[kevinkevin]

   This is a little bit more of physics than chemistry but I did get the equations out of a chemistry book.  I am currently calculating the density of atomic nuclei of different atoms (I do this kinda stuff for fun, most people would think its lame, but who cares, science is the best).  I have run into a problem as I arrived at a heavier element, aluminum.  Everything was good as I went from hydrogen to boron, the density stayed relatively the same and then increased noticeably at carbon.(the change was noticeable because I was using only two Sig figures, but I did check my math on the problems without rounding off in the middle of the problem to see more in depth how the density was increasing).  I then skipped ahead because I was running out of white board space and when I calculated the density of aluminum it came out to be less than the density of a hydrogen nuclei.  I have done the math many times and the math seems correct but I cant conceptualize in my head why the density would get smaller as the number of protons and neutrons increases. Is this a error?  Any help would be great. Thanks!

[fledarmus]

It would help if you described the method you used for calculating the density of the nucleus. It's kind of hard to determine whether you've done anything wrong if we don't know what you've done...

[kevinkevin]

   I agree.
   I was using four equations in the process of calculating the density.  First, since the nucleus is close to a sphere I was using the equation for the volume of a sphere which is V=(4/3)pi r^3.  But before I can calculate the volume I need to calculate the radius which I got from using the equation r=(cube root of the atomic number) x (1.2x10^-15 m).  The second part of that equation is an experimental value for the size of a nucleus, which I don't quite get because it seems to me as the nucleus obtains more protons and neutrons the number would change.  I then take the radius and plug it back into the volume equation to get the volume of the nucleus.  Since density is equal to mass/volume all I need to do is calculate the mass of the nucleus now.  I get this by taking the atomic mass of whatever element I'm dealing with and multiply it by one atomic mass unit (1.66x10^-27 kg).  I can then use all this info to calculate the density.  I use this same progression and equations for all the nuclei I did, and as I said before Aluminum is where my problem arrived.  I skipped some atoms but I only skipped a couple I was not expecting a drastic change in density like I saw, which lead me to thinking something is wrong.  Thanks again!     

[DrCMS]

You've got your maths wrong somewhere.

[gippgig]

The radius needs to be measured rather than estimated. You're just calculating the density of your estimate which depends on how you estimate and has nothing to do with the actual density. (I believe (and I could be wrong) nuclear density tends to be constant, altho some exotic "halo" nuclei have much lower densities.)

[fledarmus]

Where did you get the equation r=(cube root of the atomic number) x (1.2x10^-15 m)? This completely ignores the contribution of neutrons. It should be the atomic mass number, the number of protons + neutrons. And this equation is derived by assuming that the density of a stable nuclei is a constant and calculating the increased radius of the nuclei from the number of nucleons based on that assumption. It is only valid for stable nuclei, and the densities of all stable nuclei caluculated in this manner should be the same, due to the assumptions made to generate the equation.

[kevinkevin]

   Sorry that was typo I was suppose to say the cube root of the atomic mass number.  If I don't do it this way is there no other way to do it? Or do the nucleus of all atoms have the same density?  If I don't use the experimental value in calculating the atomic radius is there any other way or number I could use? 

[fledarmus]

    If I don't use the experimental value in calculating the atomic radius is there any other way or number I could use? 

There is a difference between an experimental value and a calculated value. If you have an experimental value for the atomic radius, you don't need to calculate it from the equation you gave. You just use it to calculate the volume (the volume of a sphere is probably a good approximation), and then divide it into the measured atomic mass, which you can get from high resolution mass spec data.

I'm not sure how nuclear radii can be measured, however. Perhaps a nuclear chemist could enlighten us on that point?

[Borek]

I'm not sure how nuclear radii can be measured, however. Perhaps a nuclear chemist could enlighten us on that point?

Rutherford experiment?

It probably gives only an upper estimate though. But increasing energy of alpha particles we can get to the point when the upper estimate is quite close to the reality.

[kevinkevin]

  I have been thinking about that idea you suggested about using Rutherford's experiment to calculate a radius of a nucleus.  How to you suppose it would work?  Because I cant even really imagine how the radius would be calculated either. This may be a far stretch but I have a possible hypothesis, I would love some insight if this could possible work.  Rutherford's original experiment was to find out more about the nucleus of the atom.  I don't need to talk about what the results were and what he was able to conclude from them (besides everyone on here knows about it more than I do). But what about this, what if we get the energy of an alpha particle just right so that we can swing it right by the nucleus's of the desired atoms and measure the angel of deflection that the alpha particle undergoes as it passes by the nucleus.  Wont the alpha particle be pushed away the hardest when it approaches the imaginary line vertical with the nucleus in the direction that the alpha particle is traveling. Therefore with enough precise data we could measure the point in which the alpha particle is deflected with the greatest angel or path that would tell us where the center of the nucleus is, and with persistence, when the alpha particles travel by the nucleus wont this be right next to, if not on the imaginary line vertical with the center of the atomic radius?  Just throwing that out there.  It is harder than I thought to explain it in words.  But is this reasonable you think? 
  Just a side question, would quantum rules prevent us from getting the alpha particle exactly where we want it?(Right under, on top, or underneath the nucleus of the desired element so that it passes by it and past it so we can measure the angel with which the alpha particle repels away.)         

[Borek]

From the energy of the alpha particle, and from the charges involved, you can calculate how close alpha particle that is deflected exactly back could approach the nucleus. From what I know that's how Rutherford estimated the upper limit to the nucleus size.

[fledarmus]

It looks like Rutherford's upper limit turned out to be a little less than an order of magnitude higher than the actual size; he got 2.7 x 10_-14, instead of 7.3 x 10_-15 (http://en.wikipedia.org/wiki/Rutherford_scattering). The reason, of course, is that he was firing positively charged alpha particles at a positively charged nucleus - the repulsion of the positive charges added additional force to deflect the alpha particles.

Apparently nuclear sizes were refined using electron scattering rather than alpha particles. http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/elescat.html#c1 - electron scattering

Some recent work has used laser measurements to explore a "neutron halo", a less dense, one-neutron-thick area around the dense part of the nucleus in some nuclei. http://www.uni-mainz.de/eng/13031.php

An interesting discussion - I had never really considered how nuclear size was measured.

[kevinkevin]

   That makes sense that the two positive charges, one of the alpha particle and one of the nucleus, would repell eachother and add on a noticable force that would deflect the alpha particles.  So I understand now why his number would be high. 

From the energy of the alpha particle, and from the charges involved, you can calculate how close alpha particle that is deflected exactly back could approach the nucleus. From what I know that's how Rutherford estimated the upper limit to the nucleus size.

  I still don't quite get what is being said here.  Are you saying that if we know the role the charges will play in the repulsion and the energy of the alpha particle we can..?  I'm a little confused after this point, I am really trying to visualize this.  Because if the alpha particle and the nucleus never hit and are just repulsed by one another, the alpha particle will never hit the nucleus and just get close to it then fly away, which will always yield a higher value for the nuclear size right?  This is where I am getting a little confused.

[Borek]

I still don't quite get what is being said here.  Are you saying that if we know the role the charges will play in the repulsion and the energy of the alpha particle we can..?

Knowing energy of alpha particle we can calculate how close it got to the nucleus.

Because if the alpha particle and the nucleus never hit and are just repulsed by one another, the alpha particle will never hit the nucleus and just get close to it then fly away, which will always yield a higher value for the nuclear size right?

Yes, It doesn't give us an exact value of the radius, just an upper estimate - knowing how closely alpha particles can get we know nucleus radius must be smaller. Increasing energy of the alpha particles we will get to the moment when they will be no longer deflected but rather fused into the nucleus - that will give the smallest upper estimate this method can give.

Note that nucleus is not a rigid ball with a well defined boundary. Just like atomic radius nucleus radius depends on how it is defined/measured.

[kevinkevin]

  Excellent, thanks for all the info. I feel like I got a lot out of this post! =)