Let's solve **Question 1(c): What is the velocity in the center of the band in the M direction?** --- ### **Step 1: Write the energy and find the velocity expression** The energy dispersion is: \[ E = E_1 + A \left[ 1 - \cos(k_x a) \right]/2 + A \left[ 1 - \cos(k_y a) \right]/2 \] Given \( E_1 = \), so: \[ E = \frac{A}{2}\left[ 2 - \cos(k_x a) - \cos(k_y a) \right] \] The velocity is given by: \[ v = \frac{1}{\hbar} \nabla_k E \] --- **Explanation:** The velocity of an electron in a band is the gradient of energy with respect to \( \vec{k} \), divided by \( \hbar \). --- ### **Step 2: Evaluate at the center (Γ point) in the M direction** At the **center of the band** (Γ point), \( k_x = , k_y = \): - \( \cos() = 1 \) - So, \( E(,) = \frac{A}{2}[2 - 1 - 1] = \) Velocity components: \[ v_x = \frac{1}{\hbar} \frac{\partial E}{\partial k_x} = \frac{1}{\hbar} \frac{A}{2} \cdot a \sin(k_x a) \] \[ v_y = \frac{1}{\hbar} \frac{\partial E}{\partial k_y} = \frac{1}{\hbar} \frac{A}{2} \cdot a \sin(k_y a) \] At \( k_x = , k_y = \), \( \sin() = \): - \( v_x = \) - \( v_y = \) --- **Explanation:** At the center of the band (Γ), both velocity components are zero because the slope of the band is zero (extremum point). --- ### **Final Answer** **The velocity in the center of the band in the M direction is \( \boxed{} \).**

# Solution to Question 1 ## Part 1: Velocity in the Center of the Band in the M Direction ### **Step 1: Write the Energy and Find the Velocity Expression** The energy dispersion relation is given by: \[ E = E_1 + A \left[ 1 - \cos(k_x a) \right]/2 + A \left[ 1 - \cos(k_y a) \right]/2 \] Given that \( E_1 = 0 \), we can simplify this to: \[ E = \frac{A}{2}\left[ 2 - \cos(k_x a) - \cos(k_y a) \right] \] The velocity \( \vec{v} \) is derived from the energy dispersion as: \[ \vec{v} = \frac{1}{\hbar} \nabla_k E \] ### **Explanation:** The velocity of an electron in a band is determined by the gradient of the energy with respect to the wave vector \( \vec{k} \), normalized by \( \hbar \). --- ### **Step 2: Evaluate at the Center (M Point) in the M Direction** At the **M point**, we need to determine the wave vector values. Assuming M corresponds to specific values \( k_x \) and \( k_y \): - \( k_x = \frac{\pi}{a} \) - \( k_y = 0 \) Substituting these values into the energy equation: \[ E\left(\frac{\pi}{a}, 0\right) = \frac{A}{2}\left[ 2 - \cos\left(\frac{\pi}{a} a\right) - \cos(0) \right] \] This simplifies to: \[ E\left(\frac{\pi}{a}, 0\right) = \frac{A}{2}\left[ 2 - (-1) - 1 \right] = \frac{A}{2}(2 + 1 - 1) = \frac{A}{2}(2) = A \] Now, the velocity components are computed as follows: \[ v_x = \frac{1}{\hbar} \frac{\partial E}{\partial k_x} = \frac{1}{\hbar} \left(\frac{A}{2} a \sin(k_x a)\right) \] \[ v_y = \frac{1}{\hbar} \frac{\partial E}{\partial k_y} = \frac{1}{\hbar} \left(\frac{A}{2} a \sin(k_y a)\right) \] At the M point, substituting \( k_x = \frac{\pi}{a} \) and \( k_y = 0 \): - \( \sin(k_x a) = \sin(\pi) = 0 \) - \( \sin(k_y a) = \sin(0) = 0 \) Thus, we find: \[ v_x = \frac{1}{\hbar} \cdot 0 = 0 \] \[ v_y = \frac{1}{\hbar} \cdot 0 = 0 \] ### **Explanation:** At the M point, both velocity components are zero because the energy has a maximum value, indicating no net motion of electrons. --- ### **Final Answer** **The velocity in the center of the band in the M direction is \( \boxed{0} \).** --- # Solution to Question 2 ## Part 1: Direction of Current with Electric Field ### **Step 1: Determine the Direction of Current** When an electric field \( \vec{E} \) is applied along the \( +x \) direction, the current \( \vec{J} \) will flow in the same direction as the electric field. ### **Explanation:** Current flow is always in the direction of the applied electric field for a positive charge carrier. --- ### **Final Answer** The direction of the current when an electric field is applied along the \( +x \) direction is \( \boxed{+x} \). --- ## Part 2: Angle of Electric Field for Y-Direction Current ### **Step 1: Determine the Required Angle** To make the current flow in the \( +y \) direction while an electric field is applied, we need to find the angle \( \theta \) such that the effective component of the electric field in the \( y \) direction equals that in the \( x \) direction. Assuming the electric field \( \vec{E} \) makes an angle \( \theta \) with the \( x \)-axis, we can express the components: - \( E_x = E \cos(\theta) \) - \( E_y = E \sin(\theta) \) For the current to flow in the \( y \)-direction, \( E_x \) must equal zero, which implies: \[ \cos(\theta) = 0 \implies \theta = 90^\circ \] ### **Explanation:** To achieve current flow in the \( y \)-direction, the electric field must be directed entirely along the \( y \)-axis, which corresponds to an angle of \( 90^\circ \) from the \( x \)-axis. --- ### **Final Answer** The angle of the electric field needed to make the current flow in the \( y \)-direction is \( \boxed{90^\circ} \).