As a purely intellectual exercise in relation to air and alpha radiation I tried calculating how many molecules of air I’d encounter moving in a straight line through the few centimeters of air required to stop such radiation.
I’m hoping someone will find the idea interesting enough to comment on my calculations.
I’m aiming at finding the number of molecules in my way through 1.0 cm of air and for simplicity’s sake I’m doing the calculation for nitrogen.
To not break my mind I need to go via some sort of volume and I go with 0.1 L, so 0.1m * 0.1m * 0.01m.
The density of N[sub]2[/sub] is 1.251 g/L, so my volume has 0.1251 grams of N[sub]2[/sub].
Nitrogen has 1/(281,6610[sup]-27[/sup]) molecules per gram, which is 2.1510[sup]25[/sup]. So my 0.1 L has 2.6910[sup]24[/sup] molecules.
An estimate for the size of a N[sub]2[/sub] molecule is (54*10[sup]-12[/sup])[sup]2[/sup]8, which the internet tells me is twice the triple bond radius of N as width and four times the triple bond radius of N as length. 2,3310[sup]-20[/sup] m[sup]2[/sup].
Looking at my 0.1 L volume it has an area of 0.01 m[sup]2[/sup]. The nitrogen molecules have a total area of 500000 m [sup]2[/sup], so there would be 50 million of them along the 1.0 cm thickness of my volume.
This has lots of room for error, but I think it’s basically a decent estimate. Anyone have the time on their hand to check it.
I only went via a given volume to simplify the math and my thinking. If I haven’t messed up somewhere it indicates any straight line of one cm passes through 50 million air molecules.
Avogadro’s constant (the number of molecules in one mole) is 6.022·10[sup]23[/sup]
The volume of one mole of gas at STP (standard temperature and pressure) is 22.4 dm³
Thus, one dm[sup]3[/sup] of nitrogen contains 6.022·10[sup]23[/sup]/22.4 = 2.69·10[sup]22[/sup] molecules, not 2.69·10[sup]24[/sup]
Definitely the way to go here is to note that 22.4 L of any gas as standard temperature and pressure has Avogadro’s number of molecules in it. A 1 cm x 1 cm x 1 cm cube (0.001 L) thus has 6.02x10[sup]23[/sup] / (22.4 L) * (0.001 L) N[sub]2[/sub] molecules, or 2.69x10[sup]19[/sup] molecules. The cube root of that number will be the approximate number along a single edge of the cube (or a line through the middle) :
(2.69x10[sup]19[/sup])^(1/3) = 3 million
Multiply by two if you want to count nitrogen atoms instead of nitrogen molecules.
ETA: Note that this approach never brings in grams (which in your version you introduce (and then later cancel away) through the density and the molar mass). You could further eliminate the need for any specific volume (0.001 L, say) by noting that each molecule “owns” this much volume: 22.4 L / (Avogadro’s number). The cube root of that gives the molecular spacing. Divide 1 cm by that spacing to get 3 million molecules.
Yes, and then I did more calculations to get a totally different number. This number might be meaningless, but it certainly doesn’t represent the number of molecules in .1L.
I’m interested in how much space they “actually” cover, not how much they “own”. I’m fairly sure the area or volume of the molecule has to enter into the calculation at some point.
Corrected for the three orders of magnitude slip, I get 500 000 molecules in a 1 cm stretch of air.
The alpha particle is going to see more air molecules than occupy the volume its path sweeps out. If it had zero size it would still hit some molecules. You care about whether the air molecule centers and the alpha particle center come as close to each other as the sum of their radii. You don’t care if the air molecule center actually enters the alpha particle’s volume.
and have it evaluate it: 6300. (Google is a handy calculator.) Also, I’m not sure where the m[sup]3[/sup] part of your answer comes from. The above expression is unitless.
