Figuring out Bravais lattice from primitive basis vectors

[spaghetti3451]

Homework Statement

Given that the primitive basis vectors of a lattice are $\mathbf{a} = \frac{a}{2}(\mathbf{i}+\mathbf{j})$, $\mathbf{b} = \frac{a}{2}(\mathbf{j}+\mathbf{k})$, $\mathbf{c} = \frac{a}{2}(\mathbf{k}+\mathbf{i})$, where $\mathbf{i}$, $\mathbf{j}$, and $\mathbf{k}$ are the usual three unit vectors along cartesian coordinates, what is the Bravais lattice?

Homework Equations

The Attempt at a Solution

There are seven different crystal systems and fourteen different Bravais lattices.

The common length of the primitive unit cell is $a$, so the crystal system is either cubic or trigonal (rhombohedral).

Furthermore, the basis vectors are oriented at $90°$ to each other, so the crystal system must be cubic.

Finally, the Bravais lattice is face-centred cubic, because, if the basis vectors originate from one corner of the primitive unit cell, then they point to lattice sites at the centre of three adjacent (to the corner) faces of the primitive unit cell.

Is my answer correct?

 

[unscientific]

It's an FCC. Why? (Hint: Draw out the vectors)

Answer the following questions:
1. How many lattice points does FCC have, and what are their locations?
2. Using the info above, write out the lattice vectors.

 

[spaghetti3451]

In my mind, I translated from the origin (at the corner of one primitive unit cell) by the the lattice vector $\mathbf{R} = n_{1} \mathbf{a} + n_{2} \mathbf{b} + n_{3} \mathbf{c}$, where I used $(n_{1}, n_{2}, n_{3}) = (0,0,1),$ $(n_{1}, n_{2}, n_{3}) = (0,1,0),$ and $(n_{1}, n_{2}, n_{3}) = (1,0,0)$. Each time, I found that the system is invariant under translation.

 

[unscientific]

You are over-complicating things.

1. All three lattice vectors have the same length - What does this tell you about its structure? (Cubic, orthorhombic, tetragonal ...)

2. Using your answer to part 1, draw a unit cell.

3. Now draw 5 more neighbouring unit cells.

4. What is the bravais lattice type?

 

[spaghetti3451]

1. FCC has four lattice points. They are located one at the corner of a chosen unit cell, and the other three at the centres of each of the three faces which intersect at the corner.

2. The lattice vectors are $\mathbf{R} = n_{1} \mathbf{a} + n_{2} \mathbf{b} + n_{3} \mathbf{c}$, where
$(n_{1}, n_{2}, n_{3}) = (0,0,0)$, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.

 

[unscientific]

1. FCC has four lattice points. They are located one at the corner of a chosen unit cell, and the other three at the centres of each of the three faces which intersect at the corner.

Yes. Good. Now using the corner as (0,0,0) how do you get to the other three lattice points? Hence write down the lattice vectors.

2. The lattice vectors are $\mathbf{R} = n_{1} \mathbf{a} + n_{2} \mathbf{b} + n_{3} \mathbf{c}$, where
$(n_{1}, n_{2}, n_{3}) = (0,0,0)$, $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$.

Wrong. You just told me the lattice points. What are the lattice vectors?

 

[spaghetti3451]

The lattice vectors are $\mathbf{R} = \mathbf{0}$, $\mathbf{R} = \mathbf{a}$, $\mathbf{R} = \mathbf{b}$, and $\mathbf{R} = \mathbf{c}$.

 

[unscientific]

The lattice vectors are $\mathbf{R} = \mathbf{0}$, $\mathbf{R} = \mathbf{a}$, $\mathbf{R} = \mathbf{b}$, and $\mathbf{R} = \mathbf{c}$.

No. Using the corner as (0,0,0) how do you get to the other three lattice points? Hence write down the lattice vectors.

 

[spaghetti3451]

In that case, the three lattice vectors are $\mathbf{R} = \mathbf{a}$, $\mathbf{R} = \mathbf{b}$, and $\mathbf{R} = \mathbf{c}$.

 

[unscientific]

In that case, the three lattice vectors are $\mathbf{R} = \mathbf{a}$, $\mathbf{R} = \mathbf{b}$, and $\mathbf{R} = \mathbf{c}$.

Oops. That's right. To ensure you get all the marks, draw out the cubes to show that the repeating motif is indeed an FCC.