ch2

ch2 - SOLUTIONS CHAPTER 2 2.1 Of the two materials whose...

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SOLUTIONS CHAPTER 2 2.1. Of the two materials whose E-K diagrams are shown in Figure P2.1, which will have the lowest effective mass for electrons? Which will have the lowest effective mass for holes? Explain how you arrived at your conclusion. E E C E V E K E C E V K The effective mass is proportional to the reciprocal of the curvature of the band. The band with the greatest curvature therefore has the smallest effective mass. The E-K diagram on the right has a higher curvature in the conduction band, thus the effective mass for electrons in the conduction band is smaller than for the material on the left. For holes, the material on the left has higher curvature and thus the effective mass for holes is smaller for the material on the left. 2.2. Assume a material has an E-K diagram given by E K ( 29 c o n d u c t i o n = E C + E 1 s i n 2 ( K a ) E ( K ) v a l e n c e = E V - E 2 s i n 2 ( K a ) Let a=0.5nm , E 1 =5eV, and E 2 =4eV. a) Sketch the E-K diagram for the first Brillouin zone. Label the axes completely. Anderson & Anderson 1 2/15/04 Solutions Chapter 2
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E K π / a - π / a E 5 e V 4 e V C V E b) What is the effective mass for an electron near the bottom of the conduction band? We use the expression for the effective mass : m * = h 2 d 2 E d K 2 - 1 K = 0 Taking the derivatives: m * = h 2 d 2 d K 2 E C + E 1 s i n 2 ( K a ) ( 29 - 1 K = 0 = h 2 E 1 d d K 2 a s i n ( K a ) c o s ( K a ) ( 29 - 1 K = 0 = h 2 E 1 2 a - a s i n 2 ( K a ) + a c o s 2 ( K a ) ( 29 [ ] - 1 K = 0 = h 2 2 E 1 a 2 = 6 . 6 3 × 1 0 - 3 4 J s 2 π 2 2 5 e V 1 . 6 × 1 0 - 1 9 J / e V ( 29 5 × 1 0 - 1 0 m ( 29 2 = 2 . 8 × 1 0 - 3 2 k g = 0 . 0 3 m 0 c) What is the effective mass for holes near the top of the valence band? Anderson & Anderson 2 2/15/04 Solutions Chapter 2
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m * = h 2 d 2 d K 2 E V - E 2 s i n 2 ( K a ) ( 29 - 1 K = 0 = h 2 E 2 d d K - 2 a s i n ( K a ) c o s ( K a ) ( 29 - 1 K = 0 = h 2 E 2 2 a + a s i n 2 ( K a ) - a c o s 2 ( K a ) ( 29 [ ] - 1 K = 0 = - h 2 2 E 2 a 2 = - 6 . 6 3 × 1 0 - 3 4 J s 2 π 2 2 4 e V 1 . 6 × 1 0 - 1 9 J / e V ( 29 5 × 1 0 - 1 0 m ( 29 2 = - 3 . 5 × 1 0 - 3 2 k g = - 0 . 0 4 m 0 This is the effective mass for electrons, which is negative as expected. The effective mass for holes is positive, or m h * = + 0 . 0 4 m 0 2.3. Explain in your own words why the effective mass of an electron is different than its mass in vacuum. The effective mass includes the effects of the periodic potential of the crystal, so that Newtonian or classical mechanics can be used. If a given external force is applied, the effect on the electron is the sum of the external force plus all the internal forces due to the varying potential energy in the crystal. The effective mass can be used to predict the behavior of an electron in the crystal in reaction to the external force. 2.4. For the material of Figure P2.2, there are two different effective masses.
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