## Identification of a particle

This is a very fine image. However, the explanation is not correct. That you have observed is a naturally occurring incoming proton (mass m1) colliding with a stationary proton (mass m0). The stationary proton comes from a water molecule (one of its two hydrogen atoms), or, if you have an alcohol condensing vapor, one of the hydrogen atoms in the alcohol molecule from a nucleus of hydrogen. The incoming proton (m1) was produced in the shower (cascade) of particles coming from the interaction of a primary cosmic ray in the upper atmosphere (mostly at 90%, protons). In fact, the incoming proton you observed was created a few meters away from your chamber by a neutron or another proton (knocked off from a nucleus). It moves at a certain speed, I can say that it’s energy is about 5 or 10 MeV because the length in the cloud chamber is below 40 cm. This give a speed of 0,1c or 30 000 km/second with a Lorentz factor Y of 1,005. According to relativistic physics, it has a mass of m1=Ym0 where m0 is the rest mass of the proton. In other terms, this proton weight much heavier than a stationary proton (the weight of an object increase with the speed and the Lorentz factor give the amount). What happens in the cloud chamber is that a « heavy » proton m1 collide with a normal proton (at rest mass). This is a typical elastic scattering of 2 particles where m1>m2 (m2 is the stationary proton). The resulting angle is inferior at 90°. If the particle had the same mass (m1=m2) you would observe the 90° angle like in billiard game. It can happen if the incoming proton have a very low energy (< 1 MeV). Anyways, that’s a nice event you have recorded. Regards, Cloudylabs.

Thank you very much for your reply. I’ve always been uneasy about suggesting the track was caused by a primary H2 cosmic ray because it didn’t seem likely that it could survive to the surface. Yet in comparison to other known tracks, its brightness and density strongly hint that it was created by a doubly charged particle. A singly charge proton wouldn’t produce as many ionizations and therefore the track it produced would be lighter and appear similar to a beta particle, though because of its mass be straighter. Can you explain how a singly charged proton can create a track as dense as a doubly charged alpha particle? I want to make it clear that I am not an expert in particle tracks in cloud chambers and that I am ready, willing and happy to correct my mistake in this video. My asking for more information is not meant as a challenge to your comment, but to clarify what you’re saying in my own mind so I can word my correction clearly. Thank you for your time in helping me understand what’s going on with this track.In the video I mentioned that this was a very rare event. Looking over the video frame by frame I had estimated that out of approximately 3,500 tracks, 3,400 were beta particles, 100 were alphas and one would be as pictured in the video. Over all the tracks recorded, I’ve only seen two of these.

Hi, sorry for delay but here is your answer : This is a rare event, but I managed to record such event but 2x rarer. For example (double scattering of a proton) https://www.cloudylabs.fr/wp/experiences/#altitude. Protons are naturally occurring around us, but it’s only about 1% of what is observable. The rest of particle are mostly electrons, positons, alphas from radon and a large amount of muons. Muons are very scarce and are impossible to distinguish in the chamber from high energy electrons and interact poorly with matter. They come only from a vertical direction and it’s very unlikely you will observe one in your sea level cloud chamber because you observe an horizontal plane in the chamber. What you observed is not an alpha particle.

Why? first, we can see an alpha particle at 00:50 in the video. It comes from natural radon Rn222. The alpha have about 5,5 MeV of energy and thus make a track of 3-5 cm according to the dE/dx (the amount of energy loss per distance). The event appear few second later so it’s length is about 20-30 centimeters according to the scale. An alpha which would make such track would need about 20-25 MeV of energy and it’s purely impossible to generate this alpha particle from a decay of a Rn222 even from forbbiden quantum state. But this alpha may come from a spallation (a big interaction of a neutron or proton outside the chamber with a nucleus) but we don’t observe any other remnant of such reaction, the particle is alone so a spallation can’t be also an explanation.

Thus simple observations say that this is very very unlikely that’s an alpha. Ok let’s consider that the incoming particule is an alpha particle anyways. Alpha particle doesn’t exist alone as atom. I reject the presence of He gaz, it’s concentration is too low to consider it seriously. So you should agree that the incoming alpha have met another particle which in this case can only be a proton from an atom and not a stationnary alpha or naturraly occuring He nucleus. You can also say that it may be a nuclear reaction, that’s to say the incoming alpha collided with a nucleus then expelled a part of the atom in form of another alpha. But you don’t observe any recoil track of the atom left in your picture so this explanation is excluded. It only « can be » the collision of an incoming « alpha » to a proton. Lets calculate that.

We will consider head-on collision to simplify (that’s to say that the particles makes a perfect collision). In reality, the latter result are modulated by a factor cos² but we don’t need it to understand. The classical scattering formula between 2 particles and for a head-on collision say that KE2=4*(m2m1)/[(m1+m2)²]KE1. Where m1 is the mass of your alpha and m2 the mass of the proton. KE2 is the kinetic energy obtained by the proton after the scattering, KE1 the initial kinetic energy of your alpha. So we have KE2=0,64KE1 assuming m1=4 (mass of an alpha) and m2=1 to simplify. This mean that after the collision, the proton should receive a maximum of 64% of the initial kinetic energy of the alpha (assuming it’s about 20 MeV as said previously). And this mean that we should observe a secondary track of about 0,64*20 MeV= 13 MeV which should correspond to a path of a proton of 1250 cm but that we see is that the secondary track terminate few centimeter away after the collision. To prove that if it was an incoming alpha you should observe the same historical event recorded in 1932 : http://www.cloudylabs.fr/wp/wp-content/uploads/2014/02/alpha-scattering-on-nucleus-1024×488.png

In the picture at far left is an Alpha particle scattering in H2 gas in a 25 cm cloud chamber. A collision appear between an alpha and a H nucleus : case m1>m2.The particle going in right is the proton as it ionize less and make a very long track. But we dont observe that in your video, we observe two tracks of nearly equal length after the collision. The picture in middle is what you wish it was : Mid (1925) : Alpha particle scattering in Helium gas. A collision appear between an incoming alpha and a He nucleus : case m1=m2. Picture in right is the collision with a heavier Nitrogen nucleus (m2>m1) .

I exclude the middle case for you because 1) it’s not an incoming alpha particle and 2) because the natural concentration of He are too low to pretend that the « incoming alpha particle » collided with a lone He nucleus (5,24 part per million) ! I know the lenght of tracks because I simulated them (https://www.cloudylabs.fr/wp/wp-content/uploads/2014/01/range-particle-air.png). So the calculus don’t match the reality.

Lets consider it’s an incoming proton. The range of 20-30 cm give KE1 = 5MeV and m1 is 1,005 due to the Lorentz factor. m2=1 (a stationary proton from an atom, at rest mass). We obtain KE2=0,999KE1 which mean that nearly all the energy of the incoming particle goes to the stationary proton. But as we observe a non-head-on collision because we observe an angle, it mean that the target will receive after the collision approx. half of the kinetic energy of the projectile, so KE2=0,5 KE1. We should observe two track of approximately the same length, that we observe. To prove that, consider a non-head-on collision between a « heavy » moving electron and another stationary : http://www.cloudylabs.fr/wp/wp-content/uploads/2019/09/halfenergy-electron.png (we observe two resulting track of same length because they get the same energy from the collision. It’s your event (but with protons). To conclude, I guess that your cloud chamber is dry iced because there is some turbulence (moving gas) which indicate that it’s height is high.

Remember that there is only a small active layer where the supersaturated layer is. The proton appear « brighter » or « denser » than your common alpha particle because it passed by luck, right in the supersaturated layer where the density of unstable vapor is very high ! if it passed a little above the supersaturated layer, it would appear far less brighter than an alpha. Anyways that’s a clear picture and good event that you recorded. Scattering of proton at sea level is uncommon to see.