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260 CHAPTER16. LIVEWIRESANDDEADSTARS

Accordingto(16.31),

R=

(

2 mL^2

̄h^2 π^2

EF

) 1 / 2

(16.34)

sothenumberofelectrons,intermsofEF,becomes

N=

π

3

L^3

( 2 mE

F

̄h^2 π^2

) 3 / 2

(16.35)

SolvingfortheFermienergy,weget

EF=

̄h^2 π^2

2 m

( 3 N

πL^3

) 2 / 3

(16.36)

Butthenumberdensityofelectrons,ne,inthesolidisjust

ne=

N

L^3

(16.37)

Therefore, we find that the Fermienergy just depends on the electronmass and
density(andnotonthesizeofthebox)

EF=

̄h^2 π^2

2 m

( 3

π

ne

) 2 / 3

(16.38)

Next,approximatethesumoverallstateswithenergieslessthanEFbyanintegral

ET =

∑

n 1

∑

n 2

∑

n 3

2 En 1 n 2 n 3

≈

∫

"n·"n≤R^2

d^3 n 2 En 1 n 2 n 3

=

h ̄^2 π^2

mL^2

1

8

∫

d^3 n%n·%n

=

̄h^2 π^2

8 mL^2

4 π

∫R

0

dnn^4

=

̄h^2 π^3

10 mL^2

R^5 (16.39)

wherethefactorof 1 / 8 infrontoftheintegralcomes,again,becausethesitesliein
theoctantofthespherewithpositiveintegercoordinates.Nextfrom(16.33)

R=

( 3 N

π

) 1 / 3

(16.40)

UsingL=V^1 /^3 ,whereV isthevolumeofthesolid,wegetfinally

ET=

̄h^2 π^3

10 m

(

3 N

π

) 5 / 3

V−^2 /^3 (16.41)