3
$\begingroup$

I've been watching an MIT course which explains the way that a nuclear bomb works. I am not a physicist or physics student, however; I am a computer scientist. As such, I am not looking for a level of detail as required from someone who studies physics. I also had a look at this question From what I understand so far can be summarized as follows:

  1. Uranium-235 is an isotope which, when the individual atoms receive a neutron, the atoms will split, release "energy," and give off amongst other things, more neutrons flying through the air.

  2. If there is some density and mass of Uranium-235 atoms all in close proximity, it follows that the above action will cause a chain reaction because once one atom splits and sends out neutrons, those neutrons will in turn hit other atoms which will split and do the same.

  3. When atoms split, neutrons will often fly out and not actually cause fission in other atoms... Thus, many neutrons are "lost." To combat this, there is a shield put around a U-235 mass in one of the production bombs to help reflect neutrons back in toward other atoms to induce more fission.

    My question is two-fold. If the complexity of answering these 2 questions with medium to low detail is too high for one thread, please let me know and I will remove one and separate it to another thread.

Question #1: Does subcritical mass mean that say, I could have a tennis-ball sized chunk of Uranium-235 in my hand, and NO atom-splitting will be occurring at all because it is not at critical mass? My confusion here lies in that I am picturing many many U-235 atoms in this ball, and the probability that one of those atoms will receive a neutron from the surrounding environment or another of the atoms, is relatively high. Is this an incorrect assumption here?

Question #2: How exactly would taking 2 such balls (for example), both at subcritical mass, and slamming them into each other, suddenly bring them at critical mass and fire off neutrons in a way that simply having 1 ball wouldn't already achieve if a neutron happened to hit it? My misunderstanding here probably has to do with critical mass/density. Say for example, 5,000 atoms inside of one of my U-235 balls split and then the process stops because it's not at critical mass... Would this energy emission be noticeable? It's hard for me to conceptualize just how much energy each individual atom gives off as it splits, and the relate that back to something that a human could perceive. Obviously, when one of these balls explodes in a bomb, it's safe to say that it is VERY perceptible to many people. But, where is the threshold for that? How many atoms must split for heat or some emission to be "felt?"

$\endgroup$
2

3 Answers 3

8
$\begingroup$

Question #1: Does subcritical mass mean that say, I could have a tennis-ball sized chunk of Uranium-235 in my hand, and NO atom-splitting will be occurring at all because it is not at critical mass?

No. Critical mass is required for a self-sustained chain reaction, but individual fission reactions can and certainly do occur below critical mass. All it takes is a urainum-235 nucleus coming in contact with a neutron. Cosmic rays produce around a few hundred neutrons per square meter per second, so there's likely to be at least a bit of fission taking place. Since the mass is sub-critical, the probability of one fission reaction triggering another fission reaction is pretty low, though.

How exactly would taking 2 such balls (for example), both at subcritical mass, and slamming them into each other, suddenly bring them at critical mass and fire off neutrons in a way that simply having 1 ball wouldn't already achieve if a neutron happened to hit it?

What matters is more clearly stated as the total number of nuclei a neutron encounters on its way out. If a neutron has a certain probability of interacting with every U235 nucleus along its path, then the total probability of a neutron triggering fission somewhere in the ball goes up with the radius of the ball. And that quantity (the probability of the neutron triggering fission somewhere) is what matters in terms of sustaining a chain reaction, as a neutron is only "lost" if it doesn't trigger fission in any of the nuclei in its way.

Say for example, 5,000 atoms inside of one of my U-235 balls split and then the process stops because it's not at critical mass... Would this energy emission be noticeable?

It depends on what you mean by "noticeable". Each U235 nucleus releases something like 200 MeV when it undergoes fission. 5,000 of these reactions would release 1 TeV of energy. This is quite a bit of energy from the perspective of a sensitive radiation detector, and will be very easily noticeable in that context. However, in macroscopic terms, it's not much. It's roughly the kinetic energy of a mosquito in flight, or enough energy to melt a microscopic speck (0.5 ng) of ice.

Obviously, when one of these balls explodes in a bomb, it's safe to say that it is VERY perceptible to many people. But, where is the threshold for that?

Depends on what you mean by "perceptible". The physics determining perception of sound, light, etc. are very highly dependent on the environment around the explosion, and the design of the explosive itself.

If we assume that all of the energy released is released as sound, and that the sound is all directed toward a particular listener, and that that listener will notice if the sound is about as loud as a gunshot (100 dB, corresponding to 0.01 J of acoustic energy), then you would need somewhere around 300 million fission reactions to even get anywhere close to a chance of detecting it. In reality, most of the energy of a nuclear explosion doesn't go into sound, and the sound is dispersed over a large solid angle, with intensity decreasing with distance, so any kind of realistic number is likely far higher.

$\endgroup$
0
4
$\begingroup$

With regard to "feeling" nuclear processes:

The spinthariscope (https://en.wikipedia.org/wiki/Spinthariscope) consists of a source of radiation (often alpha-particles), a fluorescent screen (zinc sulfide), and a magnifying eyepiece for viewing the screen.

After suitably dark-adapting the eye, one can see flashes of light as individual alpha particles hit the screen and yield up their energy as multiple photons of visible light. The human eye can detect the light energy from one individual nuclear decay.

This is a natural decay, not a fission, but the energy range is similar.

$\endgroup$
-2
$\begingroup$

Q(1) - You have the correct idea. Some decays would create neutrons that would hit other another nucleus and cause it to decay, thus a chain reaction. The normally quoted half-life for uranium-235 would assume that there were no chain reactions.

Think of each decay as having 1/2 chance of creating another decay. So on average a given decay will create one more decay. $1 = \sum 1/2 + 1/4 + 1/8 +...$

Q(2) The notion of a critical mass is that the chain reactions happen so fast that all of the U-235 is converted "instantaneously" which is about a microsecond for a nuclear explosion.

Think of each decay as creating 2 more decays. So on average a given decay will create an infinite number of decays. $\infty = \sum 2, 4, 8, 16, 32, 64, ...$

$\endgroup$
1
  • $\begingroup$ No, a critical mass will have a self-sustaining reaction where each set of fissions during some time interval ( e.g. 1 ms) results in the same number of fissions during the next 1 ms. An increase in the fission rate requires a super-critical mass with a K>1.0000. If a large K (say 1.5 or 2 or 3 or 10) can be maintained for 500 ms, then you have the fast explosion. K<1 is sub-critical (decreasing power), K=1 is critical (steady state power), K>1 (increasing power). $\endgroup$
    – Bill N
    Commented May 29, 2020 at 1:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.