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Nuclear fission reactors are used as an energy source in many parts of the world, but it is only recently that the use of nuclear fusion as a possible power source is achieving some encouraging results - Leaving Cert Physics - Question 8 - 2012

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Nuclear fission reactors are used as an energy source in many parts of the world, but it is only recently that the use of nuclear fusion as a possible power source i... show full transcript

Worked Solution & Example Answer:Nuclear fission reactors are used as an energy source in many parts of the world, but it is only recently that the use of nuclear fusion as a possible power source is achieving some encouraging results - Leaving Cert Physics - Question 8 - 2012

Step 1

Distinguish between nuclear fission and nuclear fusion.

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Answer

Nuclear fission involves the splitting of large atomic nuclei into smaller fragments, releasing energy in the process. In contrast, nuclear fusion occurs when small nuclei combine to form a larger nucleus, also resulting in energy release. Fission typically involves uranium or plutonium isotopes, while fusion involves light isotopes like deuterium and tritium.

Step 2

What are the advantages of fusion over fission in terms of fuel sources and reaction products?

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Answer

Fusion offers several advantages over fission:

  1. Fuel Abundance: Fusion fuel sources, such as deuterium, are more abundant than uranium. Deuterium can be extracted from water, making it widely available.
  2. Less Radioactive Waste: The byproducts of fusion reactions are generally less harmful and produce significantly less long-lived radioactive waste compared to fission, which creates hazardous waste products.

Step 3

How much energy is produced when a deuterium nucleus (²H) combines with a tritium nucleus (³H)?

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Answer

To calculate the energy produced in the fusion of deuterium and tritium, we use the mass-energy equivalence principle. The reactants are:

  • Deuterium (²H): Mass = 2.014102 u
  • Tritium (³H): Mass = 3.016049 u
  • Helium (⁴He): Mass = 4.002603 u
  • Neutron (n): Mass = 1.008665 u

We sum the masses of the reactants:

extTotalmassofreactants=2.014102+3.016049=5.030151u ext{Total mass of reactants} = 2.014102 + 3.016049 = 5.030151 \, u

The mass of products:

extTotalmassofproducts=4.002603+1.008665=5.011268u ext{Total mass of products} = 4.002603 + 1.008665 = 5.011268 \, u

Now, the mass defect (mass lost) is:

Δm=5.030151u5.011268u=0.018883u\Delta m = 5.030151 \, u - 5.011268 \, u = 0.018883 \, u

Using Einstein's equation E=mc2E = mc^2, where 1 u corresponds to 931 MeV, the energy produced is:

E=0.018883u×931MeV/u=17.59MeVE = 0.018883 \, u \times 931 \, MeV/u = 17.59 \, MeV

Step 4

Calculate the force of repulsion between a deuterium and a tritium nucleus when they are 2 mm apart in free space.

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Answer

The force of repulsion between two positively charged nuclei can be calculated using Coulomb's law:

F=kq1q2r2F = \frac{k \cdot |q_1 q_2|}{r^2}

Where:

  • k=8.99×109Nm2/C2k = 8.99 \times 10^9 \, N m^2/C^2
  • q1q_1 and q2q_2 are the charges of the deuterium and tritium nuclei, each having a charge of approximately +1e, which is 1.602×1019C1.602 \times 10^{-19} C.
  • Distance, r=2×103mr = 2 \times 10^{-3} \, m

Now substituting the values, we have:

F=(8.99×109)(1.602×1019)2(2×103)2=576.60NF = \frac{(8.99 \times 10^9) \cdot (1.602 \times 10^{-19})^2}{(2 \times 10^{-3})^2} \\ = 576.60 \, N

Step 5

Fusion can only take place at very high temperatures. Explain why.

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Answer

Fusion requires extremely high temperatures (millions of degrees Celsius) to provide the necessary energy for overcoming the electrostatic repulsion between positively charged atomic nuclei. At these high temperatures, particles move at sufficiently high speeds for collisions to occur with enough force for fusion to take place. In essence, the kinetic energy of the particles at high temperatures is vital for achieving the conditions necessary for fusion.

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