Question

The primitive translation vectors of a crystal lattice are given by :
a₁=(√(3)a/2)i + (a/2)j
a₂=-(√(3)a/2)i + (a/2)j
a₃= ck
To calculate the volume of the primitive cell and obtain the primitive translation vectors of the reciprocal lattice.

Answer 1

Volume of primitive cell is given by

$V=a1.(a2\times a3)$

$$a2\times a3=\def\arraystretch{1.5} \begin{array}{c:c:c} a & b & c \\ \hline d & e & f \\ \hdashline g & h & i \end{array}$$

Where a=i, b=j, c=k

From vector a2 and a3, we have

d= -(√(3)a/2), e=(a/2), f=0

g=0, h=0, i=c

Now solve for determinants, we get cross product of a2 and a3=a(ei-fh)-b(di-fg)+c(dh-eg)

=a(ei)-b(di)

=i(a/2xc)+j(√(3)a/2xc)

Volume of primitive cell

V=a1.(a2xa3)={(√(3)a/2)i + (a/2)j}.{(a/2xc)i+(√(3)a/2xc)j}

V=a²c√3/2

And

the primitive translation vectors of the reciprocal lattice is given by

$b1=(2\pi/a√3)i+(2\pi/a)j$

$b2=(-2\pi/a√3)i+(2\pi/a)j$

$b3=(2\pi/c)k$