# How did E=mc^2 contribute to the atomic bomb?

1. Jan 28, 2016

### Isaac0427

Before I got into physics, I thought that $e=mc^2$ was the atomic bomb equation. I now know that the equation itself has nothing to do with the atomic bomb. So, I'm wondering, how did $e=mc^2$ contribute to the development of nuclear/atomic weaponry/power?

Edit: I do not know if this is in the right forum.

2. Jan 28, 2016

### SteamKing

Staff Emeritus
It serves to estimate how much energy can be liberated from the conversion of a tiny amount of matter. By making some back of the envelope calculations, one can conclude that even a crude atomic device can release as much energy as the detonation of many tons of conventional explosives. OTOH, the peaceful release of this same amount of energy in a controlled fashion could provide significant amounts of heat with which one can generate electricity for consumption.

3. Jan 28, 2016

### Isaac0427

I'm not completely clear on what you mean by "the conversion of a tiny amount of matter". I understand the concept of U-235 and Pu-239 fission, but I don't know anything else about how atomic bombs work.

4. Jan 28, 2016

### SteamKing

Staff Emeritus
Well, there are plenty of articles online which discuss the mechanism by which energy is released when an atom of U-235 or Pu-239 is fissioned.

https://en.wikipedia.org/wiki/Nuclear_fission

There are plenty of articles and books on the design of the first atomic weapons, including one called "The Los Alamos Primer", which is a collection of technical notes for a series of lectures given at Los Alamos during the war by one of the scientists who investigated building the Bomb:

https://en.wikipedia.org/wiki/Los_Alamos_Primer

You should understand that in E = mc2, E is the amount of energy released, m is the amount of mass converted to energy, and c is the speed of light, which is approximately 300 million meters per second. You square 300 million, you've got a pretty huge number.

5. Jan 28, 2016

### Staff: Mentor

E=mc2 really didn't contribute to the atomic bomb, however, it does describe an equivalence between energy and matter.

When a heavy atom fissions, the sum of the masses of the fission products (and the two or three neutrons released) are less than the mass of the original excited fissile atom. The difference in mass is manifest as the kinetic energy of the two fission products, which have nuclei that are more tightly bound than the U or Pu atom before it fissions.

The energy relates to the binding energy of the nuclei.

Similarly, when two light atoms fuse, they reconfigure into more tightly bound nuclei. The sum of the masses of the product nuclei are less than the sum of the masses of the reacting nuclei, and the mass difference is manifest in the kinetic energy of the product nuclei.

http://hyperphysics.phy-astr.gsu.edu/hbase/nucene/nucbin.html#c2

6. Jan 28, 2016

### Staff: Mentor

I would say that it contributed insofar as it described and helped enable calculation of the energy released. It's just that that's a fairly minor part of the effort required to make one, which took a lot more basic physics and engineering. By comparison, the energy stored in a compressed gas tank is in its simplest form E=PdV, but that doesn't tell you much of anything about what really happens when you compress/decompress air or how to do it.

7. Jan 29, 2016

### Isaac0427

Ok, thanks guys. I do understand $E=mc^2$, my real question was on how the mass got converted to energy, and I understand that a lot better now.

8. Feb 4, 2016

### Staff: Mentor

Note that this conversion is not limited to nuclear reactions. Chemical reactions have the same conversion, but the fraction of mass that can get converted is much smaller.
The mass difference between uranium and its fission products is about 0.1%, something a kitchen scale can measure. The mass differences in chemical reactions is of the order of 0.0000001%, which is really challenging to measure.

9. Feb 7, 2016

### Garlic

How and why does mass change in chemical reactions?

10. Feb 7, 2016

### Staff: Mentor

The energy content changes. Exactly the same as for nuclear reactions.
The mass of electron plus proton is slightly higher than the mass of a hydrogen atom.
2 molecules of hydrogen plus one molecule of oxygen have a slightly higher mass than two molecules of water.

11. Feb 8, 2016

### gmax137

As mentioned by others, the equation itself isn't really used in design of the weapons; however, if you're clever enough to know the binding energies (masses) of the inputs and outputs you can estimate the power of the device. That's likely the basis for Einstein's famous letter to FDR that encouraged the beginning of the Manhattan Project during WWII. The Hungarians (Szilard, Teller, and Wigner) convinced Einstein to sign the letter; the urgency was due to the large output power expected (via $e=mc^2$). See

https://en.wikipedia.org/wiki/Einstein–Szilárd_letter

for more on the story. If you're really interested in the history, there are a lot of good books out there. Richard Rhodes' "Making of the Atomic Bomb" is probably one of the best.

12. Feb 25, 2016

### joema

Despite having read about fission and fusion, the OP was confused about how matter was converted to energy. Alas, this is typically poorly explained at the popular or even secondary school level. There are always diagrams and descriptions of fission chain reactions and comparisons to dominoes and mousetraps, but how exactly this liberates so much energy is usually not well articulated. Here is a typical example: dominoes and lots of time drawing on a chalk board but the underlying source of the energy is not explained:

The average person could read the Wikipedia article on nuclear fission and still not understand it. There is a subheading "Origin of the active energy", but it's worded in a way that few people would readily grasp it. Eventually deep into that subheading they finally state "the origin of this energy is the nuclear force", but most readers will have given up by then.

Astronuc explained this more clearly. I wish there was some well-vetted boilerplate description provided to secondary school teachers. I have talked to many high school students with two years (each) of chemistry and physics who still don't know.

IMO one of the sources is the viewpoint that physics should be "dumbed down" or restricted to pre-20th century developments in secondary school education. I recently saw this short video that respectfully explains the situation:

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