I was reading my bro’s physics book about stuff(ok, I was skipping to the good parts about how nukes work) and was kind of confused about something. It describes how 4 protons come together to make a helium nucleus and loose some mass in the process.(Sounds like a simplified version of course.) Anyway what confused me was that I remember reading neutrons are more massive than protons. So I’m curious is it correct that you can take 4 hydrogen nuclei and fuse them into helium and come out with less mass? Also if that’s right why is that configuration less massive anyway?
I think an H-bomb works with Deuterium, which is an isotope of Hydrogen. It has one proton and one neutrons.
I don’t know if that’s the same situation for stellar fusion.
I think I remember that the sun uses catalyzed fusion. Basically taking a Carbon-12 nucleus and building on it by adding one proton after another until eventually this nucleus fissles back to C-12 and a helium-4 nucleus.(But I guess I still have the same question how this results in a mass loss.)
Very simplified answer is it takes less energy to hold the helium nucleas together than the total energy available before they fused. This excess energy is released (by the sun, or a hydrogen bomb, etc.) and therefore less mass is left because E=mc^2. Since c is a constant, the mass has to go down if energy is given away. In other words, mass and energy are two forms of the same thing.
I am sure someone else can eleborate much more scientifically…
The atomic weight of hydrogen is 1.00794. Four hydrogen nuclei weigh 4.03176. The atomic weight of helium is 4.02602. So when the 4 protons fuse into helium there is excess weight that comes out as energy.
This is simplified, of course, and I expect a nuclear physicist will be along eventually to make it more rigorous.
When you get those protons and neutrons to go together, the fused group has less rest mass than the separate parts, yes. The reason is that it’s in a more stable configuration, and it releases energy putting them together. That energy is called “binding energy”. Heavier elements are unstable, and release energy when breaking apart (hence fusion works with light elements, fission with heavy ones). Search google for “fusion binding energy” for more info.
Fusion isn’t about converting mass to energy its about converting potential energy to kinetic and electromagnetic energy. In other words it’s about changing one type of energy into another. The remnant of the nucleus has less internal energy and as a result of this decrease in internal energy it has less mass, not vice versa.
And, as usual, I must add my normal caveat – the mass of the entire system does not change
How do you figure that? Do you not agree with David Simmons’s figures?
IIRC the fusion battle is lost at iron. At that point it takes more energy to fuse into the next element than is returned. It doesn’t mean you can’t fuse elements heavier than iron…you just need to provide gobs of energy to do so (a supernova explosion is one good way).
Once our sun converts its core to iron it’s all over for the sun (poor earth having been through long before that).
I’m afraid it won’t even get that far. For a low-mass star like the sun, nuclear fusion stops at carbon and oxygen. More massive stars can get their cores all the way up to iron before going supernova, though.
This has been discussed in great detail many times on this board but here are the basics. The relativistic equation that relates mass and energy is:
m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup]
m = mass
E = energy
p = momentum
(c = 1)
Since the system had zero momentum before the explosion it must have zero momentum after the explosion and therefore:
m[sup]2[/sup] = E[sup]2[/sup] - 0[sup]2[/sup] or m = E or if you want to include c as a conversion constant then m = E/c[sup]2[/sup] and the system mass does not change. Mass cannot be converted to energy.
A vault that contains nuclear weapon will weigh the same both before and after the explosion.
It must have zero total (vector) momentum, but the individual particles could have momentum, which would contribute to the kinetic energy.
I also get what you’re saying by weighing the nuclear silo, but really, using scales is a bad way to determine mass, since it would be measuring relativistic mass (I think).
“But Achernar, how can you talk about individual particles if there’s only one that comes out of the reaction, an alpha particle (or Helium nucleus)?”
Good question, Achernar. In reality, in the proton-proton I (or p-p I) reaction, there’s more than just the alpha particle. The actual reaction goes:
IN: Six protons
OUT: An alpha particle, two protons, two positrons, and two electron neutrinos
The kinetic energy of what? Were talking about the whole system here. If you want to talk about some isolated system of particles then you’ve changed the system and all bets are off.
Nope, if the momentum is zero there can be no relativistic mass. (Grinds teeth as he uses the dreaded term relativistic mass)
Well a lot of kinetic energy is carried away by neutrinos. But I guess if you’re looking at the system as a whole, then any extra KE that individual particles take on will eventually get turned into photons through collisions, and radiated away. Still, though, that’s a form of energy. Where does this energy come from if not mass loss?
And of course you meant that if the momentum is zero, then the relativistic mass is the same as the rest mass, right?
The whole fusion-fission business is explained, for the layman, quite nicely in Understanding Physics by Isaac Asimov. The whole thing can be illustrated by a quantity called"packing fraction." If you want to know about that you’ll have to buy a copy of the book or go the the library. Or get a text on nuclear physics.
In any case, if you check the link you see a minimum in the curve of packing fractions vs. mass number. Elements to the left of the minimum tend to more easily fused and those to the right tend to be easier to fission. The minimum occurs in the mass number region from about 45 to 75, or from approximately titanium to germanium and those elements are difficult to get to either fuse or fission. Iron is right in there at a mass number of about 56. In short such elements are the most stable atoms in the table.
If the system is isolated and it starts out with zero momentum, then it must always have zero momentum no matter what kind of internal changes occur. If the system has zero momentum then it must have zero kinetic energy. (As a whole)
I’m sorry but I refuse to use that horrible term again. Mass equals mass, equals rest mass, equals rest energy, equals the energy of a system that cannot be transformed away. There is no other definition of mass. m[sup]2[/sup] = E[sup]2[/sup] - p[sup]2[/sup]
Sorry, I missed this part. Nucleons in a nucleus have orbitals similar to electrons in an atom. And the nucleons in these orbitals have potential energy. When a nuclear reaction occurs the nucleus rearanges itself and the potential energy decreases (binding energy increases) and this decrease in potential energy results in kinetic and electromagnetic energy. (This is a gross simplification). (But once again the system energy and mass remains the same) See David Simmons’ last post.
Okay, then it sounds to me like you’re saying that the mass of a free proton is less than 1.007276487 amu. Am I right?
Also, are you saying that if you had, say, an electron-positron pair with zero net momentum, that the “mass of the system” would not be equal to the sum of the masses of the individual particles? Rather, it would be equal to the sum of their relativistic masses?
I don’ know what the mass of a proton is, but it has more mass as a free particle than it has in a nucleus. When protons are fused energy is liberated and this decrease in energy results in a local mass defect.
Please, please stop using relativistic mass. Mass in general is not additive. For instance two photons with antiparallel momentum have mass, but neither photon by itself has mass. However in your example the two masses are additive, because the sum of their masses equals the total energy of the system in the zero momentum frame.