Question: Known: I(v) = 2n2, %3D p(e) x e koT, E = nhv, n: any positive integer 2 p(e) E n=1 > p(e) = 1

Known: I(v) = 2n2, %3D p(e) x e koT, E = nhv,


Known: I(v) = 2n2, %3D p(e) x e koT, E = nhv, n: any positive integer 2 p(e) E n=1 > p(e) = 1 n=1 Prove that: hv hv ekBT 1 - Hint 1: E=1r" = (for 0

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