User talk:BackStabbath

Nuclear diameter
Just a minor point. The article states "208 nucleons, corresponding to a diameter of about 6 nucleons" in the 4th paragraph of the requirements section. I may be completely on the wrong track here, but it seems to me that if a nucleus is roughly spherical (which I am pretty sure is true, and if not it's pointless to talk about its diameter) then you should be able to get an approximation of the total number of nucleons with 4π/3 * r^3.

If you have a diameter of 6 nucleons (thus r = 3), that gives roughly 38 nucleons total. Which seems to disagree with the article.

If you take the diameter as 7 nucleons (r = 3.5), then you end up with more like 180, which is significantly closer to 208

Might be splitting hairs, but it was just something I noticed.

BackStabbath (talk) 01:14, 3 February 2010 (UTC)


 * If you want the ratio of the radii (or diameters) of two spheres, one of which is 208 times the volume of the other, the ratio is 208^(1/3) = 5.92. The sphere with 208 times the volume will have almost 6 times the linear dimensions of the other. The constants in front drop out of these ratios. That factor is what you're adding, but shouldn't. Welcome to Wikipedia! S  B Harris 01:56, 3 February 2010 (UTC)


 * I wouldn't have thought that the volume of 208 nucleons in a nucleus would be 208 times the volume of 1 nucleon though. 208 times the mass, for sure (neglecting binding energy), but aren't there are gaps between each of the nucleons? It doesn't seem to make sense to calculate the diameter of the whole using the volume of a single nucleon. If the nucleons are equally spaced, and the nucleus is spherical, then the total number of nucleons must be governed by 4π/3 * r^3.
 * BackStabbath (talk) 04:00, 26 February 2010 (UTC)