Quantitative Analysis of Nuclear Reactions - Learn

Quantitative Analysis of Nuclear Reactions – Learn

All quantitative analysis of nuclear reactions relies on two key principles, the Law of Conversation of Energy and Einstein’s mass-energy equivalence. Through nuclear reactions, there is an apparent loss of mass between the reactants and products. Considering the relationship between mass and energy, described by E=mc², this would mean that the conservation of energy has been violated, which is not possible. As such, this means that has mass “lost” (termed the mass deficit) is energy that has been converted into other forms, namely heat (kinetic) and light.

When undertaking quantitative analysis of a nuclear reaction it is important to remember that the atomic mass must be used. This will always be provided in the question, the values in the periodic table cannot be used.

Alpha decay

Large, unstable nucleus usually undergo alpha decay, involving the emission of an alpha particle (He-4 nucleus) from the parent nucleus. The energy released in this reaction generally increases the kinetic energy of the products. To satisfy the conservation of momentum, this results in the significantly smaller alpha particle being emitted with a much greater magnitude of velocity than the decayed nucleus. The larger nucleus is sometimes left in an excited state, emitting a gamma photon to enter a more stable configuration.

Beta decay

The energy produced in beta decay is very similar to alpha decay, although at a smaller scale. In agreement with the conservation of momentum, the beta particle is released at a very large velocity relative to the barely impacted decayed nucleus, due to the enormous difference in size. As with alpha decay, there is the potential for a gamma photon to be emitted from the nucleus if it is left in an excited state.

Gamma radiation

The energy associated through the emission of a gamma photon from an excited nucleus is comparatively quite small. The mass deficit for excited nuclei are generally not provided. Other sources of gamma radiation include particle and anti-particle annihilations, and nuclear fusion.

Nuclear fusion

Fusion reactions release more energy than any other nuclear reaction, requiring huge amount of energy input for the reaction to occur in the first place. As a result of this, it has been difficult for physicists on Earth to sustain fusion reactions in a laboratory setting. The closest source of consistent fusion reactions is the sun, where reactions like the proton-proton chain and CNO cycle occur (and later in it’s life cycle the triple-alpha process).

Nuclear fission

Just as with other nuclear reactions, there is a difference in mass between the reactants and products in a fission reaction, including any neutrons. Again, the energy released can contribute to the kinetic energy of the products and may release gamma radiation. When considering the mass of the neutrons it is important not to attribute a value of 1 atomic mass unit, as it has a larger mass and this will be provided in the question.

Comparing nuclear fission and nuclear fusion

When comparing the energy released in Example 3 (5.50 MeV) and Example 4 (173 MeV), it would make sense to say that more energy has been released in the fission reaction. This would contradict many earlier statements, so it is important to understand why fusion releases more energy than fission. When examining energy released in different reactions, it must be done on a like for like basis, so a new metric is needed.

To compare specific nuclear reactions, the energy per nucleon is determined and then these values are contrasted.

Binding energy per nucleon graphs

It is common to represent the binding energy per nucleon in a graph, which can provide scope for analysis and is regularly given as stimulus in questions. An example of this style of graph has been provided below.

The graph displays a general trend of binding energy increasing rapidly towards iron and then decreasing gradually as the number of nucleons increases. As iron contains the most binding energy per nucleon, it is considered the most stable element (a fact you may have heard elsewhere). It is also interesting to note that hydrogen-1 has a binding energy of zero, as there are no nucleons to bind.

Example 1:

Radium-222 naturally undergoes alpha decay forming polonium-218. Consider the information provided in the table below, and determine the energy released in this reaction in MeV.

Isotope	Atomic Mass (u)
He-4	4.002620
Po-218	218.00897
Ra-222	222.017576

Answer:

Nuclear equation: Ra-222 → Po-218 + He-4

Mass of reactant = 222.017576 u

Mass of products = 218.00897 + 4.00260

= 222.01157 u

Mass deficit = Mass of reactant – Mass of products

= 222.017576 – 222.01157

= 0.006006 u [1 u = 1.661 x 10^-27 kg]

= 9.98 x 10^-30 kg

Now apply the mass-energy equivalence.

$E=mc^{2}\\$

$E\ =\ 9.98\times 10^{-30} \times \left( 3\times 10^{8}\right)^{2}\\$

$\ \ \ \ =\ 8.98\times 10^{-13} \ J\ \ \ \ \ \ \ \ (eV\ =\ \cfrac{J}{q_{e}})$

$\ \ \ \ =\ 5.60\times 10^{6} \ eV\\$

$\ \ \ \ =\ 5.60\ MeV\\$

Hence, 5.60 MeV of energy has been released during this reaction. It will contribute to the kinetic energies of the polonium-218 atom and the alpha particle, in accordance with the conversation of momentum.

Example 2:

Strontium-90 naturally undergoes beta decay forming yttrium-90. Consider the information provided in the table below, and determine the energy released in this reaction in MeV.

Particle	Atomic Mass (u)
Beta particle	0.00055
Sr-90	89.90773
Y-90	89.90584

Answer:

Nuclear equation: Sr-90 → Y-90 + β^–

Mass of reactant = 89.90773 u

Mass of products = 89.90584 + 0.00055

= 89.90639 u

Mass deficit = Mass of reactant – Mass of products

= 89.90773 – 89.90639

= 0.00134 u [1 u = 1.661 x 10^-27 kg]

= 2.23 x 10^-30 kg

Now apply the mass-energy equivalence.

$E=mc^{2}\\$

$E\ =\ 2.23\times 10^{-30} \times \left( 3\times 10^{8}\right)^{2}\\$

$\ \ \ \ =\ 2.00\times 10^{-13} \ J\ \ \ \ \ \ \ \ (eV\ =\ \cfrac{J}{q_{e}})$

$\ \ \ \ =\ 1.25\times 10^{6} \ eV\\$

$\ \ \ \ =\ 1.25\ MeV\\$

Hence, 1.25 MeV of energy has been released during this reaction.

Example 3:

During the proton-proton chain, a proton (H-1) fuses with deuterium (H-2) to form helium-3. Determine the energy released during this fusion reaction.

Particle	Atomic Mass (u)
H-1	1.007825
H-2	2.014102
He-3	3.016029

Answer:

Nuclear equation: H-1 + H-2 → He-3

Mass of reactants = 1.007825 + 2.014102

= 3.021927 u

Mass of products = 3.016029 u

Mass deficit = Mass of reactant – Mass of products

= 3.021927 – 3.016029

= 0.005898 u [1 u = 1.661 x 10^-27 kg]

= 9.80 x 10^-30 kg

Now apply the mass-energy equivalence.

$E=mc^{2}\\$

$E\ =\ 9.80\times 10^{-30} \times \left( 3\times 10^{8}\right)^{2}\\$

$\ \ \ \ =\ 8.81\times 10^{-13} \ J\ \ \ \ \ \ \ \ (eV\ =\ \cfrac{J}{q_{e}})$

$\ \ \ \ =\ 5.50\times 10^{6} \ eV\\$

$\ \ \ \ =\ 5.50\ MeV\\$

Hence, 5.50 MeV of energy has been released during this reaction.

Example 4:

A neutron enters the nucleus of a uranium-235 atom, causing a fission reaction to take place. While there are a number of reactions that can occur, scientist observe rubidium-96 and caesium-137 as the products.

Particle	Atomic Mass (u)
Neutron	1.00866
Rb-96	95.93427
Cs-137	136.90709
U-235	235.04393

a) Determine the number of neutrons released in this reaction.

b) Determine the energy released in this reaction in MeV.

Answer:

a) Nuclear equation: U-235 + n → Rb-96 + Cs-137 + xn (where x is an unknown number of neutrons)

The reactants contains 236 nucleons, and the products 233 nucleons. Hence, three neutrons must be released in this reaction.

b) Nuclear equation: U-235 + n → Rb-96 + Cs-137 + 3n

Mass of reactant = 235.04393 + 1.00866

= 236.05259 u

Mass of products = 95.93427 + 136.90709 + (3 x 1.00866)

= 235.86734 u

Mass deficit = Mass of reactant – Mass of products

= 236.05259 – 235.86734

= 0.18525 u [1 u = 1.661 x 10^-27 kg]

= 3.08 x 10^-28 kg

Now apply the mass-energy equivalence.

$E=mc^{2}\\$

$E\ =\ 3.08\times 10^{-28} \times \left( 3\times 10^{8}\right)^{2}\\$

$\ \ \ \ =\ 2.77\times 10^{-11} \ J\ \ \ \ \ \ \ \ (eV\ =\ \cfrac{J}{q_{e}})$

$\ \ \ \ =\ 172.87\times 10^{6} \ eV\\$

$\ \ \ \ =\ 173\ MeV\\$

Hence, 173 MeV of energy has been released during this reaction.

Example 5:

Compare the energy released per nucleon, in the fusion of H-1 and H-2 into He-3 (Example 3) and fission of U-235 into Rb-96 and Cs-137 (Example 4).

Answer:

The fusion of H-1 and H-2 into He-3 yields 5.50 MeV. This nuclear reaction involves 3 nucleons.

$Energy\ per\ nucleon\ =\ \cfrac{Energy\ released}{Number\ of\ nucleons}$

$Energy\ per\ nucleon\ =\ \cfrac{5.50}{3}$

$Energy\ per\ nucleon\ =\ 1.83\ MeV/Nucleon$

The fission of U-235 into Rb-96 and Cs-137 yields 173 MeV. This nuclear reaction involves 236 nucleons.

$Energy\ per\ nucleon\ =\ \cfrac{Energy\ released}{Number\ of\ nucleons}$

$Energy\ per\ nucleon\ =\ \cfrac{173}{236}$

$Energy\ per\ nucleon\ =\ 0.73\ MeV/Nucleon$

Therefore, the fusion of H-1 and H-2 into He-3 releases 1.10 MeV more energy per nucleon (1.83 MeV – 0.73 MeV) than the fission of U-235 into Rb-96 and Cs-137.

Example 6:

Determine the energy released by the fusion of H-1 and H-2 into He-3 using the binding energy graph.

Answer:

Binding energy per nucleon of reactants (H-1 and H-2) is approximately 1.1 MeV and the binding energy of product (He-3) is approximately 2.9 MeV. This has been determined by visual approximation using the graph.

Therefore, the energy released is approximately 1.8 MeV (2.9 MeV – 1.1 MeV). This supports the detailed quantitative anaylsis from Example 3 and Example 5.

These graphs can be used to compare the energy yields of fusion and fission. If a vertical line cis constructed through iron-56, all the elements to the left are formed through fusion and all the elements to the right are created through fission. All the elements that form as a result of fusion can be traced back to protons, the initial building blocks of the universe.

No element larger than uranium is presented in this graph. It is the largest naturally occurring element in the universe, created in supernova, the violent deaths of large stars. Any element larger than uranium, containing more than 92 protons, does not form naturally.

Example 7:

Compare the energy released by the proton-proton chain and the fission of uranium-235 using the binding energy graph above.

Solution:

The proton-proton chain fuses Hydrogen-1 to Helium-4. From the binding energy graph, H-1 has zero binding energy, and He-4 has a binding energy of approximately 7.1 MeV per nucleon. Consequently, 7.1 MeV is released overall in this series of reactions.

In the fission of U-235, it will into two smaller nuclei. These will average out to have close to half of the number of nucleons of the original nucleus, about 117. From the graph, U-235 has a binding energy of 7.6 MeV and a nucleus with 117 nucleons with a binding energy of approximately 8.5 MeV. This means that 0.9 MeV (a similar value determined in Example 5) is released in this reaction.

Hence, from the graph, approximately 6.2 MeV more energy is released via the proton-proton chain than the fission of U-235.