Answer the following for a primitive cubic unit cell. Answers should be numerical, set r = 8.0
-edge in terms of r, the lattice pt radius:___
-face diagonal in terms of r, the lattice pt radius:___
-body diagonal in terms of r, the lattice pt radius:___
Answer the following for a face centered unit cell.Answers should be numerical, set r = 6.0.
-edge in terms of r, the lattice pt radius:___
-face diagonal in terms of r, the lattice pt radius:___
-body diagonal in terms of r, the lattice pt radius
Please help.I will truely appreciate it..
Do you mean by help that you want us to work the problems for you. Tell us what you don't understand and perhaps we can get you started. You might also explain what pt stands for.
I know that the edge for the first one is 2r so in terms of r the lattice point (pt) would = 16.
To find the face diagonal using the Pythagorean theorem it would be (2r)^2+(2r)^2= (face diagonal)^2.. But every time I would plug it in, its incorrect.
And for the body diagonal, I'm lost...
For r=6.0
6(2)= 12
12(squ root 2) = 16.97056
12(squ root 3) = 20.7846
And for r=8.0, do the same steps.
Hope that helps!
the 3rd part its
sqrt(a^2+b^2+c^2)
To solve these questions, we need to understand the relationship between the edge length, face diagonal, and body diagonal of a unit cell and the radius of the lattice points.
For a primitive cubic unit cell:
1. Edge in terms of r, the lattice pt radius:
In a primitive cubic unit cell, the edge length is twice the radius (r) of the lattice points. So, the edge length can be calculated by multiplying 2 with r.
Edge length = 2 * r
Substituting r = 8.0, we have:
Edge length = 2 * 8.0 = 16.0
2. Face diagonal in terms of r, the lattice pt radius:
The face diagonal of a primitive cubic unit cell is calculated by multiplying the edge length by the square root of 2. So, we can find the face diagonal by multiplying the edge length calculated in the previous step by √2.
Face diagonal = Edge length * √2
Substituting the previously calculated edge length:
Face diagonal = 16.0 * √2 = 22.627
3. Body diagonal in terms of r, the lattice pt radius:
The body diagonal of a primitive cubic unit cell is calculated by multiplying the edge length by the square root of 3. So, we can find the body diagonal by multiplying the edge length by √3.
Body diagonal = Edge length * √3
Substituting the previously calculated edge length:
Body diagonal = 16.0 * √3 = 27.712
Now, let's solve the questions for a face-centered cubic unit cell:
1. Edge in terms of r, the lattice pt radius:
In a face-centered cubic unit cell, each edge length is four times the radius (r) of the lattice points. So, the edge length can be calculated by multiplying 4 with r.
Edge length = 4 * r
Substituting r = 6.0, we have:
Edge length = 4 * 6.0 = 24.0
2. Face diagonal in terms of r, the lattice pt radius:
The face diagonal of a face-centered cubic unit cell is calculated by multiplying the edge length by the square root of 2. So, we can find the face diagonal by multiplying the edge length calculated in the previous step by √2.
Face diagonal = Edge length * √2
Substituting the previously calculated edge length:
Face diagonal = 24.0 * √2 = 33.941
3. Body diagonal in terms of r, the lattice pt radius:
The body diagonal of a face-centered cubic unit cell is calculated by multiplying the edge length by the square root of 4. So, we can find the body diagonal by multiplying the edge length by √4.
Body diagonal = Edge length * √4
Substituting the previously calculated edge length:
Body diagonal = 24.0 * √4 = 48.0