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Chapter 10  Nuclear Chemistry

Most of "regular" chemistry is concerned with the coulombic interactions between positive nuclei and negative electrons.

Nuclear chemistry is concerned with the nuclear forces between nucleons, positive protons and neutral neutrons.

Nuclear chemistry has the capacity to change the identity of an atom from one element to another.

Most chemists work with chemicals made of stable, non-radioactive atoms because they want the chemicals they make to stay the same; i.e. they want a constant elemental composition.

This is inherent in the use of the atomic mass as found on the periodic table.

The atomic mass is a natural abundance weighted average of the stable isotopes of each element.

Isotopes are atoms of the same element that have different numbers of neutrons.

Boron as 2 stable isotopes, boron-10 with 5 neutrons and boron-11 with 6 neutrons.

Boron has an average atomic mass of 10.8 amu (atomic mass unit).

Boron-11 is 80% of the boron found on earth and boron-10 is 20%, giving the atomic mass.

0.8(11) + 0.2(10) = 10.8 

Oxygen has 3 stable isotopes, oxygen-16, -17, and -18, but oxygen-17 and -18 are so rare on Earth that they don’t contribute much to the average mass which is just 16 amu for oxygen.

However, nuclear chemists often work with unstable, radioactive isotopes of the elements; for example, boron-9.

The nucleus of boron-9 has 5 protons (because it’s boron) and 4 neutrons.

The sum of the number of protons and neutrons, collectively called nucleons, is the mass number, A; boron-9 has a mass number, A = 9.

The number of protons is given by the atomic number, Z.

The nuclear notation would have us write boron-9 as 59B following the general format ZAXx, where Z and the element symbol, Xx, are redundant pieces of information since the number of protons defines the element.

We can tell that boron with its 5 positive protons prefers to have 5 or 6 neutrons in the nucleus because boron-10 and boron-11 are stable isotopes.

A general conclusion is that the more protons a nucleus has, the more and more neutrons it needs to be stable, presumably to overcome the repulsion between all the positive protons that are packed so tightly together in the nucleus.

For example, the only stable isotope of gold, gold-197, has 79 protons and therefore 197 − 79 = 118 neutrons.

Problem

Fluorine only has 1 stable isotope. What is the mass number of this isotope and how many neutrons does this isotope have?

Fluorine’s average atomic mass is 19.00 amu so fluorine-19, with A=19 is the stable isotope having 19−9=10 neutrons.

———————————————————————–

Problem

Is fluorine-8 a possible unstable isotope of fluroine?

No. In order for an atom to be fluorine, it must have 9 protons exactly, meaning the minimum mass number possible for fluorine is A=9. However, fluorine-9 is so unstable, having 0 neutrons, we don’t know anything about it. Fluorine-14 (with 5 neutrons) is the lowest isotope of fluorine that we know about.

———————————————————————–

Returning to our boron example, with only 4 neutrons, boron-9 is unstable, and therefore radioactive, meaning it will decay by some sort of nuclear reaction into a different nucleus.

There are 4 common modes of decay that an unstable, radioactive nucleus can undergo (there are more than 4 but we are ignoring the less common ones; the decay of boron-9 actually occurs by proton emission which isn’t one we’re going to cover).

We’ll see that which decay mode is used in the nuclear decay reaction by a radioactive isotope is determined by whether a particular nucleus has too many or too few neutrons relative to its number of protons, in addition to the overall mass number of the nucleus.

  1. alpha decay, α-decay = The emission, from the nucleus, of 2 protons and 2 neutrons all together in what amounts to a helium-4 nucleus, 42He, sometimes written 42α.
  2. beta decay, β-decay = The emission, from the nucleus, of 1 electron during the conversion of a neutron (01n) into a proton (11p).
    01n → β + 11p
  3. positron emission, β+-emission = The emission, from the nucleus, of 1 positron (antimatter electron, a "positive electron") during the conversion of a proton into a neutron.
    11p → β+ + 01n
  4. electron capture, EC = an electron from an orbital in the atom (i.e. not from the nucleus) falls down into the nucleus and is absorbed by a proton turning the proton into a neutron (even though a neutron is not made of a proton and an electron, see above about quarks).
    e + 11p → 01n
  5. gamma-ray, γ-radiation = The very high frequency light that turned Bruce Banner into The Hulk, just kidding; but it is the most frequent type of electromagnetic radiation (i.e. light) that is emitted during many nuclear reactions as a way to release energy.

The neutron to proton ratio, N/Z, is often quoted as a measure of the stability of an isotope.

For lighter elements, N/Z ≥ 1 leads to stable nuclei whereas for heavier elements, this ratio can get up to N/Z = 118/79 ≈ 1.5 using gold-197 has an example.

The "band of stability" is seen below in a plot of the number of neutrons versus the number of protons for stable nuclei shown in pink.

Problem

What is the likely decay product of fluorine-17?

Fluorine-17 has 8 neutrons, 2 less than the only stable isotope, fluorine-19.

Fluorine-17 needs more neutrons so it will make a neutron from a proton either by β+-emission or electron capture, either would produce the same product, oxygen-17 according to the following reactions.

Positron emission

917F → β+ + 817O

Electron capture

917F + e817O

———————————————————————–

Problem

What is the likely decay product of carbon-14?

Carbon-14 has a mass number 2 amu higher than the average atomic mass found in the periodic table, 12.011 amu, therefore carbon-14 has too many neutrons and will likely decay by β-decay in order to convert an extra neutron into a proton.

614C → β + 714N

———————————————————————–

10.1  Radiocarbon Dating

Carbon-14 is naturally produced in the upper atmosphere as a product of the collision of neutrons from the solar wind with 714N atoms.

01n + 714N → 614C + 11H

This carbon-14 is then oxidized by O2 (20% of Earth’s atmosphere) into isotopically labeled carbon dioxide, 14CO2.

614C + O214CO2

Plants consume CO2 in the Calvin cycle to make glyceraldehyde 3-phosphate (which can then be turned into a bunch of other things like glucose (C6H12O6) and polysaccharides (starch).

14CO2 + other stuff →plants14C3H7O6P

Presumably, a plant, anything that eats plants, and anything that eats things that eat plants would accumulate carbon-14 in their tissues, eventually reaching a constant value.

Once the organism dies, the amount of carbon-14 in its tissues will decrease because of radioactive decay.

Given an estimate of what the constant value of carbon-14 was when the organism was alive compared to the amount of carbon-14 currently seen provides an estimate of how old the tissue is, i.e. how long ago the organism died and stopped replenishing its accumulation of carbon-14.

In order to do this, you use the first-order integrated rate law since all radioactive decay follows first-order kinetics.

ln


 
[14C]f
[14C]0
 
 


= −kt 

The amount of carbon-14 even while an organism is alive is miniscule, so actually measuring the concentration is destructive to the tissue and difficult.

Instead, the rate of disintegrations can be measured by just detecting the decay particles being ejected.

Since the kinetics obey a first order rate law, r = k [14C], the first-order integrated rate law can be rewritten in terms of the rate of radioactive decay.

ln


 
rf/k
r0/k
 
 


= ln


 
rf
r0
 
 


= −kt 

The half-life of carbon-14 is 5730 years, so it hangs around for a little while allowing us to date organic objects up to about 50,000 years old.

Problem

If a living organism has 15.3 disintegrations of carbon-14 per minute per gram of total carbon and a dead piece of tissue only has 7.4 dpm/(g C), how old is the dead tissue?

First we need the rate constant which we get from the half-life.

k = 
ln(2)
t1/2
 = 
ln(2)
5730  y
 = 1.210 × 10−4  y−1 

Plugging all the data into the first-order integrated rate law:

ln


7.4
15.3
 


= −(1.210 × 10−4  y−1)t 

Solving for t tells us the tissue is 6000 years old.

t = 
ln


7.4
15.3
 


−(1.210 × 10−4  y−1)
 = 6000  y 

———————————————————————–

10.2  Uranium-Lead Dating

Radioactive uranium-238 (U-238), through a radioactive decay series, ultimately turns into stable lead-206.

The longest half-life of the many decays in the series is the first one which is an α-decay to thorium-234 of 4.5 × 109 y, which is 10000 longer than the next longest half-life of the decay series.

So, one can compare the current amount of lead-206 found in an old rock to the current amount of uranium-238 to figure out how old the rock is, assuming all of the lead-206 came from decayed uranium-238.

Problem

If a rock has 0.200 g of Pb-206 for every 1 g of U-238, how old is the rock?

The first step is to figure out the rate constant for the longest half-life of the decay series between U-238 and Pb-206.

k = 
ln(2)
t1/2
 = 
ln(2)
4.5 × 109  y
 = 1.54 × 10−10  y−1 

The second step is to convert the current mass of Pb-206 to the mass of U-238 it came from.

It is sufficient, though not "good" practice, to approximate the atomic masses of the isotopes using their mass numbers.

Do NOT use the average atomic mass of the elements found in the periodic table as those are representative of the average atomic mass of stable isotopes.

We’re not working with an average sample of an element from Earth, we know exactly which isotopes are involved, so we should use their atomic masses only.

 0.200 g Pb-206 × 
mol Pb-206
206 g Pb-206 
 × 
mol U-238
mol Pb-206
 × 
238 g U-238
mol U-238
 = 0.231 g U-238 
 

We return to a modified form of the first-order integrated rate law to determine how old the rock is, where [238U] is just replaced with an amount of U-238 in grams.

 ln


g U-238
g U-238+0.231 g U-238
 


 = −(1.54 × 10−10  y−1t 
t = 1.35 × 109  y 

———————————————————————–

10.3  Nuclear binding energy

You cannot calculate the expected mass of an isotope based on the masses of the particles it is made of.

Nitrogen-14 is made of 7 protons, 7 neutrons, and 7 electrons; but the mass of 7 protons, 7 neutrons, and 7 electrons is more than the mass of 1 nitrogen-14 atom.

This is called the mass defect, Δ m, the difference in mass between an isotope’s atomic mass and the sum of the masses of its parts.

The mass defect of nitrogen-14 is

Δ m =  7(1.00783 amu 11H) + 7(1.00867 amu 01n) − 14.00307 amu N-14  
Δ m =  14.1155 amu − 14.00307 amu N-14 = 0.11243 amu  

Note that the mass of a hydrogen atom is used because it accounts for both 1 proton and 1 electron simultaneously.

When all the particles that comprise the nucleus of an atom come together, energy is released in an amount directly related to the mass defect; the mass lost upon formation of the nucleus turns into energy, the nuclear binding energy.

It is called the nuclear binding energy because it is the amount of energy that would need to be added to the nucleus to break it back apart again because the individual nucleons would need to get their mass back.

The nuclear binding energy is related to the mass defect by Einstein’s famous equation, where c= the speed of light, 2.998 × 108 m/s.

Ebinding = Δ m c2 

In order to use this equation, our mass defect needs to be in kg/mol instead of amu.

Because 1 amu is defined to be 1/12 the mass of a C-12 atom AND 1 mole is defined to be the number of particles in 12 g of C-12 (6.02214 × 1023):

 1 g × 
mol C-12
12 g
 × 
6.02214 × 1023 C-12
mol C-12
 × 
12 amu
C-12
 = 6.02214 × 1023 amu 
 
 0.11243 amu × 
g
6.02214 × 1023 amu
 × 
6.02214 × 1023
mol
 = 0.11243 g/mol = 0.11243 × 10−3 kg/mol 
 
Ebinding =  (0.11243 × 10−3 kg/mol )(2.998 × 108 m/s)2 = 1.0105 × 1013 kg · m2/s2 · mol  
 1.0105 × 1013 kg · m2/s2 · mol = 1.0105 × 1013 J/mol = 1.0105 × 1010 kJ/mol  

Holy crap, 1.0105 × 1010 kJ/mol is A LOT of energy for 1 mol of nitrogen-14.

A more physics friendly energy unit is the mega-electronvolt, 1 MeV = 1.602 × 10−16 kJ.

An electronvolt is the energy gained/lost by an electron passing through a potential difference of 1 volt, so it is simply the product of a volt (V = J/C) and the charge of an electron expressed in coulombs:

eV = 1 V × 96485 C/mol e ÷ 6.022 × 1023 mol−1 = 1.6022 × 10−19 J 
 1.0105 × 1010 kJ/mol × 
MeV
1.602 × 10−16 kJ
 × 
mol N-14
6.022 × 1023 N-14
 = 104.74 MeV 
 

One could take the shortcut and convert equivalent mass units directly to MeV.

 0.11243 amu/N-14 × 
935.1 MeV
amu
 = 104.73 MeV/N-14 
 

Regardless if you use the chemistry friendly kJ/mol or the physics friendly MeV, nuclear binding energies are usually reported per the number of nucleons in the nucleus so that the nuclear binding energies of different elements can be compared to one another since some elements require combining more particles than others.

 1.0105 × 1010 kJ/mol N-14 × 
mol N-14
14 mol nucleons
 = 7.22 × 108 kJ/mol nucleons N-14 
 
 104.73 MeV/N-14 × 
N-14
14 nucleons
 = 7.48 MeV/nucleon N-14 
 

If you plot the nuclear binding energies for the stable isotopes of the elements you get a global maximum* around iron-56, iron-58, and nickel-62; so one might say these are the most stable isotopes as they convert the most mass into energy per nucleon found in their nuclei.

10.4  Fission and Fusion

Fission is the spontaneous breakdown of a large nucleus into smaller nuclei and nucleons; α-decay is NOT considered fission.

Fission of U-236 can be induced by neutron bombardment of U-235.

92235U + 01n → 92236U → 56141Ba + 3692Kr + 3 01n

In the presence of a critical mass of U-235, the emission of neutrons by the fission of U-236 could trigger other U-235 to undergo fission through the formation of U-236 leading to a chain reaction.

The steps of a chain reaction are

  1. initiation (e.g. by neutron bombardment)
  2. propagation (e.g. because a critical mass is present)
  3. termination (e.g. because critical mass is no longer present)

Critical mass is just the mass necessary to sustain a chain reaction.

In nuclear power plants, control rods made of not-fissionable elements like cadmium are used to control the rate of the fusion by absorbing "extra" neutrons.

As an aside, uranium enrichment via centrifuges is necessary to make a critical mass of U-235 because only 0.7% of uranium on Earth is U-235, the rest is U-238; hence why the U.S. is very concerned with the enrichment activities of authoritarian regimes as one way to limit the proliferation of nuclear weapons is to control the enrichment of fissionable materials.

The Little Boy atomic bomb dropped on Hiroshima, Japan on August 6, 1945 was a U-235 fission bomb.

Of the 64 kg of U-235 on the Little Boy bomb, it is estimated only 0.91 kg actually underwent fission before the bomb blew itself apart, releasing an estimated 63 × 109 kJ of energy.
Fusion is the combining of nuclei to make a heavier nucleus.

Unlike neutron bombardment, fusion is difficult due to repulsion between the two positive nuclei as they get closer to each other.

Molybdenum-96 undergoes fusion with deuterium (fancy name for hydrogen-2) to make technetium-97.

4296Mo + 12H → 4397Tc + 01n

All of the transuranium elements (Z > 92), which have NO stable isotopes, are synthesized by some fusion of U-238 with some other nucleus in particle accelerators which are machines that get the nuclei moving with enough kinetic energy to overcome the nuclear repulsion between the two fusing nuclei, as in the production of Californium, Cf.

92238U + 612C → 98246Cf + 4 01n

The sun, and hydrogen-bombs, utilize the fusion of hydrogen nuclei and the result is the creation of A LOT of energy.

12H + 13H → 24He + 01n

The sun undergoes thermonuclear fusion, meaning that the temperatures are so high the hydrogen nuclei have enough kinetic energy to overcome the repulsion between the fusing nuclei; the release of lots of energy only ensures continued thermonuclear fusion.
Both the fusion of light nuclei and fission of heavy nuclei are exothermic reactions.

This is rooted in the differences in nuclear binding energy per nucleon.

If we take the negative of the nuclear binding energy plot from earlier, it becomes immediately obvious why both processes are exothermic despite being opposite processes.

The most stable nuclei (Fe and Ni) are "downhill" (energy releasing, exothermic) from the fusing light nuclei and the fissing heavy nuclei.

In both cases, more mass is being converted to energy as more stable nuclei are products of both reactions.

The energy of fusion and fission reactions can be calculated by taking the difference in mass between the products and reactants, similar to the mass defect.

The fission products are rather variable, depending on the exact conditions each nucleus finds itself in.

Another possible fission of U-236 is shown below.

92235U + 01n → 92236U → 3890Sr + 54144Xe + 2 01n
IsotopeAtomic Mass (amu)
92236U236.045568
3890Sr89.907738
54144Xe143.93851
01n1.00867

The mass difference between reactants and products is −0.18198 amu which amounts to the release of 169.5 MeV of energy per U-236, or for us chemists, 6.96 × 107 kJ/g U-235.


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