Why is fusion stronger than fission?

Gecko

topic pretty much explains it. why is it that fusion releases more energy than fission?

Morbius

Gecko,

Depends on what you mean by "releases more energy" - it may or may
not. It's all the same nuclear force - whether the reaction is fission or
fusion. You really can't say fusion releases more energy than fission -
it depends on what specific reaction you are talking about. What really
matters is what the details of the reaction are - and how much mass is
turned into energy. Let's look at a couple of nuclear reactions - the
fission of U235 [ as done in nuclear reactors ] and D-T fusion [ which is
what is being worked on for producing power].

First each fission reaction gives you about 200 MeV worth of energy.
[ Of that about 10 MeV goes into neutrinos that you can't capture - but
lets use 200 MeV for round numbers ]

The nucleus that you are fissioning - U235 has a mass of 235 atomic mass
units [ 236 if you count the incident neutron ]

So fission gives you about 1 MeV / atomic mass unit of fuel.

Lets do the same with D-T fusion. [ Deuterium - Tritium ]

D-T fusion gives you 17.6 MeV of energy [ less than a fission reaction ]

However, the fuel has a mass of 5 [ 2 for the D and 3 for the T ]

So the energy per atomic mass unit of fuel is 17.6 MeV / 5 amu = 3.52 MeV / amu.

So fission gives you more energy per reaction - but fusion gives you
more energy per unit mass for these reactions.

Why the energies [ 200 MeV for fission and 17.6 MeV for fusion ] are
what they are is that they are the difference in the masses between
the reactants and products.

If you take the mass of D, add the mass of T, subtract the mass of He4
and subtract the mass of a neutron, then multiply by the square of the
speed of light [ E=mc^2] you will get 17.6 MeV.

Why the masses are what they are - that gets complicated.

Dr. Gregory Greenman
Physicist

ceptimus

When you fuse ordinary Hydrogen into Helium, 0.7% of the mass is converted to energy. When Uranium undergoes fission only about 0.1% of the mass is converted. So in mass conversion terms, you might say that fusion is about seven times better than fission.

Morbius

ceptimus,

You can only say that only for a particular reaction - like D-T fusion.

You can't say that "fusion" always releases more than "fission" - it
depends on the reaction.

There's nothing inherently 7X more powerful about fusion -
they BOTH rely on the SAME force - the strong nuclear force.

Dr. Gregory Greenman
Physicist

theCandyman

So is fusion being studied as a possible energy source because it is "cleaner" than fission even though it produces less energy per reaction? Or is it that less overall fuel is wanted for the same amount of energy?

kapton

firsly it's known as the weak nuclear force, not the strong. (w+, w- and Z particles)

Fusion has a much greater release of energy, (fission of u235 is about 10 * 8, fusion of D - T is 10 * 8.4 and Lh2/Lox is 10 * 1 per unit of mass/energy)
- (all results are mathematically rounded)

Fusion is much greater per unit of mass, fission releases less energy, this is converse to the binding energy of a nucleus.

The energy release of fission/fusion is on the lengths of (electromagnetic waves - short at a high frequency, photons, kinetically charged atoms and particles, in the form of gamma and xrays).

Fusion releases more energy than fission per unit of mass.

theCandyman

I thought that the weak nucear force was something to the order of one billionth the strength of the strong nuclear force. What is the role of strong nuclear force in these processes?

Morbius

kapton [ aka U235 ],

WRONG WRONG WRONG

When you liberate energy via nuclear fission or fusion - it's the
"strong nuclear force" that produces the energy - NOT the "weak"
force [ or "electro-weak", since there is a unification of the "weak"
force with the Coulomb force]. The "weak" force is the one that is
responsible for beta decay for example.

If you are going to do a calculation - put UNITS on your numbers so
we know what we are talking about. What is a "per unit mass/energy"
The quantity mass divided by energy is not dimensionless - it has UNITS -
and your statement above, in the absence of units; is meaningless.

What the Hell does that gobbley gook mean? "The energy release
of fission /fusion is on the lengths of..." means WHAT???

You are not even dimensionally correct. Go back to your junior high
science class and learn that you have to be consistent in your units.

Energy release is measured in a unit of energy - MeVs are convenient for
nuclear reactions, eVs are convenient for chemical reactions, and
Joules are convenient for macroscopic quantities of energy.

What is a "kinetically charged atom". You have to learn to separate
the concepts of kinetic energy and charge. They are two different
properties.

As I explained earlier - it depends on the reaction!

Both fusion and fission derive energy the same way - they both
rearrange the protons and neutrons in the nuclei of the reactants.
One fuses light nuclei together - one splits heavy nuclei.

The binding energy needed to hold together the products is less than
that needed to hold together the reactants. This differential in
binding energy is the energy that is released.

Let me re-iterate from my previous post. Let's look at fission on
U235. The fission of U235 produces about 200 MeV of energy.
Since the reactantants have a mass of 236 amu [ 235 from the uranium
and 1 from the neutron ] the energy produced per unit mass is

E/M = 200 MeV / 236 amu = 0.85 MeV/amu

Now let's look at D-T fusion. D-T fusion releases 17.6 MeV of energy
while the mass of the reactants is 5 amu [ 2 from D, 3 from T] Hence,

E/M = 17.6 MeV/ 5 amu = 3.52 MeV/amu

which is about 4.1X the analogous E/M ratio for fission above. [ So it
looks like fusion is more powerful]

Now lets look at another fusion reaction:

D + D --> He3 + n + 3.25 MeV

The energy released in this reaction is 3.25 MeV and the mass of the
reactants is 4 [ 2 from each of the two Deuterium nuclei]. Proceeding -

E/M = 3.25 MeV / 4 amu = 0.81 MeV/amu

That's LESS than the E/M ratio for U235 fission!!!

Here is a fusion reaction that gives you LESS energy per mass than
a fission.

Or let's take the following fusion reaction:

Li6 + H --> He4 + He3 + 4.0 MeV

The energy is 4.0 MeV, and the reactant mass is 7 amu. As before;

E/M = 4.0 MeV / 7 amu = 0.57 MeV / amu

That's a LOT LESS than the U235 fission reaction!!!

How much energy you get, and how much energy you get per unit mass
doesn't depend on whether you have fission or fusion - there's nothing
inherently "more powerful" in fusion than fission. The D-T fusion reaction
produces more energy per mass than U235 fission while the D-D and
the Li6-H fusion reactions above produces LESS energy per mass than
U235 fission.

It all depends on what reactions you are talking about!!
Why does that seem to be too difficult a concept to understand?

All chemical reactions rely on the Coulomb force - including all
oxidation [ burning ] reactions. If you ignite some paper or wood - you
will get a certain amount of energy per unit mass. If you ignite some
HMX high-explosive - you are going to get a LOT more energy per unit
mass. Both reactions derive their energy from the same type of
reaction - oxidation - and the same Coulomb force.

But how much energy you get and how much energy you get per unit
mass is dependent on the particular reaction - NOT the TYPE of the
reaction.

Dr. Gregory Greenman
Physicist

kapton

A UNIT OF MASS. Don't you know what a unit of mass is?
Oh, but i thought you have already completed your doctorate.
I refer to a (Kg). A Kilogram is not related to weight (because 9.81 newtons interfere - at sea level) but metric measurable mass.
1 unit of mass = 1kg.

Table for energy per kJ/g - (that means Kilojoule/gram)

10*1 - LH2/LOX (10 to the power of 1) - corrected/rounded
10*8 - U235 (10 to the power of 8) - corrected/rounded
10*8.6 - D - T (10 to the power of 8 + 6 to the power of 7) -corrected

Isp (lb-f-s/lb-m) = 144.22(e{kJ/g} 1/2

I was simply making reference to what output (in terms of subatomic particles) energy releasing from a nuclear fission or fusion reaction contains.

If we where to discuss nuclear rockets (which I hope we are, because this is a nuclear engineering thread) then the process of ICF or MCF will yield greater energy levels than that in NTR's using Solid reactor cores. The process of D - T, D - D, T - T or any more common isotopes that would be considered in man made fusion will yield much greater levels of energy than that for fission, u235 - pu239. (This is depicted clearly in a thermonuclear device, that when cross-section and other paramerters are introduced and calculated, much higher efficiencies are produced.) - 'Little boy' = 1.5% efficiency / 'Fat man' = 17% efficiency)(- efficiency refers to energy yield -)

Now we are talking about nuclear engineering, and I have never seen a NTR fly so excuse the referal to nuclear weapons as being the closest proven entity.

padawan13

please explain it?

It doesn't seem logical that in a fission reaction uranium-235 releases less electron volts per nucleon compared to the fusion reaction T-D, when uranium is around 45 times larger, obviously the size of the nucleus is not a factor. Could it be the binding energy of the atom, uranium has a higher binding energy and therefore releases less, where as the hydrogen isotopes have less binding energy, releasing more energy.

Im basically asking what determines the amount of mass converted to energy, in a fission or fusion reaction?

This is random but according to einstein's law e=mc2 the mass lost determines the amount of energy, and therefore it can be said mass is converted into energy, therefore wouldn't it be logical to say mass is energy, in a different maybe more concentrated form?

Drakkith

padawan13, this thread is 6 1/2 years old since the last post!

Anyways, the fission of one nucleus of uranium releases more energy than the fusion of two neuclei, however the mass of the fusion nuclei are much much less than Uranium.
The amount of energy released in each event is determined by the binding energy of the nuclei. The wikipedia article on Nuclear Fusion has a chart with each amount if you are intersted. Also, E=MC^2 does NOT mean that energy is converted into mass or vice versa. It means that the removal or introduction of energy is accompanied by a decrease or increase in mass. The definition of energy is the capability for a system to do work. It is a description of how two systems interact and is not something physical in itself. Mass is NOT energy. They are two completely different concepts.

There are plenty of current threads on these concepts and I recommend that you visit them and not keep posting in this old thread.

zheng89120

simply put, the binding energy curve is sharper for hydrogen (fusing) than for Uranium (fissioning)

padawan13

haha yeah i know its old, i wasn't really expecting anyone to reply, but thanks.

So you say that mass is not being converted to energy, but a decrease in mass is accompanied by the production of energy. So where exactly does the energy come from? Is it to do with the strong nuclear force, between nucleons?

Is it possible to derive the amount of mass lost for specific reactions?

An explanation of the why the strong nuclear force exists would be great.

Drakkith

The energy comes from the difference in binding energy between the different nucleons. When you fuse tritium and deuterium together, the binding energy of the helium product is more than the two fuels. This means that if you tried to pull the nucleons apart it takes more energy to pull the nucleons in helium apart than it does in the tritium and deuterium. The "extra" energy is released as heat, kinetic energy, and photons.

Imagine it like this. The binding energy of tritium and deuterium is like a ball sitting on a table. The binding energy of helium is like that same ball sitting on the ground. Lifting the ball requires more energy if you have to lift it from the floor. Similarly dropping the ball gives you more energy if you drop it to the floor than you do to the table. Hence why fusing d-t together gives you energy, it is like dropping the ball from the table to the ground.

When you look at all of the elements it looks kind of like a staircase with two tops and one bottom that forms a V shape. The least massive and most massive elements occupy the top steps. Fusing the lightest ones or splitting the heaviest ones is like dropping a ball from the top step to a lower one. As you do this again and again with the elements, it is like dropping the ball from the top to a lower step, then a step lower than that, and so on until you reach Iron and Nickel. These two elements form the bottom of the V staircase. You cannot get energy out of fusing or splitting them in the same way that you cannot drop the to a lower step. There isn't a lower step!

There is no explanation as to "why" any of the fundamental forces exist. They simply do.

padawan13

shouldn't this be the other way around, the total binding energy of the fuels is less then the helium produced. It wouldn't make sense for the binding energy to increase, because then energy would be lost.

I understand what your saying, that the greater the difference in binding energy, between the fuels and the product the more energy produced. So, the elements closer to the bottom of the V stair case have smaller differences between their fuel binding energy and their products binding energy, and therefore dont produce any energy.

so how do you determine the exact mass lost in a fusion or fission reaction?

Drakkith

I mean that Helium has more binding energy than either Deuterium or Tritium, not both of them combined. (Although it might, I'm not sure. I know helium has a very high binding energy because it has exactly 2 neutrons and 2 protons)

The mass loss can be calculated, and I think it can be measured as well.

padawan13

if the energy needed to break the atom (nuclear binding energy) increases, then the energy holding the nucleus together (ill just call it the attractive energy) must also increase.

If the attractive energy is increasing how can energy be produced?