KaSap Electronic Materials_Chapter_4_problem_solutions

KaSap Electronic Materials_Chapter_4_problem_solutions -...

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Chapter 4 Solutions to Assigned Problems 4.4 The one s and three p orbitals hybridize to form 4 ψ hyb orbitals. In Sb there are 5 valence electrons. One hyb has two paired electrons and 3 hyb has 1 electron as shown in Figure 4Q4-1. In In there are 3 electrons so one hyb is empty. This empty hyb of In can overlap the full hyb of Sb. The overlapped orbital, the bonding orbital, has two paired electrons. This is a bond between In and Sb even though the electrons come from Sb (this type of bonding is called dative bonding ). It is a bond because the electrons in the overlapped orbital are shared by both Sb and In. The other 3 hyb of Sb can overlap 3 hyb of neighboring In to form "normal bonds". Repeating this in three dimensions generates the InSb crystal where each atom bonds to four neighboring atoms as shown. As all the bonding orbitals are full, the valence band formed from these orbitals is also full. The crystal structure is reminiscent of that of Si, as all the valence electrons are in bonds. Since it is similar to Si, InSb is a semiconductor. Sb In In atom (Valency III) Sb atom (Valency V) Sb In hyb orbitals Sb ion core (+5 e ) Valence electron hyb orbitals In ion core (+3 e ) Valence electron Sb Sb Sb Sb Sb Sb Sb In In In In In In In 4.6 For a two dimensional electron gas confined within a square region of sides a we have: () 2 2 2 1 2 2 8 n n a m h E e + = Only positive n 1 and n 2 are allowed. Each n 1 and n 2 combination is an orbital state. Define a new variable n as: n 2 = n 1 2 + n 2 2 substitute: 2 2 2 8 n a m h E e = Let us consider how many states there are with energies less than E . E corresponds to n n .
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2 2 2 8 n a m h E e = 2 2 8 h E m a n e = n 1 n 2 n 1 n 2 n 1 2 + n 2 2 = n ' 2 n 1 = 1 n 2 = 3 0 1 2 3 4 5 12 3 4 56 n 1 = 2, n 2 = 2 Figure 4Q6-1: Each state, or electron wavefunction in the crystal, can be represented by a box at n 1 , n 2 . Consider Figure 4Q6-1. All states within the quarter arc defined by n have E < E . The area of this quarter arc is the total number of orbital states. The total number of states, S , including spin is twice as many, = = 2 2 2 2 8 4 1 2 4 1 2 h E m a n S e π 2 2 4 h E m a S e = The density of states g is defined as the number of states per unit area per unit energy. 2 2 2 2 2 4 4 1 1 h m h m a a E d dS a e e ππ = = = g Thus, for a two dimensional gas, the density of states is constant.
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4.9 Since Cu is in group I, its valency is also 1. The electron concentration n is then the atomic concentration multiplied by the group number, or: () ( ) 3 28 3 3 3 1 23 m 10 490 . 8 kg/mol 10 55 . 63 kg/m 10 96 . 8 mol 10 022 . 6 1 ) Valency ( × = × × × = = at A M D N n Using Equation 4.22:
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KaSap Electronic Materials_Chapter_4_problem_solutions -...

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