Mass and energy

1. Energy equation:

• Einstein’s work on special relativity led him to publish a paper in 1905. This paper suggested that energy and mass were different ways of expressing the same thing that energy and mass were interchangeable and linked using the equation,
• [math] E = m * c^2 [/math]
• Where E is energy (J)
• m is change in mass (kg)
• c is the speed of light, [math] c = 3 * 10^8 ms^{-1} [/math]
• By 1932 his ideas were proved experimentally by Cockcroft and Walton.
• The helium nucleus contains four nucleons two protons and two neutrons.
• The mass of the helium nucleus is very slightly smaller than the mass of its separate nucleons.
• As the helium nucleus forms, some mass is converted to energy and released.
• Calculating the energy released when an alpha particle is formed is straightforward:
  Mass of a proton is [math] 1.6726 * 10^{-27} kg [/math]
  Mass of a neutron is [math] 1.6749 * 10^{-27} kg [/math] mass of two protons is [math] 6.695 * 10^{-27} kg [/math]
  Measured mass of a helium nucleus is [math] 6.6337 * 10^{-27} kg [/math]
  Mass difference is [math] 6.13 * 10^{-29} kg [/math]
• Using
• [math] E = m * c^2 [/math]
• The energy released when a single alpha particle is formed is [math] 5.5 * 10^{-12} \ or 34 MeV [/math],
• A very small amount of mass may be converted into a very large amount of energy and vice versa.
• Work must be done to overcome the very strong nuclear forces that bind the nucleons together and pull a helium nucleus apart. The energy put in to do this creates the extra mass.
• A nucleus of Z protons and N neutrons has a mass that is less than the mass of the protons and neutrons that make it up.
• Mass defect is the difference in mass between individual nucleons and their mass when they from a nucleus.
• This difference in mass is called the mass defect.
• Where mass defect
• [math] ∆m = Zm_p + Nm_n – M_{\text{nucleus}}[/math]
• Measured in kg or atomic mass units (u).
• Since mass and energy are interchangeable, we can also express mass as energy. Binding energy is the energy that corresponds to the mass defect, and is related to the mass defect using
• [math] Binding energy = mass defect × c^2 [/math]
• Figure 1 Binding energy

• Binding energy is the energy that would have to be supplied to the nucleus to separate it back into its constituent protons and neutrons. The binding energy can be expressed in J or in MeV.

2. Atomic mass unit:

• The atomic mass unit (amu) is a unit of mass used to express the mass of atoms and molecules. It is defined as:
• [math] 1 amu = \frac{1}{12} \text{ the mass of a carbon – 12 atom}  \\
  1 amu= 1.66 × 10^{-27} \text{kilograms} [/math]
• The atomic mass unit is used to express the mass of atoms and molecules in a convenient and easily understandable way. It is often denoted by the symbol “u” or “amu”.
• Figure 2 Atomic mass unit of carbon isotope
• Table 1 gives some particle masses in atomic mass units. These are quoted to a large number of significant figures because their differences are small.
• 

• Example

• Calculate the mass defect (in u) and binding energy (in MeV) for an oxygen nucleus, [math] O^{16} [/math]  The mass of an oxygen nucleus is 15.9949u.
  Given data:
  Mass of an oxygen nucleus = 15.9949u
  Mass of proton =[math] m_p [/math] = 1.00728u
  Mass of neutron = [math] m_n [/math]= 1.00867u
  Number of protons = 8
  Number of neutrons = 8
• Find data:
  Mass defect in amu =?
  Binding energy =?
• Formula:
• [math]\Delta m = Z m_p + N m_n – M_{\text{nucleus}}  [/math]
• Solution:
• [math] \Delta m = Z m_p + N m_n – M_{\text{nucleus}} \\
  \Delta m = ( 8(1.00728) + 8 (1.00867) – 15.9949 \\
  \Delta m = 16.1276 – 15.9949 \\
  \Delta m = 0.1327 \, \text{u} [/math]
• The mass defect is 0.1327 u.
• Since 1 amu = 931.5 MeV, the binding energy is 123.6 MeV to four significant figures.

⇒ Conversion of units; 1 amu = 931.5 MeV

• According to Einstein mass energy equivalence is represented by
• [math] E = mc^2 \\
  \text{taking m} = 1 a.m.u = 1.66 * 10^{-27} kg [/math]
• And
• [math] c = 3 * 10^8 m/s [/math]
• Put in equation
• [math] E = mc^2 \\
  E = \left( 1.66 * 10^{-27} \right) * \left( 3 * 10^8 \right)^2 \\
  E = \left( 1.66 * 10^{-27} \right) * \left( 9 \times 10^{16} \right) \\
  E = 14.9 * 10^{-11} \\
  E = 1.49 * 10^{-10} \, \text{J}[/math]
• [math]\text{As} \, 1 \, \text{MeV} = 1.6 * 10^{-13} \, \text{J} [/math]
• [math]E = \frac{1.49 * 10^{-10}}{1.6 * 10^{-13}} \\
  E = 0.93125 * 10^3 \, \text{MeV} \\
  E = 931.25 \, \text{MeV} [/math]

⇒Binding energy per nucleon:

• By measurement, it was found that binding energy is different for different nuclei.
• The binding energy per nucleon for stable nuclei is shown in figure 3
• The average binding energy per nucleon
• [math] \text{Binding energy per nucleon} = \frac {\text{total binding energy}} {\text{number of nucleons}}[/math]
• Experimental data shows that nuclei with a high binding energy per nucleon are most stable. More energy per nucleon is needed to pull the nucleons apart. This information allows us to predict the stability of nuclei of different masses.
• Figure 3
• The graph in Figure 3 shows that:
  – Binding energy per nucleon increases rapidly with nucleon number for lighter elements and is about 8 MeV per nucleon for helium and elements heavier than lithium.
  – Helium nuclei are very stable relative to other low-mass nuclei, which explains why alpha decay is more common than proton emission.
  – Binding energy per nucleon has its highest value for [math] F^{56} [/math], at 8.79 MeV per nucleon, and decreases with increasing nucleon number for any stable nucleus heavier than [math] F^{56} [/math] .
• Binding energy has a positive value. However, stable nuclei have less nuclear potential energy than the free nucleons.

3. Fission and fusion processes:

⇒Fusion processes:

• Fusion processes involve the combination of two or more atomic nuclei to form a single, heavier nucleus.
• This process releases a vast amount of energy, as the resulting nucleus has a lower mass than the original nuclei.
• Figure 4 Nuclear fusion between tritium and deuterium
• The temperature in a star’s core is several million kelvin and the density is in the region of 150000kgm³. These very high temperatures give nuclei enough kinetic energy to overcome the electrostatic repulsion between protons in the nucleus.
• The high density inside the star’s core forces nuclei so close together that the strong force becomes involved. This attractive force acts over very short distances.
• One fusion reaction that releases energy in stars like the Sun is the fusion of deuterium and tritium to form helium and a neutron.
• [math] _{1}^{2} \text{H} + _{1}^{3} \text{H} \rightarrow _{2}^{4} \text{He} + _{0}^{1} \text{n} [/math]
• The mass difference for this reaction is the difference between the mass of the original nuclei (deuterium and tritium) and the mass of the products (helium nucleus and neutron):
• [math] \text{Mass difference} = (2.013553 \, \text{u} + 3.015500 \, \text{u}) – (4.001505 \, \text{u} + 1.008665 \, \text{u} ) \\
  \text{Mass difference} = 0.018883 \, \text{u} \\
  \text{The energy released is} \, 0.018883 \, \text{u} * 931.5 \, \text{MeV} = 17.59 \, \text{MeV} [/math]
• The fusion reactions that occur in different types of stars depend on the star’s mass, core temperature and density.
• A chain of reactions, which may include steps that seem impossible in terms of energy, can happen if conditions are suitable.
• For example, the triple alpha cycle occurs, but only in red giants and red supergiant, where the core temperatures are greater than 100 million kelvins.
• [math] _{2}^{4}\text{He} + _{2}^{4}\text{He} \rightarrow _{4}^{8}\text{Be} – 93.7 \, \text{keV} \\
  _{4}^{8}\text{Be} + _{2}^{4}\text{He} \rightarrow _{6}^{12}\text{C} + 7.367 \, \text{MeV} [/math]
• The difference in mass is converted into energy according to Einstein’s famous equation,
• [math] E= mc^2 [/math]
• There are several types of fusion processes, including:
  – Thermonuclear fusion: This is the most common type of fusion, where two nuclei combine at extremely high temperatures (about 150 million° C) to form a single nucleus.
• Figure 5 Thermonuclear fusion
• – Nuclear fusion: This type of fusion occurs at lower temperatures and involves the combination of two nuclei to form a single nucleus.
• – Neutron-induced fusion: This process occurs when a neutron collides with a nucleus, causing it to fuse with another nucleus.
• – Inertial confinement fusion: This type of fusion uses high-powered lasers or particle beams to compress and heat a small pellet of fusion fuel to the point where fusion occurs.
• – Magnetic confinement fusion: This type of fusion uses strong magnetic fields to confine and heat a plasma (ionized gas) to the point where fusion occurs.
• Fusion processes have many potential applications, including:
  – Energy production: Fusion reactions release vast amounts of energy, which could be harnessed to generate electricity.
  – Medical applications: Fusion reactions can be used to produce radioisotopes for medical imaging and treatment.
  – Space exploration: Fusion propulsion systems could potentially be used to power spacecraft.
  – Nuclear weapons: Fusion reactions are used in the development of nuclear weapons.
• Figure 6 Nuclear bomb
• Some of the benefits of fusion processes include:
  – Zero greenhouse gas emissions
  – Abundant fuel supply (usually isotopes of hydrogen)
  – High energy density
  – Low waste production
• However, fusion processes also have some challenges to overcome, such as:
  – Achieving and sustaining high temperatures and pressures
  – Confining and stabilizing the plasma
  – Developing materials that can withstand the extreme conditions
  – Improving efficiency and cost-effectiveness

⇒Fission process:

• Fission reactions are established in the nuclear fuel using neutrons travelling slowly enough to be captured when they are fired at U-235 nuclei. A U-235 nucleus that captures a neutron becomes very unstable, and splits into two or more smaller pieces, and releases energy in the form of heat.
• When it has absorbed the neutron, some people think of the nucleus as being like a wobbly jelly, which splits if it is wobbled too much (Figure 7).
• Figure 7 U-235 nucleus is very unstable if it absorbs a thermal neutron
• Each fission reaction produces two, three or sometimes four neutrons, which may be absorbed by other U-235 nuclei if the neutrons are made to travel slowly enough. There are several possible reactions (Figure 8). For example:
• [math] _{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow _{56}^{141}\text{Ba} + _{36}^{92}\text{Kr} + 3 \, _{0}^{1}\text{n} \\
  _{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow _{54}^{144}\text{Xe} + _{38}^{90}\text{Sr} + 2 \, _{0}^{1}\text{n} \\
  _{92}^{235}\text{U} + _{0}^{1}\text{n} \rightarrow _{55}^{140}\text{Cs} + _{37}^{92}\text{Rb} + 4 \, _{0}^{1}\text{n} [/math]
• Figure 8 stages in the fission of U-235
• Nuclear fission reactions can only continue in a reactor if the number of nuclei involved in the fission reaction stays constant or increases. This occurs if, on average, one or more neutrons is produced and absorbed per fission reaction.
• This type of self-sustaining reaction is called a chain reaction (Figure 9).
• Chain reactions are only sustainable with a minimum amount of fuel, called the critical mass.
• This is because neutrons lost from the surface are no longer involved in the chain reactions. The shape as well as the mass of the sample affect the critical mass.
• Figure 9 A chain reaction grows if more neutrons are produced at each stage than are absorbed

4. Appreciation that knowledge of the physics of nuclear energy allows society to use science to inform decision making.

• Understanding the physics of nuclear energy is crucial for making informed decisions about its use and management. By applying scientific knowledge, society can:
  – Develop safe and efficient nuclear power plants
  – Ensure responsible handling and storage of nuclear waste
  – Assess and mitigate potential risks associated with nuclear energy
  – Evaluate the benefits and drawbacks of nuclear energy compared to other energy sources
  – Inform policy decisions on energy production, distribution, and consumption
  – Develop new technologies and innovations in nuclear energy.
  – Enhance education and public awareness of nuclear energy and its implications.
• By leveraging the physics of nuclear energy, we can harness its potential while minimizing its risks, ultimately leading to more informed decision-making and responsible management of this powerful energy source.
• Some examples of using science to inform decision-making in nuclear energy include:
  – Risk assessment and safety analysis for nuclear power plants.
  – Development of more efficient reactor designs and fuel cycles.
  – Evaluation of nuclear waste management strategies.
  – Comparison of nuclear energy with other energy sources in terms of cost, environmental impact, and scalability.
  – Investigation of advanced reactor concepts and Generation IV reactors.
• By applying scientific knowledge, we can ensure that nuclear energy is used responsibly and sustainably, aligning with societal needs and values.