# Re: Close-packed arrangement question

Any help with the following question would be greatly appreciated!

The diameter of a rubidium atom is 4.95 Å. We will consider two different ways of placing the atoms on a surface. In arrangement A, all the atoms are lined up with one another. Arrangement B is called a close-packed arrangement because the atoms sit in the "depressions" formed by the previous row of atoms.

A) Using arrangement A, how many Rb atoms could be placed on a square surface that is 1.0 cm on a side?
B) How many Rb atoms could be placed on a square surface that is 1.0 cm on a side, using arrangement B?

I was able to get A) answer = 4.1 x 1014 atoms/cm2, but I am really confused as to how to find out B) answer should be: 4.7 x 1014 atoms/cm2.

### Re: Close-packed arrangement question

In arrangement A, the distance between the nuclei will be 4.95 Å, so it is case of calculating the number of atoms in 1cm2
In arrangement B, the second row of atoms will be slightly closer. I drew an equilateral triangle between three adjacent atoms (2 in one row and one above) The distance between the nuclei is still 4.95 Å, but the vertical distance between the nuclei is less. I will attach a diagram (Click on it to enlarge it - I hope it helps).
Using Sin rule and the angle of the triangle at 60o, I got the vertical distance (x) between atom nuclei in one row to the one above to be Sin60 = x/4.95 which calculates at 4.005 Å. I'm sure using this to work out the number of row of atoms will give the increase in atom number in B. There will be the same number of atoms from L to R  in a row as there were in A, but more rows can be fitted in. So number of atoms from L to R is found by dividing the internuclear distance (4.95 Å) into 1cm (converting this to Å).
The number of atoms from bottom to top is found by dividing the vertical distance (4.005 Å) into 1cm (converting this to Å).
Then the number of atoms in the square will be the two numbers multiplied together.
I have checked this roughly and it gives an answer of above 5 x 1014 which is more than expected. So there is perhaps a flaw in my reasoning. I don't think if you allowed 1 atom less on every second row (because of the arrangement) this would give the expected answer. (It doesn't)

(Sin of angle = side opposite/hypoteneuse) - it's 50+ years since I learned that

### Re: Close-packed arrangement question

Still not there but a little closer.
4.005A is the vertical distance between nuclei.
If we add on 1/2 of the atomic radius for that part of the atom below the nucleus in the triangle shown, and the same for that part above the single atom, we get 4.005 + 4.95 = 8.955 for the thickness of 2 vertical rows of atoms, or an average thickess of 1 row of 4.4775A.
If I use this to find the number of rows of atoms vertically (100000000/4.4775) and 100000000/4.95 for the number of atoms from left to right, I get 4.51 x 1014 atoms in B (which is closer)

I'll add an extra diagram to explain this rambling. This now part of the diagram attached to my previous post

### Re: Close-packed arrangement question

Hi Kingchemist,

I tried out your math - and I got something a little different though! When I calculated the vertical distance between atoms sin(60) = x/4.95 I got x = 4.286 Å. This actually works with the rest of the math that you did. When you take (100000000/4.286Å)  x (100000000/4.95) = 4.7 x 1014.

I guess you didn't quite need to find the average height of the rows? It made sense in your diagram, but that answer for some reason worked out perfectly. Or maybe that was just a really lucky calculation on my part? Hm.

Thank you so much for all the work you've done!! Your diagrams definitely helped me understand the concept.

### Re: Close-packed arrangement question

PS - I'm still getting 4.005 for the value of x
Sin(60) = 0.809
So x = 0.809 * 4.95 = 4.005

It must be my calculator?

### Re: Close-packed arrangement question

Is your calculator in the degree mode? I don't know if you would get that value if you were in radians, maybe that's what's happening?

I tried it on another calculator, just to see if it was mine - but I got the same thing on another one when in degrees.

OH I just tried it grads! I think that's the mode your calculator is in, because that's when I get 0.809 for sin(60).

### Re: Close-packed arrangement question

Thanks -
never used grads before. (Can't teach an old dog new tricks)