What is the mathematical relationship between the atomic radius (r) and the lattice parameter (a0) in the BCC structure? a. b. c
.
2 answers:
Answer: Lattice parameter, a = (4R)/(√3)
Step-by-step explanation:
The typical arrangement of atoms in a unit cell of BCC is shown in the first attachment.
The second attachment shows how to obtain the value of the diagonal of the base of the unit cell.
If the diagonal of the base of the unit cell = x
(a^2) + (a^2) = (x^2)
x = a(√2)
Then, diagonal across the unit cell (a cube) makes a right angled triangle with one side of the unit cell & the diagonal on the base of the unit cell.
Let the diagonal across the cube be y
Pythagoras theorem,
(a^2) + ((a(√2))^2) = (y^2)
(a^2) + 2(a^2) = (y^2) = 3(a^2)
y = a√3
But the diagonal through the cube = 4R (evident from the image in the first attachment)
y = 4R = a√3
a = (4R)/(√3)
QED!!!
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Answer:
The other endpoint is (-33, 17)
Step-by-step explanation:
The rule of the mid-point of a segment whose endpoints are
(,
) and (
,
) is
In our question
∵ The coordinates of the endpoints of a segment are (-15, 13) and (x, y)
∴ = -15 and
= x
∴ = 13 and
= y
∵ The coordinates of the mid-point of this segment are (-24, 15)
∴ = -24 and
= 15
→ Use the rule of the mid-point to find x and y
∵
→ Multiply both sides by 2
∴ -48 = -15 + x
→ Add 15 to both sides
∴ -33 = x
∵
→ Multiply both sides by 2
∴ 30 = 13 + y
→ Subtract 13 from both sides
∴ 17 = y
∴ The other endpoint is (-33, 17)