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1217_Physics ProblemsTechnical Physics

# 1217_Physics ProblemsTechnical Physics - 558 P43.33...

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558 Molecules and Solids P43.33 E n EN E dE e av = z 1 0 af At T = 0, NE = 0 for EE > F ;. Since fE = 1 for EF < and = 0 for > F , we can take NE CE m h E e == 12 32 3 82 π E n CE dE C n Ed E C n E e E e E e av F 52 FF = zz 5 0 0 . But from Equation 43.24, C n E e = 3 2 F , so that E E av F = F H G I K J F H G I K J = 2 5 3 2 3 5 . P43.34 Consider first the wave function in x. At x = 0 and xL = , ψ = 0. Therefore, sin kL x = 0 and x = πππ ,,, 23 . Similarly, sin y = 0 and y = sin z = 0 and z = = F H G I K J F H G I K J F H G I K J A nx L ny L nz L x y z sin sin sin . From + + =− 2 2 2 2 2 22 2 ψψψ xyz m UE e = a f , we have inside the box, where U = −−− F H G I K J n L n L n L m E x y ze 2 2 2 ψψ = E mL nnn e x y z =+ + = = 2 222 2 123 ej ,, , , , . Outside the box we require = The minimum energy state inside the box is === 1, with E e = 3 2 2 = P43.35 (a) The density of states at energy E is gE CE a f = . Hence, the required ratio is g g C C 850 700 110 . . . . . eV eV a f . (b) From Eq. 43.22, the number of occupied states having energy E is CE e EE kT B bg = + 1 F . Hence, the required ratio is N N e e kT B B 1 1 700 700 850 700 . . . . .. eV eV = + + L N M M O Q
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