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## Length of primitive cell

1. The problem statement, all variables and given/known data
Knowing that sodium has a BCC structure, a density of 970 kg/m^3 and each atom has an atomic mass of 22.98 amu, determine the length of the primitive cell.

2. Relevant equations
Not sure.

3. The attempt at a solution
I calculated how many atoms of Na there is in 1m^3, namely around $2.55 \times 10 ^{28}$ atoms. But I have a huge problem. I know that each primitive cell contain 9 sodium atoms. The problem appears when I start with 1 primitive cell and I put another primitive cell close to the first one, I don't obtain 18 atoms but 9+9-4=14 atoms because both cells share 4 atoms.
I can imagine other configurations of primitive cells where adding another primitive cell only gives 1 more atom for example. So it all depends how in space the primitive cells are.
Thus I'm stuck at calculating how many atoms there are per primitive cells "in average" or something like that.
Has someone an idea on how to go further? Thanks in advance!
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 Recognitions: Homework Help The primitive cell is not the same as the elementary cell. The primitive cell is the smallest repeating parallelepiped, contains only one unit. The elementary cell shows the symmetry of the lattice, it is a cube with one atom in the centre and also atoms on the corners , which are shared between 8 cells in case of BCC structure. So the BCC cell contains 1+8/8=2 atoms. ehild

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 Quote by ehild The primitive cell is not the same as the elementary cell. The primitive cell is the smallest repeating parallelepiped, contains only one unit. The elementary cell shows the symmetry of the lattice, it is a cube with one atom in the centre and also atoms on the corners , which are shared between 8 cells in case of BCC structure. So the BCC cell contains 1+8/8=2 atoms. ehild
Ah ok thanks a lot. I'm trying to self study some solid state theory. If I translate litterally the question, it didn't state "length of primitive cell" but "net parameters" which I now assume they mean the length of the elementary cell. (and not sure what else they can ask for). According to wikipedia what they are asking for is "lattice constant". In all cases I think I know what I have to calculate, I'm just unable to do it.

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## Length of primitive cell

 Quote by fluidistic Ah ok thanks a lot. I'm trying to self study some solid state theory. If I translate litterally the question, it didn't state "length of primitive cell" but "net parameters" which I now assume they mean the length of the elementary cell. (and not sure what else they can ask for). According to wikipedia what they are asking for is "lattice constant". In all cases I think I know what I have to calculate, I'm just unable to do it.
They can ask the parameters of the primitive cell, but I also think that the lattice constant is the question, the length of one side of the cubic elementary cell.

ehild
 Recognitions: Gold Member Ok so basically my problem is that I know that there are about $2.55 \times 10 ^{28}$ atoms in 1 m^3 of sodium but I don't know for what shape of sodium (almost a linear chain or a macroscopic cube?) this number holds. Because as I stated/implied, if you give me let's say 1000 atoms of sodium, I could place them as several elementary cells close to each other so as to form a linear chain (with depth and large a, the lattice parameter) I'd get a volume V_0. While if I place those atoms as to form several elementary cells so as to form a large cube, I'd get a total volume V_1 such that V_1
 Recognitions: Homework Help Assume it as a large cube. Have you seen a macroscopic body in form of a chain of one atom wide? ehild

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 Quote by ehild Assume it as a large cube. Have you seen a macroscopic body in form of a chain of one atom wide? ehild
2 atoms wide actually. Well no.
But the cube wasn't an intuitive choice to me. I worked out the almost linear chain because it looks much easier to me.
For such shape, adding an elementary cell is equivalent to add 5 atoms in average. It's very close to 5 because only 2 elementary cells have 4 more atoms than all the other ones. For a macroscopic cubic shape I have no idea how to calculate how many atoms in average there are when I had 1 elementary cell. It's between 1 and 5 but not as close as 5 as with the case of the almost linear chain.
Using this data, I get $a \approx 5.81 \times 10^{-10}m$.
According to http://www.infoplease.com/periodictable.php?id=11 the order of magnitude is correct but instead of 5.81 I should have gotten something around 4.23.
Details of my algebra (I think it's worth showing because for the macroscopic cubic shape I should get a larger lattice constant):
In total I have $\frac{1}{5} \cdot 2.55 \times 10 ^{28}$ elementary cells. This makes a length for the chain of $L=5.1 \times 10 ^{27} a$. But L is also worth $\frac{1m^3}{a^2}$. Hence $a^3 = \frac{1}{5.1 \times 10 ^{27}} \approx 1.96 \times 10 ^{-28} m^3$ and thus $a\approx 5.81 \times 10 ^{-10}m$.
I find counter intuitive that the macroscopic shape has some influence on the lattice constant, i.e. on the distance between atoms. It seems required to keep the density constant, but that's pretty weird.
 Recognitions: Homework Help The average number of atoms in a cell change with size and shape because of the atoms at the surface of the crystal. In case when you have about 1028 atoms, the surface does not count. Moreover, the crystal is built up from atoms, not from cells. The atoms are arranged in a way that you can find a repeated unit. It is 2 atoms in case of a BCC structure. See http://departments.kings.edu/chemlab/animation/bcc.html ehild

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 Quote by ehild The average number of atoms in a cell change with size and shape because of the atoms at the surface of the crystal. In case when you have about 1028 atoms, the surface does not count. Moreover, the crystal is built up from atoms, not from cells. The atoms are arranged in a way that you can find a repeated unit. It is 2 atoms in case of a BCC structure. See http://departments.kings.edu/chemlab/animation/bcc.html ehild
Thank you very much for all the information.
Using 2 atoms per elementary cell I get a $\approx 4.28 \times 10 ^{-10}m$, very close to the value given in the website.

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 Quote by fluidistic Thank you very much for all the information. Using 2 atoms per elementary cell I get a $\approx 4.28 \times 10 ^{-10}m$, very close to the value given in the website.
The atomic radius of Na is 1.86˙10-10 m. Supposing closely packed spheres, so as the atoms touch each other, the diagonal of the elementary cell is 4 times the radius, so the side length is 1.86*4/√3=4.30˙10-10 m.

ehild
 You don't need any information besides what is given in the problem. And is not necessary to know the size and shape of the sample, as long as it is macroscopic (you can define density). You just use the definition of density: ρ=m/V but apply is for one conventional cell (BCC in this case). For a BCC cell, the volume is a^3 where a is the side of the cubic conventional cell. The same BCC cell contains two atoms so the mass is 2*22.98 amu 970kg/m^3=2*22.98amu/a^3 Solve for a and convert to desired units.
 Recognitions: Gold Member Thank you guys, I solved the problem as you can see in post 9. Nice way of solving ehild, though one would need the atomic radius of sodium (isn't given in the problem statement), but I understand the method. I didn't realize there was only 2 atoms per elementary cell, I thought there was 5 at first. As I said, using 2 atoms per cell I reached the desired result so thanks a lot.
 Recognitions: Gold Member Sorry to "bump" this thread guys but let's say I'm asked to find the Fermi energy of the sodium knowing the given data in the problem + that for each atom of sodium there's 1 free electron. My thoughts: Since the Fermi energy is the highest energy that electrons may have at T=0K, I assume I have to consider a 3 dimensional box with volume 1 m^3, calculate the energy levels of electrons and fill them (up to $2.55 \times 10 ^{28}$ electrons) according to Pauli exclusion's principle. Is the approach correct? If so, then $E_F=\frac{\hbar ^2 }{2m} \cdot 3 \pi ^2 \cdot \left ( \frac{2.55 \times 10 ^{28}}{1m^3} \right )^{2/3} \approx 1.5659 \times 10 ^{-18}J \approx 3.986 eV$. But according to hyperphysics it should be more like 3.24 eV. So I guess I made a mistake somewhere but I don't know where.
 It should be $(3\pi^{2})^{2/3}$ instead of $3\pi^{2}$

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 Quote by naftali It should be $(3\pi^{2})^{2/3}$ instead of $3\pi^{2}$
Oh, well spotted thanks! Actually this makes my result smaller by a factor of ~1/3. So I get $E_F \approx 1.29 eV$. The approach seems correct though?
 Using n=2.55 x10^28 (m^-3) I get E_F=3.2 eV. The value given in Kittel (Introduction to solid state physics) is 3.23 eV. Maybe you typed something wrong in your calculator.

 Quote by ehild The atomic radius of Na is 1.86˙10-10 m. Supposing closely packed spheres, so as the atoms touch each other, the diagonal of the elementary cell is 4 times the radius, so the side length is 1.86*4/√3=4.30˙10-10 m. ehild
This may just show that the "atomic radius" was estimated by dividing the nearest neighbor distance (measured experimentally) by two. So it is a little circular.