Plot the density of states for the lowest three bands of the one dimensional KronigPenney model discussed in Section 1 2, for h 2 2m 1, a 1, b 0, and U 0 b 1 You will need to solve for E k numerically, as we had to do for E(k) in the KronigPenney model ...

With 2m 1 a 1 b0 and Vob 1 Eq 126 reduces to 1 2K sin K cos K cos k because we can drop E in K2 sinc ...

Plot the density of states for the lowest three bands of the one-dimensional KronigPenney model discussed in

Question:

Plot the density of states for the lowest three bands of the one-dimensional Kronig–Penney model discussed in Section 1.2, for h̄²/2m = 1, a = 1, b = 0, and U₀b = 1. You will need to solve for ∂E/∂ k numerically, as we had to do for E(k) in the Kronig–Penney model.

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