What is the lattice constant of a zincblende structure with two types of atoms?

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The discussion focuses on calculating the lattice constant, volume density of valence electrons, and surface density of B atoms in a zincblende structure with two types of atoms. The lattice constant is determined to be approximately 4.73 Å using the formula involving the radii of the atoms. The volume density of valence electrons is calculated as 7.56 x 10^22 cm^-3, while the surface density of B atoms on the (110) plane is found to be around 6.32 x 10^14 atoms/cm^2. Clarifications on the formulas used for these calculations are also discussed, emphasizing the correct units and relationships between the parameters. Accurate calculations are critical for understanding the properties of materials in this structure.
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Homework Statement


A material contains two types of atoms (A and B) in a zincblende structure. Atom A is group V with a hard sphere radius of 2.5 Å. Atom B is group III with a hard sphere radius of 1.6 Å.
(a) What is the lattice constant of this material assuming that nearest neighbour atoms touch? (working on this one right now)
(b) What is the volume density of valence electrons in this material?
(c) What is the surface density of B atoms on the (110) plane in this material?

Homework Equations

The Attempt at a Solution



a) [/B]Since it's a face centered structure it should have 8X 1/8 + 6* 1/2 atoms, so 4
Is it correct to say ( r1+r2 ) *4 = my lattice constant for A?
 
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orangeincup said:

Homework Statement


A material contains two types of atoms (A and B) in a zincblende structure. Atom A is group V with a hard sphere radius of 2.5 Å. Atom B is group III with a hard sphere radius of 1.6 Å.
(a) What is the lattice constant of this material assuming that nearest neighbour atoms touch? (working on this one right now)
(b) What is the volume density of valence electrons in this material?
(c) What is the surface density of B atoms on the (110) plane in this material?

Homework Equations

The Attempt at a Solution



a) [/B]Since it's a face centered structure it should have 8X 1/8 + 6* 1/2 atoms, so 4
Is it correct to say ( r1+r2 ) *4 = my lattice constant for A?

No it is not right. If there are 4 units in a cube, how can the side of the cube 4 times the interatomic distance?
this helps http://www.ehow.com/how_8721263_determine-lattice-parameter-zincblende.html
 
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a) (4/3)^1/2*(1.6+2.5) = 4.73 A

b) Volume density of valence electrons is 8/((4.73*10^-8)^3) = 7.56*10^22 cm^3

c) 4/(sqrt2(4.73*10^-8)^2) = 1.26*10^15 atoms/cm^3
 
orangeincup said:
a) (4/3)^1/2*(1.6+2.5) = 4.73 A

b) Volume density of valence electrons is 8/((4.73*10^-8)^3) = 7.56*10^22 cm^3
How many valence electrons are there in one AB unit? How many units are there in the unit cell?
orangeincup said:
c) 4/(sqrt2(4.73*10^-8)^2) = 1.26*10^15 atoms/cm^3
The surface density is asked, it has atoms/cm^2 unit.
How did you get that formula?
 
ehild said:
How many valence electrons are there in one AB unit? How many units are there in the unit cell?

The surface density is asked, it has atoms/cm^2 unit.
How did you get that formula?
4 *(1/4) + 2*(1/2) + 2*1=4?
ehild said:
How many valence electrons are there in one AB unit? How many units are there in the unit cell?

The surface density is asked, it has atoms/cm^2 unit.
How did you get that formula?

I confused it with a volume density formula.

4/(4.73*10^-8)^3) = 3.78*10^22 cm^3 = volume density

sqrt2/((4.73*10^-8)^2)=6.32*10^14 surface density
 
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