When we talk about nuclear energy, understanding nuclear fusion is essential. It helps us gain insight into how a massive amount of energy is released when two light nuclei combine to form a single larger nucleus.
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Some examples of nuclear fusion:
^{1}_{1}H + ^{1}_{1}H → ^{2}_{1}H + e^{–} + v + 0.42 MeV … (1)
Here two protons combine to form a deuteron and a positron releasing 0.42 MeV of energy. |
^{2}_{1}H + ^{2}_{1}H → ^{3}_{2}He + n + 3.27 MeV … (2)
Here two deuterons combine to form the light isotope of Helium releasing 3.27 MeV of energy. |
^{2}_{1}H + ^{2}_{1}H → ^{3}_{1}H + ^{1}_{1}H + 4.03 MeV … (3)
In this case, two deuterons combine to form a triton and a proton releasing 4.03 MeV of energy. |
Note that in all these examples, two positively charged particles combine to form a larger nucleus. Further, the Coulomb repulsion hinders the process of fusion since it acts to prevent two positively charged particles from getting within the range of their attractive nuclear forces.
Now, the height of the Coulomb barrier depends on the charges and the radii of the two interacting nuclei. For example, the barrier height for two protons is around 400 keV. Also, higher charged nuclei have a higher barrier height. The temperature at which the protons in a proton gas have enough energy to cross the coulomb’s barrier is around 3 x 10^{9} K.
Now, to generate any useful amount of energy, nuclear fusion must occur in bulk matter. To achieve this, the temperature of the material can be raised until the particles have enough energy to cross the coulomb barrier due to their thermal motions. This is thermonuclear fusion.
Browse more Topics under Nuclei
- Atomic Mass and Composition of Nucleus
- Mass-Energy and Nuclear Binding Energy
- Nuclear Force
- Nuclear Energy – Nuclear Fission
- Radioactivity – Law of Radioactive Decay
- Radioactivity – Types of Radioactive Decay
Nuclear Fusion in the Sun
The temperature of the core of the Sun is around 1.5 x 10^{7} K. Hence, even in the sun fusion occurs only when protons having energies above the average energy are involved. In simple words, for thermonuclear fusion to occur extreme temperature and pressure conditions are needed. This is only possible in the interiors of the Sun and other stars.
In the Sun, nuclear fusion occurs in a multi-step process where hydrogen is burned into helium. Here, hydrogen is the fuel and helium, the ash. The process is represented as follows:
^{1}_{1}H + ^{1}_{1}H → ^{2}_{1}H + e^{–} + v + 0.42 MeV … (4)
e^{–} + e^{+} → γ + γ + 1.02 MeV … (5)
^{2}_{1}H + ^{1}_{1}H → ^{3}_{2}He + γ + 5.49 MeV … (6)
^{3}_{2}He + ^{3}_{2}He → ^{4}_{2}He + ^{1}_{1}H + ^{1}_{1}H + 12.86 MeV … (7)
For the fourth reaction to occur, the first three need to occur twice. Thereby, two light helium nuclei unite to form a normal helium nucleus. This four-step process can be summarised as:
4^{1}_{1}H + 2e^{–} → ^{4}_{2}He + 2v + 6γ + 26.7 MeV … (8)
Hence, we can see that four hydrogen nuclei combine to form a helium nucleus and release 26.7 MeV of energy.
Some Trivia
Hydrogen has been burning in the Sun’s core for around 5 x 10^{9} years! According to calculations, this will continue for another 5 x 10^{9} years in the future too. In about 5 billion years, the Sun’s core will be primarily helium. This is when it will start to cool down and collapse under its own gravity. Eventually, this will raise the core temperature and cause the outer envelope to expand. The sun will then look like a red giant! Once the core temperature increases to 10^{8} K, energy production will resume through fusion again. However, this time helium will be burned to make carbon.
Controlled Thermonuclear Fusion
On November 01, 1952, the USA exploded a nuclear fusion device generating energy equivalent to 10 million tons of TNT. Just to put it into perspective, one ton of TNT produces 2.6 x 10^{22} MeV of energy on explosion. This was the first thermonuclear reaction on earth. It is very difficult to achieve a sustained and controlled source of fusion power. However, it is regarded as the power source of the future and most countries are pursuing it vigorously.
Solved Question for You
Question: How long can an electric lamp of 100W be kept glowing by fusion of 2.0 kg of deuterium? Take the fusion reaction as:
^{2}_{1}H + ^{2}_{1}H → ^{3}_{2}He + n + 3.27 MeV
Solution: We know that 1 mole or 2 grams of deuterium contains 6.023 x 10^{23} atoms. Therefore, 2 kg of deuterium will contain:
{(6.023 x 10^{23})/2} x 2000 = 6.023 x 10^{26} atoms.
From the given reaction, we know that 3.27 MeV of energy is released when 2 deuterium nuclei fuse. Therefore, the total energy released per nucleus is:
(3.27/2) x 6.023 x 10^{26} MeV
= (3.27/2) x 6.023 x 10^{26} x 1.6 x 10^{–19} x 10^{6} Joules
= 1.576 x 10^{14} J
Power of the lamp is 100 W = 100 J/s. So, in one second the lamp utilizes 100 J. Therefore, the total time for which the lamp will burn is the time taken to utilize 1.576 x 10^{14} J. Here is the calculation:
(1.576 x 10^{14}/100) seconds
= (1.576 x 10^{14}/100 x 60 x 60 x 24 x 365) years
= 4.9 x 10^{4} years
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