At noon in Rajasthan, each square metre of a rooftop solar panel receives about a kilowatt of sunlight. The Sun has poured out light like this for about 4.6 billion years. The oldest rocks and fossils show that it has shone steadily all that time. No chemical fire could last a hundred-thousandth as long. If the Sun were coal burning in oxygen, it would have burned out within a few thousand years.
Four protons make helium
The steep left side of the binding energy curve promises the largest gains per nucleon. The Sun is mostly hydrogen, and in its core it joins protons, step by step, into helium-4. The route is the proton-proton chain, in three steps.
In step 1 two protons join into a deuteron. One proton must turn into a neutron, so a positron \(e^+\) and a neutrino \(\nu\) are given out. In step 2 the deuteron catches another proton and becomes helium-3, with a gamma ray. In step 3 two helium-3 nuclei, each made by steps 1 and 2, join into helium-4 and hand back two protons.
The energy follows from atomic masses. Four hydrogen atoms have \(4 \times 1.007825 = 4.031300\,\text{u}\), and one helium atom has \(4.002603\,\text{u}\). The difference is \(0.028697\,\text{u}\), so \(Q = 0.028697 \times 931.5 = 26.7\,\text{MeV}\). Atomic masses take care of the positrons: the two leftover electrons of the four hydrogen atoms annihilate them. The neutrinos escape the Sun with a little of the energy.
The Sun gives out \(3.9 \times 10^{26}\,\text{W}\). Each helium-4 nucleus formed releases \(26.7\,\text{MeV} = 4.27 \times 10^{-12}\,\text{J}\). Find (a) the number of helium nuclei formed each second and (b) the mass the Sun loses each second.
Power is energy per second, so the rate of helium formation is the power divided by the energy per nucleus.
Rate \(= P/Q\). Mass lost each second \(= P/c^2\), from \(E = mc^2\).
Rate \(= 3.9 \times 10^{26}/4.27 \times 10^{-12} = 9.1 \times 10^{37}\) per second. Mass lost \(= 3.9 \times 10^{26}/(3 \times 10^8)^2 = 4.3 \times 10^9\,\text{kg}\) each second.
The Sun makes \(9.1 \times 10^{37}\) helium nuclei and loses \(4.3 \times 10^9\,\text{kg}\) every second. Its mass is \(2 \times 10^{30}\,\text{kg}\), so this can go on for billions of years. Values used: \(1\,\text{u} = 931.5\,\text{MeV}/c^2\), \(c = 3 \times 10^8\,\text{m/s}\), \(1\,\text{MeV} = 1.6 \times 10^{-13}\,\text{J}\), \(m_{\text{H}} = 1.007825\,\text{u}\), \(m_{\text{He}} = 4.002603\,\text{u}\).
Stars hotter and heavier than the Sun use another route, the carbon-nitrogen cycle. A carbon-12 nucleus catches four protons one by one, passing through isotopes of nitrogen and oxygen. At the end it throws out a helium-4 nucleus and becomes carbon-12 again. The carbon is not used up; like a catalyst in chemistry, it only helps. The net result and the energy are the same as in the proton-proton chain.
Use \(m_{\text{H}} = 1.00783\,\text{u}\) and \(m_{\text{He}} = 4.00260\,\text{u}\) to find the energy released, in MeV, when four hydrogen atoms become one helium-4 atom. Take \(1\,\text{u} = 931.5\,\text{MeV}/c^2\).
What is the role of carbon-12 in the carbon-nitrogen cycle?
The Coulomb barrier
Why does hydrogen on the Earth not fuse? Two nuclei must come within reach of the nuclear force, about \(2\,\text{fm}\) for two protons. Both are positive, so they repel. At \(2\,\text{fm}\) their electric potential energy is \(ke^2/r = 1.44/2 = 0.72\,\text{MeV}\), or \(1.15 \times 10^{-13}\,\text{J}\). This hill of potential energy is the Coulomb barrier, the same one an alpha particle climbed in the chapter on the Bohr atom.
The protons in a hot gas have an average kinetic energy \(\tfrac{3}{2}k_BT\), where \(k_B = 1.38 \times 10^{-23}\,\text{J/K}\) is Boltzmann's constant, as the chapters on heat show. Setting the average equal to the barrier gives the temperature. \[T = \dfrac{2 \times 1.15 \times 10^{-13}}{3 \times 1.38 \times 10^{-23}} = 5.6 \times 10^{9}\,\text{K}\] Fusion that runs on heat is called thermonuclear fusion.
The Sun's core is at \(1.5 \times 10^{7}\,\text{K}\). A proton with the average kinetic energy there heads straight for another proton. Where does it turn back?
The Sun's core is at only \(1.5 \times 10^{7}\,\text{K}\). There the average kinetic energy is \(\tfrac{3}{2}k_BT = 3.1 \times 10^{-16}\,\text{J} = 1.9\,\text{keV}\), about 370 times too small. Try the chips in the figure above. Only at \(5.6 \times 10^{9}\,\text{K}\) does an average proton reach the top. Yet the Sun certainly fuses hydrogen. Two effects together make it possible.
- The fast few. \(\tfrac{3}{2}k_BT\) is only the average. A tiny fraction of the protons move several times faster than average, and the core holds an enormous number of protons.
- Tunnelling. A proton is also a wave, as the chapter on Matter Waves showed. Its wave reaches a little way into the barrier, so there is a small chance of appearing on the far side without climbing over. This is quantum tunnelling.
George Gamow worked out tunnelling through a Coulomb barrier in 1928, to explain alpha decay. The chance of tunnelling falls very steeply as the energy drops below the top. The fusion rate is a trade between the rare fast protons and the easier tunnelling of faster ones. For any single proton in the Sun's core, fusing with another takes billions of years on average. That is why the Sun burns slowly enough to last.
A deuteron and a triton each carry charge \(+e\). Their nuclei touch when their centres are \(3.24\,\text{fm}\) apart. Take \(ke^2 = 1.44\,\text{MeV fm}\). What is the height of their Coulomb barrier, in MeV?
The Sun's core is far cooler than the \(5.6 \times 10^9\,\text{K}\) that the average proton would need. Which of these help fusion happen there?
A star in a bottle
On the Earth, the easiest fuel is deuterium with tritium, \({}^{2}\text{H} + {}^{3}\text{H} \to {}^{4}\text{He} + n\), with \(Q = 17.6\,\text{MeV}\). Its barrier is lower than that of two protons, and it needs no slow weak step like the Sun's step 1. The neutron carries off \(14.1\,\text{MeV}\) of the energy, as the lesson on the Q value showed.
The fuel for one reaction has a mass of about \(5\,\text{u} = 8.3 \times 10^{-27}\,\text{kg}\). So one kilogram of fuel gives this much energy. \[E = \dfrac{17.6 \times 1.6 \times 10^{-13}}{8.3 \times 10^{-27}} = 3.4 \times 10^{14}\,\text{J}\] That is about four times what one kilogram of uranium-235 gives by fission.
The deuterium-tritium reaction releases \(17.6\,\text{MeV}\) from fuel of mass \(8.3 \times 10^{-27}\,\text{kg}\). Fission of 1 kg of uranium-235 gives \(8.2 \times 10^{13}\,\text{J}\). How many times more energy does 1 kg of deuterium-tritium fuel give?
The fuel must be heated to over \(10^{8}\,\text{K}\), ten times hotter than the Sun's core. The Earth cannot squeeze it the way the Sun's gravity does. At that temperature the atoms are torn into bare nuclei and electrons, a plasma. No wall can touch it. The plasma must also be dense enough, and held long enough, to give back more energy than was spent heating it.
A tokamak holds the plasma in a ring-shaped magnetic bottle. Charged particles spiral along the magnetic field lines and cannot reach the walls. The largest tokamak ever built, ITER, is being assembled at Cadarache in France. It aims to give out 500 MW of fusion power from 50 MW of heating.
India is one of the seven members of ITER, with the European Union, China, Japan, Korea, Russia and the United States. ITER-India works from the Institute for Plasma Research at Gandhinagar, which has built its own tokamaks, Aditya and SST-1. Indian industry built one of ITER's largest parts, the cryostat. It is a steel vacuum chamber about 30 m wide and 30 m tall, which surrounds the whole machine.
Why take so much trouble? Deuterium is plentiful: about one hydrogen atom in 6400 in sea water is deuterium. The main product, helium, is harmless. A fusion reactor holds only a little fuel at a time, so a runaway reaction like a supercritical fission chain is impossible. The difficulties are real too: the extreme temperature, the confinement, and the fast neutrons that make the walls radioactive.
The Sun forms \(9.1 \times 10^{37}\) helium-4 nuclei every second, each from four hydrogen atoms of \(1.007825\,\text{u} = 1.67 \times 10^{-27}\,\text{kg}\). Estimate the mass of hydrogen it turns into helium each second, in kilograms.
Which is the main reason a fusion power station is harder to build than a fission one?
- Thinking the average thermal energy must reach the barrier for fusion to happen.The fastest few particles and quantum tunnelling let fusion run far below that temperature, slowly.
- Forgetting the positrons' annihilation when finding \(Q\) of the proton-proton chain.Use atomic masses for four hydrogen atoms and one helium atom; the annihilation is then counted.
- Calling carbon the fuel of the carbon-nitrogen cycle.Carbon-12 is returned at the end. It is a catalyst; hydrogen is the fuel.
- Taking the Coulomb barrier as a wall at the nucleus's surface only.The repulsion \(ke^2/r\) acts at every distance; the barrier is a hill that falls off as \(1/r\).