Sciencemadness Discussion Board

A Cloud Chamber

blogfast25 - 11-4-2013 at 12:31

I finally got round to building a simple cloud chamber for the detection of ionising radiation (full story here: cloud chamber) Here’s what she looks like:


I used methanol instead of the popularly recommended isopropanol (IPA) and first tested the chamber with a 2 % thoriated (as ThO2) tungsten welding rod as a source of ionising particles:

Th90232 == > He24 + Ra88228 (the Th and Ra are my excuse for parking the story in ‘Chemistry ‘ ;-)), with a half life of just over 14 billion (!) years.

With the rod in place it takes 10 – 15 minutes to see the first cloud trails of α particles to appear to shoot off from the thoriated tungsten:






The flat discs are neodymium magnets but they didn't actually do much at all, not even locally.

After removing the rod, it took a very considerable time (about ½ hour, at a guess) before the first cosmic particles revealed themselves. They are however hard to photograph and my multi-shot autotimer shooting mode left me down, so that I haven’t got any decent photos yet. The dry ice (2.5 kg bought) ran out yesterday, so good photos will now have to wait…


[Edited on 11-4-2013 by blogfast25]

violet sin - 11-4-2013 at 12:52

awesome! nice project.

Endimion17 - 11-4-2013 at 13:25

Ooooh, methanol. :)

I expect a video, soon. Preferably with something hotter. ;)

neptunium - 11-4-2013 at 20:34

when you get tired of going to pick up more and more dry ice try with an old (or new) AC unit...
i bought one some years ago and re assigned it for that very purpose! it worked great! unfortunately no pictures ...

nice work nonetheless!

blogfast25 - 12-4-2013 at 03:49

Having dry ice 'on tap' could certainly be useful...

MrHomeScientist - 12-4-2013 at 06:17

Very cool! I've wanted to build one of these for quite a while, and I just so happen to have the same thoriated welding rods as you!

By the way, are you still writing on your personal blog or have you switched to this new one? Is that something related to your work?

blogfast25 - 12-4-2013 at 08:04

Quote: Originally posted by MrHomeScientist  
Very cool! I've wanted to build one of these for quite a while, and I just so happen to have the same thoriated welding rods as you!

By the way, are you still writing on your personal blog or have you switched to this new one? Is that something related to your work?


Thanks, it's well worth doing, IMHO.

Now, you're the physicist, so explain to me why my neomagnets didn't do anything, huh? Very disappointing that was. What kind of magnetic field strength would be needed to get these alphas to pirouette?

Re. the thoriated rod, the photos don't do it much justice: to the naked eye there's a lot more activity going on.

I now mainly write my science stuff on the blog that came with the WP estore software, combining business with pleasure as it were. ;)

[Edited on 12-4-2013 by blogfast25]

MrHomeScientist - 12-4-2013 at 11:23

Let me get back with you on that. I've got some thoughts, but it's the end of the day here so I'm a bit antsy to head home :P

Such an interesting question, I pulled out one of my old text books to research this weekend! That probably sounds strange, but the internet is my usual source.

blogfast25 - 12-4-2013 at 11:29

I had a quick look on the Tinkerwebs too but could only find generic references to the Lorentz Force. No actual practical values of B.

Endimion17 - 12-4-2013 at 11:49

I think the energies and short paths involved are too high/short for any visible effect to occur. Calculations should be made to verify that...

blogfast25 - 12-4-2013 at 13:18

Quote: Originally posted by Endimion17  
I think the energies and short paths involved are too high/short for any visible effect to occur. Calculations should be made to verify that...


That's what I wanted to do, but it's harder than it sounds... There's some really small numbers involved.

platedish29 - 12-4-2013 at 13:29

Its IS realy magic just to worn a cloud chamber as in fact you can show your neighbors and fellows more than just falling synthetic snow but also how the world is filled with awesome high energy particles no matter where you hide!
For instance, congratulations on your project. You and other who moved on to propel this old fashion is of utmost significance for the young physicist and, as a prayer, a fortuit aknownledgement for the visionary chemist!
Now, did you manage to have it working on cosmic rays too? I heard they are detectable too, just grab a very nice picture with a good camera.
Curiously though, even if I had never seen the true facts behind radiation, I would call that an artificial snow and leave your apartment with an inner feeling that you ... well... nevermind :D
See the link for the video of a well-fit chamber:
http://www.youtube.com/watch?v=ajV44zJlBgQ

blogfast25 - 13-4-2013 at 07:18

Quote: Originally posted by platedish29  
Now, did you manage to have it working on cosmic rays too? I heard they are detectable too, just grab a very nice picture with a good camera.


Yes. It took longer to be able to see them clearly. Presumably the right temperature gradient has to be established to obtain a suitable degree of supersaturation.

The problem is the electronic detection part of the device: in plain English; the camera! A decent camera and tripod, positioned well and not seeing through inevitable condensation that occurs on the outside, lower part of the 'box', combined with some photographic skills are really needed.

The type of chamber with the felt at the top around the edges, with a clear top allowing a bird's eye view is probably better for good photography.

Great video, BTW. His is positively teeming, I could see some decays too. I suspect he's using a lot more alcohol because it positively rains in there. Must try again with thicker felt/more alcohol!



[Edited on 13-4-2013 by blogfast25]

12AX7 - 13-4-2013 at 07:18

An alpha passing directly over the NdFeB magnet I think should deflect visibly, umm...
http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_8...
q = 2
B = 1.5T (or less even at magnet surface, dropping off steeply only a short distance away)
l = 0.01m (or so, whatever the magnet diameter is)
Ek = 4MeV (give or take)
m_0 = 4 amu
theta ~= (2)(1.5)(0.01) / (0.144((4M)(4))^0.5)
52 x 10^-6 rad = 0.18 arc-minutes

Okay, that's why you won't see anything...

You might get lucky enough to see a little hook shape if the end of a trail lands on a magnet, but it'll be hard to see.

Tim

blogfast25 - 13-4-2013 at 07:29

Quote: Originally posted by 12AX7  
An alpha passing directly over the NdFeB magnet I think should deflect visibly, umm...
http://www.kayelaby.npl.co.uk/atomic_and_nuclear_physics/4_8...
q = 2
B = 1.5T (or less even at magnet surface, dropping off steeply only a short distance away)
l = 0.01m (or so, whatever the magnet diameter is)
Ek = 4MeV (give or take)
m_0 = 4 amu
theta ~= (2)(1.5)(0.01) / (0.144((4M)(4))^0.5)
52 x 10^-6 rad = 0.18 arc-minutes

Okay, that's why you won't see anything...

You might get lucky enough to see a little hook shape if the end of a trail lands on a magnet, but it'll be hard to see.

Tim


Thanks Tim, much appreciated, nice link too! Yes, I guess that explains it.

I'll see if a 'mat' of much larger neomagnets glued (or screwed?) to a piece of wood (that'll be fun!) could do something. Generating a strong enough external magnetic field is outside of my capability envelope for now.

MrHomeScientist - 14-4-2013 at 13:42

I did the calculation another way, with a few different assumptions, but arrived at essentially the same result. Looks like the hard way, compared to 12AX7's method!

Constants:
- m = 6.6x10-27kg
- q = 2e = 3.204x10-19C

Assumptions:
- 3/8", N52 magnet (eyeballed the size, assumed highest strength). According to kjmagnetics.com for this magnet, surface field B = 0.4105 T
- E = 6 MeV = 9.6x10-13J, based on an alpha particle energy spectrum chart for thorium-227 that I found in my Modern Physics textbook (Tipler). This is the high end of the range.
- Assume the best case scenario: alpha flies directly over the magnet right at its surface, and assume the field is uniform across the whole magnet's surface

1. Find velocity from kinetic energy: KE = 1/2mv2
v = sqrt(2E / m) = sqrt(2 * 9.6x10-13 / 6.6x10-27) = 1.7x107 m/s

2. Find force exerted from eq. for force on a charged particle in a magnetic field: F = qv x B (ignoring vectors here)
F = (3.2x10-19)(1.7x107)(0.4105) = 2.2x10-12N

3. Find acceleration from Newton's 2nd law: F = ma
a = F / m = 2.2x10-12 / 6.6x10-27 = 3.4x1014 m/s2

4. Find time for particle to fly over the 3/8" (0.0095 m) magnet: x = vt
t = x / v = 0.0095 / 1.7x107 = 5.6x10-10s

5. Find horizontal displacement from equations of motion (where V0 = 0, X0 = 0, i.e. starting from the edge of the magnet with straight line motion): y = (a/2)t2
y = (3.4x1014 m/s2 / 2) * (5.6x10-10)2 = 5.3x10-5 m = 0.002 in

So roughly two thousandths of an inch of deflection, likely to be a good bit less as B falls off rapidly with distance.

I think the two key factors here are that, relatively speaking, (1) alpha particles are heavy, and thus have a good bit of inertia, and (2) they are slow-moving and thus won't be affected by a field as much as, say, a beta particle.

Disclaimer: It's been a while since I've done such a calculation, so things may be off! Fun exercise though. Once I build my own, I've got some extremely powerful magnets that I'll try out.

12AX7 - 14-4-2013 at 17:05

If you take the ratio of displacement to travel, it equals the deflection angle in radians (tan x ~= sin x ~= x for small x). 0.053mm / 10mm = 0.0053, which is 0.3 degrees, or 18 minutes.

Still not detectable, but interesting that it's off by a factor of 100 -- yours may very well be correct, I was going off the equation, and equations with pre-cooked constants always make me nervous. Or maybe we're both wrong! :)

Tim

MrHomeScientist - 14-4-2013 at 19:22

That is really interesting! I assumed a larger kinetic energy and smaller magnetic field than you did as well. I think the fact that we both got the same digits is too much of a coincidence - whenever this happened in my physics classes, it was 99% of the time someone had done an incorrect unit conversion, but otherwise did everything right. Not sure who is right (if either of us at all), but pretty cool nonetheless. The link you used was exactly what I was searching for, but wasn't able to find.

blogfast25 - 15-4-2013 at 11:39

Thanks Mr HS. One thing astounds me:

1. Find velocity from kinetic energy: KE = 1/2mv2
v = sqrt(2E / m) = sqrt(2 * 9.6x10-13 / 6.6x10-27) = 1.7x107 m/s

Ok, that is for Th-227, here it’s Th-232, probably lower speed I’d imagine because Th-232 is relatively stable.

So that’s how fast they travel?

What kind of magnets are you thinking of? I was going to screw a larger number of larger neomagnets to a piece of wood to create some kind of ‘magnetic mat’ but that sounds futile now!

MrHomeScientist - 15-4-2013 at 12:17

I believe so! That works out to be 5.7% of c (wiki states that alphas are generally 5MeV and 0.05c, so that's some confirmation). One thing I didn't do was consider relativity, but the Lorentz factor in this case ends up being 1.0016, so we can disregard it. This wouldn't affect movement perpendicular to v, anyway.

The magnets I have are scarily powerful, meant for electric motors. I've had them for forever, so I don't remember what their actual field strength is. They are strong enough where if you get two stuck together, you practically need a (nonmetallic!) crowbar to get them apart.

You could certainly try the magnet mat, with all magnets having the same pole facing up (might be a challenge to actually do this!). That at least would increase the time the alphas are subjected to the field, so you might see some bending. You could always order one of these monsters :D

blogfast25 - 15-4-2013 at 12:52

Wow. That's a real whopper! I suppose I could try and get my money back by dissolving it and selling off some Nd salts... :D

I think I'll stick to getting the super saturation layer to work better firstly, before trying to magnetise the entire chamber! The other problem is that the alpha paths are really quite short: a few cm only.



[Edited on 15-4-2013 by blogfast25]

platedish29 - 16-4-2013 at 18:51

How can you manage to prove those "rainfalls" are not simply converging internal entropy dissipation from the system itself?
Do you have an idea how else you can prove those really are HE particles?

blogfast25 - 17-4-2013 at 06:17

Quote: Originally posted by platedish29  

Do you have an idea how else you can prove those really are HE particles?


There are plenty of ways that it can be proved that these trails are He24 nuclei, as has been done empirically in the past, countless times. There's no need to reinvent the wheel here: I reasonably assume that my rod contains Th-232, that theory states correctly that Th-232 is an alpha emitter and that said trails are due to alpha particles.

In other words, I don't have to prove this.

MrHomeScientist - 17-4-2013 at 06:31

Quote: Originally posted by platedish29  
How can you manage to prove those "rainfalls" are not simply converging internal entropy dissipation from the system itself?
Do you have an idea how else you can prove those really are HE particles?


You can really see it when you have a "point source" of radiation, like in this video:

http://www.youtube.com/watch?v=pewTySxfTQk

Skip to about 3:15. Clearly the tracks radiate outward from the tip of the needle. If the condensation was caused by random processes in the gas itself, it would be going in random directions (and almost certainly wouldn't form straight-line tracks).

Alpha particles of similar energies will also travel similar distances, so you'll also notice that the tracks are all of comparable length. It's possible to calculate how far alphas of a certain energy should travel, but I'd need to look it up again to remember the procedure.

platedish29 - 17-4-2013 at 08:41

Ok,
So if one does not want to show his neighboors to come watch your radiation source (most would dislike, some puffed up kids would want to hold it in their bare hands etc...) I wouldn't call that impressive. You'tr just providing a way to prove and want-not to prove a previous stabilished theory everyone is familiar about. I find cosmic ray particles much more interesting than just sampled radioactive decay. Sorry for not sharing your point in this thread but since its is a discussion board you must well, deal with...
Srsly,
You gotta rebuild another chamber... !

You can prove rain paths are what they are by placing two chambers apart and observing that the tracks formed in one of them is repeated in the other one, like instant holographic telephaty of some sort of thing you can even be rich predicting future and s**t with this device!!

Obviously, you would have to consider refraction along glass/ air/glass interface in the gap between your chambers. (check this analog for the refractory media)


[Edited on 17-4-2013 by platedish29]

blogfast25 - 17-4-2013 at 08:52

Quote: Originally posted by platedish29  
Srsly,
You gotta rebuild another chamber... !



[Edited on 17-4-2013 by platedish29]


No, I haven't. I find cosmic rays too more interesting than sampled radioactivity. I saw the cosmic rays very clearly, as did two other witnesses.

W/o a shadow of a doubt more felt and more methanol will improve viewing tremendously.

This is a simple demonstration, plate, nothing more. No Nobel prizes are being pursued here. Sorry if that disappoints you but I didn't build it with twits like you in mind.

As regards: "Sorry for not sharing your point in this thread but since its is a discussion board you must well, deal with...". Sorry, but no one is obliged to deal with you if they don't fancy it.



[Edited on 17-4-2013 by blogfast25]