 1

26. e C B e Positron Emission:
Loss of a positron (a particle that has the same mass as but opposite charge than an electron) e 1
6 p, 5 n
5 p, 6 n C 11 6
 B 5 + e 1

27. e C B e Positron Emission:
Loss of a positron (a particle that has the same mass as but opposite charge than an electron) e 1
6 p, 5 n
5 p, 6 n C 11 6
 B 5 + e 1
Good for increasing the ratio of neutrons to protons!

28. Gamma Emission:
Loss of a -ray (high-energy radiation that almost always accompanies the loss of a nuclear particle) 

29. Gamma Emission:
Loss of a -ray (high-energy radiation that almost always accompanies the loss of a nuclear particle) 
Sometimes shown to emphasize energy transfer, but not necessary for balancing.

30. 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

31. 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
Opposite of Beta Decay.
Good for increasing the ratio of neutrons to protons!

33. More Balancing Examples:

34. More Balancing Examples:

43. Neutron-Proton Ratios
Any element with more than one proton (i.e., anything but hydrogen) will have repulsions between the protons in the nucleus.
A strong nuclear force helps keep the nucleus from flying apart.

44. Neutron-Proton Ratios
Neutrons play a key role stabilizing the nucleus.
Therefore, the ratio of neutrons to protons is an important factor.

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

46. Neutron-Proton Ratios
As nuclei get larger, it takes a greater number of neutrons to stabilize the nucleus.

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

48. Stable Nuclei Nuclei above this belt have too many neutrons.
They tend to decay by emitting beta particles.

49. Stable Nuclei Nuclei above this belt have too many neutrons.
They tend to decay by emitting beta particles.
Good for decreasing the ratio of neutrons to protons!

50. Stable Nuclei Nuclei below the belt have too many protons.
They tend to become more stable by positron emission or electron capture.

51. Stable Nuclei Nuclei below the belt have too many protons.
They tend to become more stable by positron emission or electron capture.
Good for increasing the ratio of neutrons to protons!

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

53. Stable Nuclei Heavy atoms have too many protons repelling.
They tend to become more stable by alpha emission.
No major change in the ratio of neutrons to protons! There will be a slight increase.

54. Stable Nuclei Heavy atoms have too many protons repelling.
They tend to become more stable by alpha emission.
Example: If have 150 n and 100 p, lose an alpha particle and will have 148 n and 98 p. Ratio rises from 1.50 to 1.51.

55. Radioactive Series
Large radioactive nuclei cannot stabilize by undergoing only one nuclear transformation.
They undergo a series of decays until they form a stable nuclide (often a nuclide of lead).

56. Some Trends
Nuclei with an even number of protons and neutrons tend to be more stable than nuclides that have odd numbers of these nucleons.

57. Some Trends
Nuclei with 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82, or 126 neutrons tend to be more stable than nuclides with a different number of nucleons.

58. Some Trends
These are called Magic Numbers!
Nuclei with 2, 8, 20, 28, 50, or 82 protons or 2, 8, 20, 28, 50, 82, or 126 neutrons tend to be more stable than nuclides with a different number of nucleons.

59. Nuclear Fission
Nuclear transformations can be induced by accelerating a particle and colliding it with the nuclide.

60. Particle Accelerators
These particle accelerators can be enormous, having circular tracks with paths that are miles long.

68. Nuclear Fusion H + H  He
Combine two small atoms to make a larger one. 2
H + H  He 2 4 2 1 1

70. Kinetics of Radioactive Decay
Nuclear transmutation is a first-order process.
The kinetics of such a process, you will recall, obey this equation:
= -kt Nt N0 ln

71. Kinetics of Radioactive Decay
Nuclear transmutation is a first-order process.
The kinetics of such a process, you will recall, obey this equation:
Chemical Reaction Equivalent:
= -kt Nt N0 ln
[A]t ln
= -kt
[A]o

72. Kinetics of Radioactive Decay
Nuclear transmutation is a first-order process.
The kinetics of such a process, you will recall, obey this equation:
For the Nuclear Version, you can use the following:
Number of Nuclei
Mass of Sample
Disintegrations per second
= -kt Nt N0 ln

73. Kinetics of Radioactive Decay
The half-life of such a process is:
= t1/2
0.693 k
Comparing the amount of a radioactive nuclide present at a given point in time with the amount normally present, one can find the age of an object.

74. Measuring Radioactivity
One can use a device like this Geiger counter to measure the amount of activity present in a radioactive sample.
The ionizing radiation creates ions, which conduct a current that is detected by the instrument.

75. 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?

76. 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

77. Kinetics of Radioactive Decay
Now that we know k, 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
2230 yr

79. 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.
It would be more appropriate to think of this equation as E = mc2 .

80. Energy in Nuclear Reactions
In the types of chemical reactions we have encountered previously, the amount of mass converted to energy has been minimal (too small to detect).
However, these energies are many thousands of times greater in nuclear reactions.

81. Example Nuclear Decay:
Loss of an -particle (a helium nucleus) He 4 2 U
238 92
 Th
234 90 He 4 2 +

82. Energy in Nuclear Reactions
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

83. Energy in Nuclear Reactions
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
Notice mass must be in kg.

84. 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
Compare to chemical reactions, typically a few hundred kJ.

85. 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
In WebAssign, enter only positive energies.

86. How did we calculate this mass Change?
It appears that mass is conserved when balancing!
Total on top = 238
Total on top = 238 U
238 92
 Th
234 90 He 4 2 +

87. How did we calculate this mass Change?
We need more exact masses when calculating energy changes!
Total on top = 238
Total on top = 238 U
238 92
 Th
234 90 He 4 2 +
g g g

88. How did we calculate this mass Change?
We need more exact masses when calculating energy changes!
Total on top = 238
Total on top = 238 U
238 92
 Th
234 90 He 4 2 +
g g g
m = final mass – initial mass = g

89. Nuclear Binding Energy
A helium nucleus consists of 2 protons and 2 neutrons, so mass should be:
massHe = 2(massneutron) + 2(massproton)
massHe = 2( amu) ( amu)
Predicted massHe = amu
True massHe = amu

90. Nuclear Binding Energy
Predicted massHe = amu
True massHe = amu
Difference = amu
This difference is called the mass defect.
1 g = x 1023 amu, so:
Difference = x g
E = x J = Binding Energy

91. Nuclear Binding Energy
E = x J = Binding Energy
To fairly compare nuclei, want binding energy per nucleon, so
4.533 x J / 4 nucleons
= 1.13 x J / nucleon

92. Nuclear Binding Energy
E = x J = Binding Energy
To fairly compare nuclei, want binding energy per nucleon, so
4.533 x J / 4 nucleons
= 1.13 x J / nucleon
Note: For 1 mole of He nuclei,
binding energy = x J