How much energy is released when hydrogen is fused to produce one kilo of helium?
One kilo of helium what? Which hydrogen? :)
Hydrogen (the single-proton, no neutron kind) doesn’t fuse directly with hydrogen to form helium. It’s a multistep reaction. See, for example, The Sun’s Energy Doesn’t Come From Fusing Hydrogen Into Helium (Mostly):
1. Proton/proton fusion into deuterium accounts for 40% of the reactions by number, releasing 1.44 MeV of energy for each reaction: 10.4% of the Sun’s total energy.
2. Deuterium/proton fusion into helium-3 accounts for 40% of the reactions by number, releasing 5.49 MeV of energy for each reaction: 39.5% of the Sun’s total energy.
3. Helium-3/helium-3 fusion into helium-4 accounts for 17% of the reactions by number, releasing 12.86 MeV of energy for each reaction: 39.3% of the Sun’s total energy.
4. And helium-3/helium-4 fusion into two helium-4s accounts for 3% of the reactions by number, releasing 19.99 MeV of energy for each reaction: 10.8% of the Sun’s total energy.
Let’s take this apart, and do the same sort of “math homework” problem that is perhaps bread-and-butter when trying to teach elementary thermodynamics and also keep it interesting. We have to face facts: there’s not a lot to get excited about when talking about the heat released by iron rusting, and so to keep things exciting, physics or other science teachers might use the equation of the much more exciting thermonuclear weapon:
4 H (4 x 1.00794) -> He (4.002602 g) + E (0.02916 g)
and then use the mass-energy equivalence of E=mc² to convert that to a “final answer” of 626 tons of TNT.
OK, but that’s the energy released when you create 4.0026g of helium, so that means the total released to create one kilo of helium-4, under those assumptions, is (1000g /4.0026g * e.g. 626.38 tons of TNT or ) 156.5 kilotons of TNT, otherwise known as 6.5477 x 10¹⁴ joules or 4.0868 x 10³³ eV.
But, that’s the wrong answer- that’s the net energy that might be available to a hypothetical merger of four hydrogen atoms into an atom of helium, assuming that the electrons are stripped and restored before and after accordingly and assuming you are starting with ground-state hydrogen and finishing up with ground-state helium. I’d wager that this is probably a physics homework problem of some form, and perhaps that’s the answer you’re expected to use to fill in your multiple choice grid.
What might be a better answer? Perhaps, accounting for the chain of reactions involved might make more sense. Let’s start with equation 2 from
’s quoted excerpt at the top of this answer:Hydrogen (1.00794g) + Deuterium(2.01410g) -> Helium-3 (3.01603g) + E (0.00601g)
1000/3.01603*129.1 tons of TNT = 42.8 kilotons of TNT. It takes more helium-3 to form a “kilogram” of helium-3, but it also produces significantly less energy. Here’s the challenge when talking about the nature of fusion reactions: you need to get specific with the details, and in the question as originally worded (How much energy is released when hydrogen is fused to produce one kilo of helium?) there are significant details left out, like which hydrogen (we might assume hydrogen-1) or which helium (we might assume helium-4) but as a matter of practice, there are hundreds of competing reactions in a chain to get from point a to point b and you can’t produce “pure” helium-4 from four hydrogens.
So if the answer isn’t 156.5 kilotons of TNT, or 42.8 kilotons of TNT, what might the real answer be?
Let’s start with the simplest chain from Ethan’s article.
2 x ¹H nucleus → ²H nucleus + positron + neutrino (1.44 MeV)
¹H + ²H → ³He + gamma ray (5.49 MeV)
2 x ³He → 2x ¹H nucleus + ⁴He nucleus (12.86 MeV)
Once you look at that, you begin to see the problems with approaching this like a freshman chemistry problem.
3 ¹H → ³He + ⁴He + γ + e⁺ (6.93 MeV) (adding the first two equations)
6 ¹H → 2 ¹H + ⁴He + 2ν + 2γ + 2e⁺ (2 x 6.93 + 12.86 = 26.72 MeV) (two times above, plus the third equation)
So fusing six nuclei of hydrogen produces two hydrogen nuclei plus one helium nucleus plus an array of other byproducts. Some of that energy is left behind in the nucleus, some is carried off by the antimatter formed and the gamma rays emitted. But, let’s keep it simple- the net energy being reported here is 26.72 MeV for each helium nucleus formed. If I ignore the cost of ionization and any I might get back de-ionizing the helium nucleus (which leads to other problems- where’d those two other pesky electrons go when four atoms of hydrogen, each having one electron, turned into one atom of helium, which is a neutral atom with only two electrons.. hmm…) I could say that four moles of hydrogen nuclei form one mole of helium nuclei, and so ( 1000g / 4.00260g/mole * 26.72 x 10⁶eV/nucleus * 6.02214 * 10²³nuclei/mole) = 4.02018 x 10³³ eV or 6.44 x 10¹⁴ joules or 154 kilotons of TNT. Not too bad, pretty close to my grossest simplification, right? I should have just stopped there?
But wait- there’s more going on here. We know that at the same time helium-3 is fusing with other helium-3 to form helium-4, there’s also competing reactions fusing helium-3 and helium 4 into two helium-4 nuclei, too, and that’s an oversimplification of what’s going on:
and we know that that’s going to take at least 4 more protons (plus the 6 to form the initial helium-4) to net (adding this into the previous sum)
10 ¹H → 2 ¹H + 2 ⁴He + 4ν + 4γ + 3e⁺ (26.72 + 19.99 MeV)
Now, on a “per mole” basis, this is producing twice the number of helium nuclei for less than double the amount of energy- 23.36 x 10⁶ eV/nucleus or 3.51389 x 10³³ eV or 134.6 kilotons of TNT. Hrm. Yet another number
This is why it’s a very complicated question. In fact, when we are talking about the energy released during the production of an element, it’s not such a simple and straightforward calculation as some high-school textbooks might lead you to believe, as simple as adding up molar weights and turning the leftover mass directly into energy, though that’s not an incorrect interpretation of what’s happening!
Matter and energy are in constant flux, being interchanged in a multitude of forms, but it is absolutely 100% accurate to say that the combined mass/energy equivalents of both mass and energy are completely conserved. Some of that ‘energy” drains off into the formation of positrons. Positrons and electrons can annihilate each other to release energy.
10 ¹H → 10 p + 10 e⁻ → 2 p + 2 α + 4ν + 4γ + 3e⁺ + 10 e⁻
→ 2 ¹H + 2 ⁴He + e⁻ + 4ν + 4γ
In the case of the sun, the internal temperature is high enough to separate the protons from the electrons forming a plasma. In some cases, the resulting electrons and positrons might annihilate one another, or might go speeding off as high energy “cosmic rays”. We know there’s a huge flux of neutrinos coming off the sun even though they’re hard to detect, and those neutrinos have some (small) mass: but there are streams of protons, electrons, positrons, alpha particles, exotic heavy nuclei, and a menagerie of other “particles” and “rays” that are also formed. It’s not as simple as “add hydrogen, make helium”.
It’s also a question whose answer isn’t straightforward. Physics isn’t baking, and the energy and matter takes very different forms depending on the how and what you are baking. It would be wrong to give any one answer, but knowing that overall, a certain percentage is coming from each term helps you find a somewhat reasonable approximation of an answer.
That, and there are dozens, perhaps hundreds of reactions competing to form new products, and the rates as which those happen are going to change depending on the localized concentrations of one or the other type of nucleus. When a nuclear physicist attempts to do a simulation, they make certain broad assumptions about the concentrations and intermediates involved.
For practical nuclear fusion performed here on earth, there’s a tremendous start-up cost associated with getting the temperatures and pressures just right, and the right mix and concentrations of starting fuels. Most earth-bound fusion reactions don’t do it the way the sun does it, because that’s just too hot and too costly, but we have definitely been able to reproduce some of the features. Hopefully at some point we will get to exceed break-even, Q=1, and that would be a momentous occasion for us earth scientists. The fact that we can do this kind of modelling and math is much more important that any single translation of “how much energy do I get out when I make one kilogram of helium”. You might not get any out- you might get exotic particles of heavy mass. You might lose it all to experimental error or out the gaps in your insulating material. You might not be able to do the experiment at all, or need to add in far more energy than you ever get out.
This is also why when you see reports of “cold fusion” it’s important to be skeptical that something which clearly appears to be more complicated and more intricate is claimed to have a simple origin. That’s not to say, be a jaded skeptic who pooh-poohs any result: carefully weigh the evidence and the accounting, and check for errors. Heck, I’ll bet I’ve made a handful of math errors just in this answer alone that I haven’t found yet. It’s not simple or easy, and at a first glance it might seem to be all a bunch of eggheaded nonsense, but if you’re systematic and careful you can begin to understand better what is going on. It’s not an easy path, but it’s a rewarding one.
(For more information on the difficulties in accounting for fusion, I would point you to the very excellent blog article, Nuclear Fusion | Do the Math.)
