nuclear chemistry chapter 21 nuclear chemistry nuclear chemistry the nucleus remember that the...
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NuclearChemistry
Chapter 21Nuclear Chemistry
NuclearChemistry
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.)
NuclearChemistry
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.
Also considered to be the nuclear charge.
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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-234Uranium-235Uranium-238
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• There are three types of Radioactivity:
1. Spontaneous Decay
2. Nuclear Fission
3. Nuclear Fusion
Radioactivity
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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.
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Alpha Decay:
Loss of an -particle (a helium nucleus)
He42
U23892
Th23490 He4
2 +
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Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
U23892
Th23490 He4
2 +
NuclearChemistry
Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
U23892
Th 23490 He4
2 +
Total on top = 238
NuclearChemistry
Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
U23892
Th23490 He4
2 +
Total on top = 238
Total on top = 238
NuclearChemistry
Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
U23892
Th 23490 He4
2 +
Total on bottom = 92
NuclearChemistry
Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
U23892
Th23490 He4
2 +
Total on bottom = 92
Total on bottom = 92
NuclearChemistry
Alpha Decay:
Note how to balance a nuclear equation (even easier than a chemical equation):
U23892
Th23490 He4
2 +
Total on top = 238
Total on top = 238
Total on bottom = 92
Total on bottom = 92
NuclearChemistry
Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Ra22688
X?? He4
2 +
NuclearChemistry
Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Ra22688
X?? He4
2 +
Total on top = 226
Total on bottom = 88
NuclearChemistry
Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Ra22688
X?? He4
2 +
Total on top = 226
Total on top = 226
Total on bottom = 88
Total on bottom = 88
NuclearChemistry
Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Ra22688
X22286 He4
2 +
Total on top = 226
Total on top = 226
Total on bottom = 88
Total on bottom = 88
NuclearChemistry
Typical Balancing Problem:
Note how to balance a nuclear equation (even easier than a chemical equation):
Ra22688
Rn22286 He4
2 +
Total on top = 226
Total on top = 226
Total on bottom = 88
Total on bottom = 88
NuclearChemistry
Beta Decay:
Loss of a -particle (a high energy electron)
0−1 e0
−1or
I13153 Xe131
54 + e0
−1
n p + e10
11 -1
0
NuclearChemistry
Beta Decay:
Loss of a -particle (a high energy electron)
0−1 e0
−1or
I13153 Xe131
54 + e0
−1
53 p, 78 n
NuclearChemistry
Beta Decay:
Loss of a -particle (a high energy electron)
0−1 e0
−1or
I13153 Xe131
54 + e0
−1
53 p, 78 n 54 p, 77 n
NuclearChemistry
Beta Decay:
Loss of a -particle (a high energy electron)
0−1 e0
−1or
I13153 Xe131
54 + e0
−1
53 p, 78 n 54 p, 77 n
Good for decreasing the ratio of neutrons to protons!
NuclearChemistry
Positron Emission:
Loss of a positron (a particle that has the same mass as but opposite charge than an electron)
e01
C116
B115 + e0
1
Anti-matter!
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Positron Emission:
Loss of a positron (a particle that has the same mass as but opposite charge than an electron)
e01
C116
B115 + e0
1
p n + e11
10 1
0
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Positron Emission:
Loss of a positron (a particle that has the same mass as but opposite charge than an electron)
e01
C116
B115 + e0
1
6 p, 5 n 5 p, 6 n
NuclearChemistry
Positron Emission:
Loss of a positron (a particle that has the same mass as but opposite charge than an electron)
e01
C116
B115 + e0
1
6 p, 5 n 5 p, 6 n
Good for increasing the ratio of neutrons to protons!
NuclearChemistry
Gamma Emission:
Loss of a -ray (high-energy radiation that almost always accompanies the loss of a nuclear particle)
00
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Gamma Emission:
Loss of a -ray (high-energy radiation that almost always accompanies the loss of a nuclear particle)
00Sometimes shown to emphasize energy transfer, but not necessary for balancing.
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Electron Capture (K-Capture)
Addition of an electron to a proton in the nucleusAs a result, a proton is transformed into a
neutron.
p11 + e0
−1 n1
0
NuclearChemistry
Electron Capture (K-Capture)
Addition of an electron to a proton in the nucleusAs a result, a proton is transformed into a
neutron.
p11 + e0
−1 n1
0
Opposite of Beta Decay.
Good for increasing the ratio of neutrons to protons!
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More Balancing Examples:
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More Balancing Examples:
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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.
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Neutron-Proton Ratios
• Neutrons play a key role stabilizing the nucleus.
• Therefore, the ratio of neutrons to protons is an important factor.
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For smaller nuclei (Z 20) stable nuclei have a neutron-to-proton ratio close to 1:1.
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Neutron-Proton Ratios
As nuclei get larger, it takes a greater number of neutrons to stabilize the nucleus.
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Stable Nuclei
The shaded region in the figure shows what nuclides would be stable, the so-called belt of stability.
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Stable Nuclei
• Nuclei above this belt have too many neutrons.
• They tend to decay by emitting beta particles.
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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!
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Stable Nuclei
• Nuclei below the belt have too many protons.
• They tend to become more stable by positron emission or electron capture.
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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!
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Stable Nuclei
• There are no stable nuclei with an atomic number greater than 83.
• These nuclei tend to decay by alpha emission.
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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.
NuclearChemistry
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.
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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).
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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.
NuclearChemistry
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.
NuclearChemistry
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.
These are called Magic Numbers!
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Nuclear Fission
Nuclear transformations can be induced by accelerating a particle and colliding it with the nuclide.
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Particle Accelerators
These particle accelerators can be enormous, having circular tracks with paths that are miles long.
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Nuclear Fusion
• Combine two small atoms to make a larger one.
H + H He21 1
2 24
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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
NuclearChemistry
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
Chemical Reaction Equivalent:
ln[A]t
[A]o
= -kt
NuclearChemistry
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
For the Nuclear Version, you can use the following:
1. Number of Nuclei
2. Mass of Sample
3. Disintegrations per second
NuclearChemistry
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.
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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.
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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?
NuclearChemistry
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
NuclearChemistry
Kinetics of Radioactive Decay
Now that we know k, we can determine t:
= -kt Nt
N0
ln
= -(1.21 10−4 yr−1) t 11.615.2
ln
= -(1.21 10−4 yr−1) t ln 0.763
= t 2230 yr
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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 = mc2 .
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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.
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Example Nuclear Decay:
Loss of an -particle (a helium nucleus)
He42
U23892
Th23490 He4
2 +
NuclearChemistry
Energy in Nuclear Reactions
The mass change for the decay of 1 mol of uranium-238 is −0.0046 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
NuclearChemistry
Energy in Nuclear Reactions
The mass change for the decay of 1 mol of uranium-238 is −0.0046 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.
NuclearChemistry
Energy in Nuclear Reactions
For example, the mass change for the decay of 1 mol of uranium-238 is −0.0046 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 JCompare to chemical reactions, typically a few hundred kJ.
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Energy in Nuclear Reactions
For example, the mass change for the decay of 1 mol of uranium-238 is −0.0046 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 JIn WebAssign, enter only positive energies.
NuclearChemistry
How did we calculate this mass Change?
It appears that mass is conserved when balancing!
U23892
Th23490 He4
2 +
Total on top = 238
Total on top = 238
NuclearChemistry
How did we calculate this mass Change?
We need more exact masses when calculating energy changes!
U23892
Th23490 He4
2 +
Total on top = 238
Total on top = 238
238.0003 g 233.9942 g 4.0015g
NuclearChemistry
How did we calculate this mass Change?
We need more exact masses when calculating energy changes!
U23892
Th23490 He4
2 +
Total on top = 238
Total on top = 238
238.0003 g 233.9942 g 4.0015g
m = final mass – initial mass = -0.0046 g
NuclearChemistry
Nuclear Binding Energy
A helium nucleus consists of 2 protons and 2 neutrons, so mass should be:
massHe = 2(massneutron) + 2(massproton)
massHe = 2(1.00866 amu) + 2(1.00728 amu)
Predicted massHe = 4.03188 amu
True massHe = 4.00150 amu
NuclearChemistry
Nuclear Binding EnergyPredicted massHe = 4.03188 amu
True massHe = 4.00150 amu
Difference = 0.03038 amu
This difference is called the mass defect.
1 g = 6.022 x 1023 amu, so:
Difference = 5.045 x 10-26 g
E = 4.533 x 10-12 J = Binding Energy
NuclearChemistry
Nuclear Binding EnergyE = 4.533 x 10-12 J = Binding Energy
To fairly compare nuclei, want binding energy per nucleon, so
4.533 x 10-12 J / 4 nucleons
= 1.13 x 10-12 J / nucleon
NuclearChemistry
Nuclear Binding EnergyE = 4.533 x 10-12 J = Binding Energy
To fairly compare nuclei, want binding energy per nucleon, so
4.533 x 10-12 J / 4 nucleons
= 1.13 x 10-12 J / nucleon
Note: For 1 mole of He nuclei,
binding energy = 2.730 x 10+12 J
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