IB DP Physics: HL复习笔记7.2.3 Nuclear Fission & Fusion

Nuclear Fusion

• Fusion is defined as:

The joining of two small nuclei to produce a larger nucleus

• Low mass nuclei (such as hydrogen and helium) can undergo fusion and release energy

The fusion of deuterium and tritium to form helium with the release of energy

• For two nuclei to fuse, both nuclei must have high kinetic energy
  • This is because the protons inside the nuclei are positively charged, which means that they repel one another

• It takes a great deal of energy to overcome the electrostatic force of repulsion.
  • This can only be achieved in an extremely high-energy environment, such as star’s core
• The final step of fusion is where a nucleus of tritium fuses with a nucleus of deuterium to form a helium nucleus and a neutron. It also releases energy.

Nuclear Fission

• Fission is defined as:

The splitting of a large atomic nucleus into smaller nuclei

• High mass nuclei (such as uranium) can undergo fission and release energy

The fission of a target nucleus, such as uranium, to produce smaller daughter nuclei with the release of energy

• Fission can be induced by firing neutrons at a nucleus
• When the nucleus is struck by a neutron, it splits into two, or more, daughter nuclei
• During fission, neutrons are ejected from the nucleus, which in turn, can collide with other nuclei which triggers a cascade effect
• This leads to a chain reaction which lasts until all of the material has undergone fission, or the reaction is halted by a moderator
• Nuclear fission is the process which produces energy in nuclear power stations, where it is well controlled
• When nuclear fission is not controlled, the chain reaction can cascade to produce the effects of a nuclear bomb

Significance of Binding Energy per Nucleon

• The number of protons in the nucleus is denoted by the symbol A
• At low values of A (a small number of protons):
  • Attractive strong nuclear forces between nucleons dominate over repulsive electrostatic forces between protons
  • In the right conditions, nuclei undergo fusion

• During fusion, the mass of the nucleus that is created is slightly less than the total mass of the original nuclei
  • The mass defect is equal to the binding energy that is released, since the nucleus that is formed is more stable

• At high values of A (a large number of protons):
  • Repulsive electrostatic forces between protons begin to dominate, and these forces tend to break apart the nucleus rather than hold it together
  • In the right conditions, nuclei undergo fission

• During fission, an unstable nucleus is converted into more stable nuclei with a smaller total mass
  • This difference in mass, the mass defect, is equal to the binding energy that is released

• The binding energy is equal to the amount of energy released in forming the nucleus, and can be calculated using:

E = (Δm)c²

• Where:
  • E = Binding energy released (J)
  • Δm = mass defect (kg)
  • c = speed of light (m s^–1)

• The daughter nuclei produced as a result of both fission and fusion have a higher binding energy per nucleon than the parent nuclei
• Therefore, energy is released as a result of the mass difference between the parent nuclei and the daughter nuclei

Part (a)

Step 1: Balance the number of protons on each side using the proton number

92 = (2 × 46) + xn_p (where n_p is the number of protons in c)

xn_p = 92 – 92 = 0

• • • Therefore, c must be a neutron since the proton number is 0

Step 2: Balance the number of nucleons on each side using the nucleon number

235 + 1 = (2 × 116) + x

x = 235 + 1 – 232 = 4

• • • Therefore, 4 neutrons are generated in the reaction

Part (b)

Step 1: Find the binding energy of each nucleus

Total binding energy of each nucleus = Binding energy per nucleon × Mass number

Binding energy of ⁹⁵Sr = 8.74 × 95 = 830.3 MeV

Binding energy of ¹³⁹Xe = 8.39 × 139 = 1166.21 MeV

Binding energy of ²³⁵U = 7.60 × 235 = 1786 MeV

Step 2: Calculate the difference in energy between the products and reactants

Energy released in reaction 1 = E_Sr + E_Xe – E_U

Energy released in reaction 1 = 830.3 + 1166.21 – 1786

Energy released in reaction 1 = 210.5 MeV

Part (c)

• • Since reaction 1 releases more energy than reaction 2, its end products will have a higher binding energy per nucleon
    • Hence they will be more stable

• • This is because the more energy is released, the further it moves up the graph of binding energy per nucleon against nucleon number (A)
    • Since at high values of A, binding energy per nucleon gradually decreases with A

• • Nuclear reactions will tend to favour the more stable route, therefore, reaction 1 is more likely to happen

Part (a)

Step 1: Determine the binding energies on each side of the equation

Binding energy = Binding Energy per Nucleon × Mass Number

• • • Binding energy before (U) = 235 × 7.59 = 1784 MeV
    • Binding energy after (Tc + In) = (112 × 8.36) + (122 × 8.51) = 1975 MeV

Step 2: Find the difference between the energies

• • • Energy released = 1975 – 1784 = 191 MeV

Part (b)

Method 1

Step 1: Convert the energy released from MeV to J

• • • 1 MeV = 1.60 × 10⁻¹³ J
    • Energy released = 191 × (1.60 × 10⁻¹³) = 3.06 × 10⁻¹¹ J

Step 2: Write down the equation for mass-energy equivalence

E = Δmc²

• • • Where c = speed of light

Step 3: Rearrange and determine the mass defect, Δm

Δm = 3.4 × 10⁻²⁸ kg

Method 2

Step 1: Convert the energy released from MeV to u

Step 2: Calculate the mass defect, Δm

• • • 1 u = 1.66 × 10⁻²⁷ kg

Δm = 0.205 × (1.66 × 10⁻²⁷) = 3.4 × 10⁻²⁸ kg