An orthorhombic cell with a =0.82nm, b= 0.94nm and c= 0.75nm.calculate the separation of the plane using millers indices given below.
1.if the separation of (a) (133) and (b) (399) planes
2.if the separation of (a) (122) (b) (244) planes
3. What is your conclusion on 1 and 2
To find the separation of planes using Miller indices, we can use the formula:
d(hkl) = a / √(h^2/a^2 + k^2/b^2 + l^2/c^2)
1.
a) For (133) plane:
h = 1, k = 3, l = 3
d(133) = 0.82 / √(1/0.82^2 + 9/0.94^2 + 9/0.75^2)
d(133) = 0.82 / √(1/0.6724 + 9/0.8836 + 9/0.5625)
d(133) = 0.82 / √(1.4881 + 9.6051 + 15.9844)
d(133) = 0.82 / √(26.0776)
d(133) = 0.82 / 5.107
d(133) ≈ 0.1608 nm
b) For (399) plane:
h = 3, k = 9, l = 9
d(399) = 0.82 / √(9/0.82^2 + 81/0.94^2 + 81/0.75^2)
Following the same steps as above, we get:
d(399) ≈ 0.1260 nm
2.
a) For (122) plane:
h = 1, k = 2, l = 2
d(122) = 0.82 / √(1/0.82^2 + 4/0.94^2 + 4/0.75^2)
Following the same steps as above, we get:
d(122) ≈ 0.1395 nm
b) For (244) plane:
h = 2, k = 4, l = 4
d(244) = 0.82 / √(4/0.82^2 + 16/0.94^2 + 16/0.75^2)
Following the same steps as above, we get:
d(244) ≈ 0.1620 nm
3.
Comparing the results from 1 and 2, we can see that the planes in (133) and (399) have a smaller separation distance compared to the planes in (122) and (244). This indicates that the planes in 133 and 399 are closer together in the crystal lattice compared to planes in 122 and 244. This suggests that the crystal structure has a closer packing in the former case.