# In nuclear fusion reaction, what is the percentage of mass converted to energy?

I read somewhere that it is about one percent of the mass, but I find this too high. Also I have done some calculations, for example, the Tsar Bomba was 50 MT bomb and weighed about 27 tons. Although I don't know how much exactly fusion fuel was used, I think it is safe to assume that a large percentage of this 27 tons was fusion fuel. So, if we want to get the same amount energy from mass-energy equation, we would need something with a mass of 2.2 kg. Which proves what I mean : if fusion reaction converted one percent of the fuel mass to energy, then we would need only 100*2.2 kg = 220 kg of fusion fuel to make the Tsar Bomba, which I find much much lower than the actual number.

Tell me what you think, please.

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WolframAlpha is the perfect tool for such calculations:

1. Fraction of mass converted to energy, result is 0.0037681, less than 0.4%.

2. How much is mass defect of 50 MT of TNT, result is 2.3 kg.

As for Tsar Bomba the fusing material is of course only small part of the actual device.

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I'm sorry, but how could you get the fraction of mass converted to energy that way ? – Abanob Ebrahim Aug 1 '13 at 15:56
@AbanobEbrahim: Click on the link. We take reaction D + T -> He-4 + n and subtract mass of reaction products from the mass of reactants. The result would be the mass defect (~17MeV/c^2). Dividing it by the mass of fusion products we receive the fraction of mass converted to energy. – user23660 Aug 1 '13 at 16:14

In nuclear fusion reaction, what is the percentage of mass converted to energy?

First a note.

A fusion bomb generates a lot of its energy by fission

Briefly the sequence is something like this:

1. Set of a small fission bomb.
This generates X-rays and neutrons.
2. The x-rays are used with a Teller Ulam device to compress a fusion part. This will contain some fuel such as lithium deuteride. (Not Tritium because it is rare and it has an annoying short half live of 12.3 years. And its decays product is He3 which likes to absorb neutrons).
3. More fusion fuel is generated by the neutrons.
Neutron + Li6 -> neutron + He + tritium.
Alternatively Li7 can be used (as unexpectedly discovered during Castle Bravo).
4. The compressed tritium and deuterium reacts, producing some energy and more neutrons.
5. The extra neutrons react with the remaining fission material (either in the initial device, or in the spark plug, or with the tamper which can be made from U238.

I read somewhere that it is about one percent of the mass, but I find this too high. Also I have done some calculations, for example, the Tsar Bomba was 50 MT bomb and weighed about 27 tons.

Well, you know the yield of the bomb (somewhat over 50 Megaton TNT).
Thus you can look up how much energy is liberated by 50 Megaton TNT
Then convert that number to mass.

Although I don't know how much exactly fusion fuel was used, I think it is safe to assume that a large percentage of this 27 tons was fusion fuel.

I am not that sure. The Tzar bomba was a 100MT design, which used a uranium tamper. Later it was downgraded to 50MT to avoid a lot of pollution. So in the initial design at least half the yield would not have been directly generated by fusion but by fission. Also U238 or lead (as alternative tamper) are very heavy, while the fusion fuel is likely lithium deuteride.

(Granted, the neutrons to release the energy from the tamper are mostly generated by the fusion reaction).

So, if we want to get the same amount energy from mass-energy equation, we would need something with a mass of 2.2 kg. Which proves what I mean : if fusion reaction converted one percent of the fuel mass to energy, then we would need only 100*2.2 kg = 220 kg of fusion fuel to make the Tsar Bomba, which I find much much lower than the actual number.

When fission fuel gets used it does not completely change to energy. Instead the atom is smashed apart and there will the several pieces. Typically this will be 2 or 3 neutrons and 2 smaller atoms (e.g. 92U to 50Tin, 42Molybdenum and neutrons).

It is just that the sum of the mass of these fragments is slightly smaller.
It is not the case that a while atom just disappears and somehow releases energy.
Part of that energy will be the new fragment, which usually move at a very high speed. And this will result in collisions, x-rays and, well, heat. Lots of it.

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I realize this is two years old, but would still like to point this out:

When saying that 1% of the mass of fuel is converted to energy, it only means the percentage of the mass of what fuel actually undergoes fusion.

Thus, your 2.2kg of fuel converted to energy would represent 2200kg (edit: oops, 220kg) of fuel which actually underwent fusion, NOT the amount which was contained in the bomb. THAT number is going to be highly variable, dependent on the efficiency of the design and construction of the device. In this instance, it would appear that approximately 1% of the fuel (depending on how much was fuel and not bomb structure + fission trigger) reacted, and the other 98+% just got blown outward at high velocity.

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Your first part of the answer is correct and I agree with it. But "it would appear that approximately 1% of the fuel reacted" is not correct. As you mentioned this question was two years ago, and I found that the fusion fuel efficiency is around 25-50% of the available fuel. So to get 2.2 kg of mass-energy equivalent, we would need 220 kg of fuel to undergo fusion, and this 220 kg would represent 25-50% of the fuel in the bomb so we would actually need to have somewhere between 440 and 880 kilograms of fuel there to end up with 2.2 kg of mass converted to energy. – Abanob Ebrahim Dec 10 '15 at 20:27