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Please go through question c) in detail.

I have no idea how to find bravais lattice vectors a1 a2 b1 and b2.

We assume a tight-binding model in which the electron hops between neighboring atoms. The lattice structure is as shown in Fig. 1. We denote the spacing between neighboring atoms by \( a \). Figure 1: The solid lines indicate the crystal structure of graphene. The basis has two atoms, labeled \( A \) and \( B \). The Bravais lattice (consider, e.g, the lattice formed by the \( A \) atoms shown by dashed lines) is triangular with a Bravais lattice spacing \( 2 \times \sin 60^{\circ} \times a=\sqrt{3} a \), where \( a \) is the spacing between neighboring atoms. a. There are two atoms per unit cell so \( \quad \) will be filled. b. The Bravais lattice is the same as the lattice formed by all the \( A \) atoms, say. As shown in Fig. 1 this is a with lattice spacing \( \sqrt{3} a \). c. From Fig. 1, we see that two basis vectors of the Bravais lattice are \[ \mathbf{a}_{1}=\sqrt{3} a(1,0), \quad \mathbf{a}_{2}=a(\sqrt{3} / 2,3 / 2) . \] The Bravais lattices, \( \mathbf{b}_{1}, \mathbf{b}_{2} \), are defined such that \[ \mathbf{b} \cdot \mathbf{a}_{j}=2 \pi \delta_{i j} . \] To determine them, either consider the formulae discussed in class for three dimensions, with the third basis vector \( a_{3}=\hat{z} \), a unit vector in the \( z \) direction, or just figure it out. The answer is \[ \begin{array}{l} \mathbf{b}_{1}=2 \pi \frac{\hat{\mathbf{z}} \times \mathbf{a}_{2}}{\left|\mathbf{a}_{1} \times \mathbf{a}_{2}\right|}=\frac{2 \pi}{3} \frac{1}{a}(\sqrt{3},-1), \\ \mathbf{b}_{2}=2 \pi \frac{\hat{\mathbf{z}} \times \mathbf{a}_{1}}{\left|\mathbf{a}_{1} \times \mathbf{a}_{2}\right|}=\frac{4 \pi}{3} \frac{1}{a}(0,1) . \end{array} \]

Here given the Bravais lattice is triangular with the spacing of 3a So the x-component of nearest solid atom is 3a and the y-component is 0 So the loc