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20085ee2_1_EE2_HW3_sol

# 20085ee2_1_EE2_HW3_sol - Q1 In Fig.14 the range of k is...

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Unformatted text preview: Q1 In Fig.14 the range of k is equal to a a a k π π π 2 ) ( =-- = Δ (Page 29) Also from “Periodic Boundary Condition” (Page 12) we know that “k” can only be discrete values, i.e. x x x n L k π 2 = (17) (Page 12) Thus the distance between state to state on the k axis will be x x L dk π 2 = Thus Number of quantum states = a L L a dk k x x x x 2 / 2 / 2 2 2 = ⋅ = Δ ⋅ π π The factor 2 accounts for spin of electron. Now a= 10 Angstrom = 10 x 10-10 meter L= 1 micron = 1 x 10-6 meter Thus, number of state = 2000. Q2 E(k)= 1-coska = -(coska)+1 Thus if you want to draw the graph manually, First, draw y=coska. Then flip the graph to get y=–(coska) Final step is to shift the whole curve along positive y direction by 1 to get y=-(coska)+1. And it will look like a π a π- Q3 Total length of the 1-D crystal = L x The length of 1 unit cell of the 1-D crystal = a Thus, Number of unit cells in the crystal = a L x Now we have one electron per unit cell. Thus # of electrons in the crystal = a L x x 1 = a L x And we also know from Q1 that # of energy states in the crystal...
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20085ee2_1_EE2_HW3_sol - Q1 In Fig.14 the range of k is...

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