Os vectores primitivos de uma rede hexagonal são:
\vec{a}_{1}=a \frac{\sqrt{3}}{2} \hat{\mathbf{x}}+a \frac{1}{2} \hat{\mathbf{y}}, \vec{a}_{2}=a \frac{\sqrt{3}}{2} \hat{\mathbf{x}}+a \frac{1}{2} \hat{\mathbf{y}}, \vec{a}_{3}=c \hat{\mathbf{z}}.
(a) Mostre que o volume da célula primitiva é a^{2} c \frac{\sqrt{3}}{2}.
(b) Calcule os vectores primitivos da rede recíproca.

Passo 1

O enunciado dá os vectores primitivos de uma rede hexagonal
\vec{a}_{1}=a\,\frac{\sqrt{3}}{2}\,\hat{\mathbf{x}}+a\,\frac{1}{2}\,\hat{\mathbf{y}},\quad \vec{a}_{2}=a\bigl(-\frac{\sqrt{3}}{2}\,\hat{\mathbf{x}}+\frac{1}{2}\,\hat{\mathbf{y}}\bigr),\quad \vec{a}_{3}=c\,\hat{\mathbf{z}}
e pede (a) o volume da célula primitiva e (b) os vectores primitivos da rede recíproca.
Para (a) vamos usar o produto misto
V=\vec{a}_{1}\cdot\bigl(\vec{a}_{2}\times\vec{a}_{3}\bigr)
e para (b) usaremos a definição
\vec{b}_{i}=2\pi\,\frac{\vec{a}_{j}\times\vec{a}_{k}} {\vec{a}_{1}\cdot(\vec{a}_{2}\times\vec{a}_{3})},
com (i,j,k) ciclo de (1,2,3).

Passo 2

Calculemos primeiro o produto vetorial \vec{a}_{2}\times\vec{a}_{3}. Como
\vec{a}_{2}=a\Bigl(-\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{x}}+\tfrac{1}{2}\,\hat{\mathbf{y}}\Bigr), \quad \vec{a}_{3}=c\,\hat{\mathbf{z}},
temos
\vec{a}_{2}\times\vec{a}_{3} = a\,c\Bigl(-\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{x}}+\tfrac{1}{2}\,\hat{\mathbf{y}}\Bigr)\times\hat{\mathbf{z}} = a\,c\Bigl(\tfrac{1}{2}\,\hat{\mathbf{x}}+\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}\Bigr),
pois \hat{\mathbf{x}}\times\hat{\mathbf{z}}=-\hat{\mathbf{y}} e \hat{\mathbf{y}}\times\hat{\mathbf{z}}=\hat{\mathbf{x}}.

Passo 3

Agora fazemos o produto escalar \vec{a}_{1}\cdot(\vec{a}_{2}\times\vec{a}_{3}). Sabendo
\vec{a}_{1}=a\Bigl(\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{x}}+\tfrac{1}{2}\,\hat{\mathbf{y}}\Bigr), \quad \vec{a}_{2}\times\vec{a}_{3}=a\,c\Bigl(\tfrac{1}{2}\,\hat{\mathbf{x}}+\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}\Bigr),
temos
\vec{a}_{1}\cdot(\vec{a}_{2}\times\vec{a}_{3}) = a^2\,c\Bigl(\tfrac{\sqrt{3}}{2}\cdot\tfrac{1}{2} + \tfrac{1}{2}\cdot\tfrac{\sqrt{3}}{2}\Bigr) = a^2\,c\,\frac{\sqrt{3}}{2}.

Passo 4

Concluímos que o volume da célula primitiva é
V=a^{2}\,c\,\frac{\sqrt{3}}{2}.

Passo 5

Para obter os vectores da rede recíproca usamos
\vec{b}_{1}=2\pi\,\frac{\vec{a}_{2}\times\vec{a}_{3}}{V},\quad \vec{b}_{2}=2\pi\,\frac{\vec{a}_{3}\times\vec{a}_{1}}{V},\quad \vec{b}_{3}=2\pi\,\frac{\vec{a}_{1}\times\vec{a}_{2}}{V}.

Passo 6

Já temos \vec{a}_{2}\times\vec{a}_{3}=a\,c\bigl(\tfrac{1}{2}\,\hat{\mathbf{x}}+\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}\bigr), logo
\vec{b}_{1} =2\pi\,\frac{a\,c\bigl(\tfrac{1}{2}\,\hat{\mathbf{x}}+\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}\bigr)} {a^{2}c\,\tfrac{\sqrt{3}}{2}} = \frac{4\pi}{a\sqrt{3}}\Bigl(\tfrac{1}{2}\,\hat{\mathbf{x}}+\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}\Bigr) = \frac{2\pi}{a\sqrt{3}}\,\hat{\mathbf{x}}+\frac{2\pi}{a}\,\hat{\mathbf{y}}.

Passo 7

Calculamos \vec{a}_{3}\times\vec{a}_{1}. De modo análogo a antes,
\vec{a}_{3}\times\vec{a}_{1} = a\,c\Bigl(\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}-\tfrac{1}{2}\,\hat{\mathbf{x}}\Bigr),
portanto
\vec{b}_{2} =2\pi\,\frac{a\,c\bigl(\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}-\tfrac{1}{2}\,\hat{\mathbf{x}}\bigr)} {a^{2}c\,\tfrac{\sqrt{3}}{2}} = \frac{4\pi}{a\sqrt{3}}\Bigl(\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{y}}-\tfrac{1}{2}\,\hat{\mathbf{x}}\Bigr) = -\frac{2\pi}{a\sqrt{3}}\,\hat{\mathbf{x}}+\frac{2\pi}{a}\,\hat{\mathbf{y}}.

Passo 8

Por fim,
\vec{a}_{1}\times\vec{a}_{2} =a^{2}\,\frac{\sqrt{3}}{2}\,\hat{\mathbf{z}} \quad\Longrightarrow\quad \vec{b}_{3} =2\pi\,\frac{a^{2}\,\tfrac{\sqrt{3}}{2}\,\hat{\mathbf{z}}} {a^{2}c\,\tfrac{\sqrt{3}}{2}} = \frac{2\pi}{c}\,\hat{\mathbf{z}}.

Resposta

O volume da célula primitiva é
V=a^{2}\,c\,\frac{\sqrt{3}}{2}.
Os vectores da rede recíproca são
\vec{b}_{1} =\frac{2\pi}{a\sqrt{3}}\,\hat{\mathbf{x}}+\frac{2\pi}{a}\,\hat{\mathbf{y}},\quad \vec{b}_{2} =-\frac{2\pi}{a\sqrt{3}}\,\hat{\mathbf{x}}+\frac{2\pi}{a}\,\hat{\mathbf{y}},\quad \vec{b}_{3} =\frac{2\pi}{c}\,\hat{\mathbf{z}}.

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