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A-Bomb Physics

Updated on June 21, 2008

Target Practice

Before I get into an explanation of what a nuke is, I need make an analogy. Pretend you're on a range firing a pistol. It's bad to say you're aiming at the target, but one is out there. What are some ways to increase your odds of hitting the target? There are two ways of increasing your odds of hitting a target: putting more targets out there and moving the targets closer to you.

So what happens in atomic bombs that relates to that? Patience, grasshopper. You see when matter is converted to energy, a little bit of mass converts to a lot of energy (Famous equation E=MC^2). When a uranium nucleus splits into two smaller nuclei, a little mass is lost in the overall reaction - mass that is converted into energy. When a uranium nuclei splits, it also releases some neutrons. Tricky thing about these neutrons is that when they collide with another nuclei that nuclei will split as well and release more neutrons. They're like bullets aiming for other targets, only when you hit a target it starts shooting bullets as well.

My favorite bomb picture.
My favorite bomb picture.

Critical Mass or Critical Density?

I've always wondered about the term "critical mass". I think it should be refered to as "critical density". In the analogy above about shooting blindly at targets, I mentioned the two ways to increase your chances of a hit are to either bring the targets closer or to add more targets. Believe it or not, Hiroshima used one tactic while Nagasaki used the other.

Critical mass is when the nuclear explosion has been achieved. You got the core explosive that sits there until it is detonated. Critical mass is when the splitting of nuclei accelerates like an avalanche so that in one millionth of a second ten percent of the nuclei in a good bomb split, releasing a shit-ton of energy.

Hiroshima's bomb (Little boy) was a gun-type bomb. It brought two chunks of uranium together at incredible speeds - adding more targets on the firing range. Nagasaki's bomb (Fat Man) was imposion-type, which was bringing the targets in closer. Now I mentioned above that critical mass was when the splitting of nuclei was like an avalanche. You need to quit thinking of a chunk of uranium as a chunk of metal and think of it as a big collection of uranium atoms. Billions of trillions of uranium atoms. On a rare event, a uranium atom spontaneously splits and releases neutrons and energy. Usually when this happens, its neutrons might cause a small avalanche of splitting but in the end the majority of neutrons are lost out into the atmosphere without being able to propagate the avalanche very far - critical mass is not achieved. When you add more uranium, those neutrons that would be lost to the atmosphere are now finding greater odds of hitting another nucleus - increasing the chances of a propagating avalanche. Critical mass occurs when enough uranium is in the core that those spontaneous splittings will trigger a propagating avalanche that consumes a significant fraction of the atoms available.

What a Sabot Depleted Uranium Round Looks Like

Implosion-Type and Depleted Uranium

Implosion-Type bombs are ones where the uranium or plutonium is compacted to incredibly small volumes, bringing the nucei of the atoms closer together and thus increasing the odds of a propagating avalanche. Implosion bombs must have conventional explosives places precisely around the core so that when they detonate, a spherical pressure wave hits the core evenly at the exact same time and compresses it to critical mass. They talk about nuclear geometrists in the movie "Sum of all Fears"... a nuclear geometrist makes sure all the conventional explosives are shapes right and placed right to create that spherical wave.

You know, I've always wondered what the big hype was about depleted uranium shells in combat. What most people don't know is that uranium has multiple isotopes. Uranium 235 is the isotope used in bombs and fuel and makes up less than one percent of the uranium on the planet. It is available to pretty much any country on the planet. What they lack is the technology to isolate the 235 isotope - isolating the 235 is called enriching. Depleted uranium is the useless leftovers - the 99%+ uranium that can't be used in bombs and fuels. IT IS NOT RADIOACTIVE. When all the 235 is taken out of the natural uranium, what's left is safer than copper in terms of radioactivity. Since the useless uranium is the most dense natural element on the planet, it makes a great armor-piercing round and to this day they have not discovered a better tank-killer for its size.

Now one bit that I will say about the hype... it's not completely unfounded. When people hear of a depleted-uranium bomb or they see pictures of charred babies and other grotesque things... That is actually a dirty bomb. That is where the 235 has been left in the uranium in significant amounts, usually from spent fuel rods from power plants or enriching facilities. The US government does not make rounds that contain a significant amount of 235 - at least I pray to god they don't.


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    • Science Guru profile imageAUTHOR

      Science Guru 

      8 years ago

      Stan brought up a point that emphasizes something about my style of writing. I exaggerate a lot. If you are writing a paper for class or business, you shouldn't be using hubpages as scholarly sources for your information.

      DU is essentially pure uranium 238. Depending on how pure it is determines how safe it is. If it's 100% pure, it's 100% safe. Now is it exactly 100% pure? No. How pure is it exactly? I don't know but I know it's greater than 99.3%.

      I have never heard of the tungsten-cobalt alloy. Tungsten is incredibly dense so it makes sense to make an armor piercing round with it. It also holds its shape at incredibly high temperatures, making it favorable when friction heat would normally destroy the AP round.

    • profile image


      8 years ago

      ''is safer than copper in terms of radioactivity'' Are you ignorant? How do you account for the surge of cancer in Iraq and even among US and British troops.

      On a side note, in fact there is a more superior AP than DU; it's a tungsten-Cobalt alloy, but ironically it's extremely carcinogenic(almost 99% of the cases), even far more dangerous than DU.

    • Science Guru profile imageAUTHOR

      Science Guru 

      8 years ago

      Energy is defined in joules, which can be re-written many ways. The units always remain the same. Force*Distance is a common one. Mass*Acceleration*Distance is another. C^2 has the same units as Acceleration*Distance. Anyway, E=MC^2 is short for Energy = Mass*(Speed of Light)^2.

      Since the speed of light is constant, the only change that can give off energy is a change in mass. There are charts that show a relative amount of mass lost for various types of nuclear reaction. Combine data from this chart with E=MC^2 and you'll get a change of mass in a nuclear reaction, if you know the amount of energy given off. Change in mass over original mass leads to a percentage. Just find the original mass and that data I was talking about.

      One common misconception is that the entire atom is destroyed. When a uranium atom splits, there are a bunch of smaller fractions left over. A lot of people think the entire mass gets converted into energy, which is incorrect. keep that in mind.

    • profile image


      8 years ago

      Science Guru thank you for your answer. What raised the question was a conversation with a friend. He stated that the % of fuel used was 2%. From E=MC^2, it would seem to lead to the conclusion that all available fuel would be converted. What is your take on this?

    • Science Guru profile imageAUTHOR

      Science Guru 

      8 years ago

      DEmmett, I can't say I have an answer for you. In the article above, I said 10% of the uranium atoms split... but that was an estimate I heard years ago from a source I can't confirm. Truth is I don't know the needed mass or the exact efficiency of the bombs. I know Plutonium requires less fuel to reach critical mass. I know the efficiency is not as high as you'd think.

      Terminology I've heard implies that you can actually get a detonation from a range of efficiencies. "High-yield" I believe is refering to the efficiency as well as the size of the explosion. I do not know the minimum efficiency required or what the maximum attainable efficiency is.

    • profile image


      8 years ago

      I am trying to find out what % of the uranium is converted to energy when an A-bomb is exploded. Also, what is the % of the mass is coverted in all types of types of A-bombs and H-bombs. In other words how efficient are these bombs?

    • ram_m profile image


      9 years ago from India

      Very nice and informative hub.


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