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Nuclear fusion is a process in which two or more light atomic nuclei combine to form a heavier nucleus. This process releases a tremendous amount of energy because the mass of the resulting nucleus is slightly less than the sum of the original masses. The "missing" mass is converted into energy according to Einstein's famous relation \( E = mc^2 \).
Unlike nuclear fission, where heavy nuclei split apart, fusion joins light nuclei. Fusion naturally occurs in stars, including our Sun, where hydrogen nuclei fuse to produce helium and energy that supports life on Earth. Harnessing fusion on Earth offers the promise of a nearly limitless, clean energy source that could revolutionize power generation.
In this section, we will explore the basic principles of nuclear fusion, the conditions required for fusion to occur, and how energy is released in fusion reactions. We'll use specific examples and practical calculations to deepen our understanding.
The core of fusion is the reaction where two light nuclei come close enough to combine and form a heavier nucleus. The most studied and experimentally achievable fusion reaction uses isotopes of hydrogen:
These nuclei fuse under the right conditions to form a helium nucleus (alpha particle), a free neutron, and a large amount of energy:
Reaction:
\[ \, ^2_1\mathrm{D} + \, ^3_1\mathrm{T} \rightarrow \, ^4_2\mathrm{He} + \, ^1_0\mathrm{n} + \text{Energy} \]
Here, the fusion produces a helium-4 nucleus, one neutron, and a large amount of kinetic energy mainly carried by the neutron. The energy released is millions of times larger than chemical reactions, explaining why fusion powers stars so effectively.
Fusion does not happen easily because the nuclei are positively charged and repel each other due to electrostatic (Coulomb) forces. To overcome this repulsion and bring the nuclei close enough for the strong nuclear force to bind them, extremely high temperatures and pressures are needed.
Let's understand these conditions stepwise:
The probability of nuclei overcoming the Coulomb barrier and fusing peaks at a certain energy called the Gamow peak, which balances the decreasing number of high-energy particles with increasing fusion reaction probability.
In stars like our Sun, fusion occurs naturally under core temperatures of about 15 million kelvin (MK). Experimental fusion reactors on Earth aim to replicate or exceed these conditions to sustain fusion reactions.
The energy produced in fusion comes from the mass defect - the difference between the sum of the masses of the original nuclei and the mass of the fused nucleus. This missing mass is converted into energy. Understanding this is critical to grasp why fusion is a powerful energy source.
The mass defect is calculated as:
The energy released \( E \) is then found by Einstein's equation:
Let's look at binding energy per nucleon for various nuclei to understand why fusion releases energy only for light nuclei:
| Nucleus | Mass Number (A) | Binding Energy per Nucleon (MeV) | Stability |
|---|---|---|---|
| Deuterium (D) | 2 | 1.1 | Low |
| Tritium (T) | 3 | 2.8 | Low |
| Helium-4 (He) | 4 | 7.1 | Higher |
| Iron-56 (Fe) | 56 | 8.8 | Highest |
| Uranium-235 (U) | 235 | 7.6 | Lower |
This table shows that lighter nuclei have smaller binding energy per nucleon, so fusing them into heavier nuclei with higher binding energy per nucleon releases energy. Conversely, very heavy nuclei release energy when split (fission), but fusion of heavy nuclei is not energetically favorable.
Step 1: Calculate total mass before fusion:
\[ m_{initial} = m_D + m_T = 2.014 + 3.016 = 5.030 \,u \]
Step 2: Calculate total mass after fusion:
\[ m_{final} = m_{He} + m_n = 4.0026 + 1.0087 = 5.0113 \,u \]
Step 3: Find mass defect:
\[ \Delta m = m_{initial} - m_{final} = 5.030 - 5.0113 = 0.0187 \,u \]
Step 4: Convert mass defect to energy using 1 u = 931.5 MeV:
\[ E = 0.0187 \times 931.5 = 17.4 \, \mathrm{MeV} \]
Answer: The D-T fusion reaction releases approximately 17.4 MeV of energy.
Step 1: Calculate Coulomb potential energy at 1 fm ( \( r = 1 \times 10^{-15} \) m ):
\[ U = \frac{1}{4 \pi \epsilon_0} \frac{Z_1 Z_2 e^2}{r} \]
For deuterium nuclei, \( Z_1 = Z_2 = 1 \), and \( e = 1.6 \times 10^{-19} \, C \). The constant \( \frac{1}{4 \pi \epsilon_0} = 9 \times 10^{9} \, Nm^2/C^2 \), so:
\[ U = 9 \times 10^{9} \times \frac{(1)(1)(1.6 \times 10^{-19})^2}{1 \times 10^{-15}} = 9 \times 10^{9} \times \frac{2.56 \times 10^{-38}}{10^{-15}} = 9 \times 10^{9} \times 2.56 \times 10^{-23} = 2.3 \times 10^{-13} \, J \]
Step 2: Relate this energy to temperature via kinetic energy of particles (~thermal energy):
\[ K = k_B T \Rightarrow T = \frac{U}{k_B} = \frac{2.3 \times 10^{-13}}{1.38 \times 10^{-23}} \approx 1.67 \times 10^{10} \, K \]
Answer: Approximately \(1.7 \times 10^{10}\) kelvin is needed classically to overcome the Coulomb barrier. In practice, fusion occurs at much lower temperatures (~108 K) due to quantum tunneling effects.
Step 1: Calculate number of moles burned per second:
Molar mass of D-T mixture ~ \( 5 \) g/mol (approximate average of deuterium and tritium).
\[ n = \frac{1 \; \mathrm{g}}{5 \; \mathrm{g/mol}} = 0.2 \, \mathrm{mol/s} \]
Step 2: Calculate number of fusion reactions per second:
\[ N = n \times N_A = 0.2 \times 6.022 \times 10^{23} = 1.2044 \times 10^{23} \, \mathrm{reactions/s} \]
Step 3: Energy per reaction in joules:
1 eV = \( 1.6 \times 10^{-19} \) J
\[ E_r = 17.6 \times 10^6 \times 1.6 \times 10^{-19} = 2.816 \times 10^{-12} \, J \]
Step 4: Calculate total power output:
\[ P = N \times E_r = 1.2044 \times 10^{23} \times 2.816 \times 10^{-12} = 3.39 \times 10^{11} \, W \]
(a) Power output: \(3.39 \times 10^{11} \, W = 339 \, \mathrm{GW}\)
Step 5: Convert power to MW:
\(1 \, \mathrm{GW} = 1000 \, \mathrm{MW}\), so power = \(339,000 \, \mathrm{MW}\), an enormous amount.
Step 6: Calculate daily fuel cost:
Fuel burned per day = \(1 \mathrm{g/s} \times 86400\, \mathrm{s} = 86400\, \mathrm{g} = 86.4 \, \mathrm{kg}\)
Cost per gram = INR 50,000
Daily cost = \(86400 \times 50,000 = 4.32 \times 10^9 \, \mathrm{INR}\)
Answer:
(a) Power output is approximately 339 GW.
(b) Equivalent to 339,000 MW.
(c) Daily fuel cost is about INR 432 crores, highlighting current fuel cost challenges.
Though a single fission reaction releases about 200 MeV, much larger than the 17.6 MeV from fusion, fusion reactions release energy per unit mass more efficiently due to higher energy density and abundant fuel (hydrogen isotopes).
Fusion produces less long-lived radioactive waste and fuel is more plentiful but requires more extreme conditions.
Answer: Fusion yields less energy per event but has advantages in fuel supply and waste; fission produces more energy per reaction but involves challenges like nuclear waste and fuel scarcity.
Step 1: Use the half-life decay formula:
\[ N = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{T_{1/2}}} \]
where \(N\) is remaining amount, \(N_0\) initial amount, \(t = 5 \, \mathrm{years}\), and \(T_{1/2} = 12.3 \, \mathrm{years}\).
Step 2: Calculate fraction remaining:
\[ \frac{N}{N_0} = \left(\frac{1}{2}\right)^{5/12.3} \approx \left(\frac{1}{2}\right)^{0.406} \approx 0.75 \]
Answer: About 75% of tritium remains after 5 years. This decay means careful inventory management is needed to ensure fuel availability over reactor lifetimes.
When to use: Identifying fusion nuclei in reaction equations.
When to use: Calculations of energy from mass defect.
When to use: Comparing stability and energy output in nuclear reactions.
When to use: Estimating or evaluating fusion ignition conditions in numerical problems.
When to use: Verifying answer correctness before final submission.
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