Embed Size (px)
Citation preview
Number of states up to E : N =V
6! 22m!2
"#$
%&'
3/2E3/2 .
Density of states:!!!!!!!!!!! dNdE
=Vm3/2
2! 2!3 E1/2
Fermi energy: !!!!!!!!!!!!!!!!!!!!EF =(3 /! )
2/3h2
8mne
2/3
Total energy = Ne E
Ne E = Ne
E dNdE
dE0
E f
!dNdE
dE0
E f
!!!!!!!!=!!!!N
E3/2 dE0
E f
!
E1/2
dE0
E f
!=
35NeEF
Total zero point electron energy:
x =xP(x)dx!P(x)dx!
ETOT =35NEFExercise: Show!
Total energy = Ne E =35NeEF
EF =h2
8m3Ne
!V"#$
%&'
2/3
Total energy = 35NeEF =
35Ne
h2
8m3Ne
!V"#$
%&'
2/3
=3h2
40m3!V
"#$
%&'
2/3
Ne5/3
Copper has a density of 8.9 gm/cm3 and molecular weight of 63.5!Assume that each copper atom has two free electrons.!
At temperature 0 K !
• Calculate the number of copper atoms/cm3.!
• Calculate the Fermi energy.!
• Calculate the average energy.!
• Calculate the total electron energy.!
Homework!Due Friday, Oct. 23 !
Application of zero point energy!
to astrophysics.!
Some aspects of the structure of a star may be understood !by considering the opposing forces of gravitational energy,!which makes the star become smaller and the Fermi energy,!which makes the star become larger.!
Energetics of white dwarf. Fermi energy.
Energy of electrons:
Ee =35NeEF
= 35Ne
h2
8me3Ne!V
"#$
%&'
2/3
Ee = 3h2
40me3!V
"#$
%&'
2/3
Ne5/3
Assume star has Ne = Np = NN / 2, where N stands for nucleons (neutrons + protons)
Ee =3h2
40me3!V
"#$
%&'
2/3Ne
5/3 = 3h2
40me3
! 43!R3"
#$%&'
"
#
$$$
%
&
'''
2/3
Ne( )5/3
!!!!!!!!!!!!!!!!!!!!Show Ee =3
4034/3
4! 2( )2/3h2
meR2 Ne5/3 = C Ne
5/3
R2
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!Ee = C Ne5/3
R2
Gravitational energy:
dEG = ! dmmGr
EG = ! dmmGr
.
0
R
"
m = #V = # 4$r3
3
dm = #dV = # 4$3d r3( ) = 4$#r2dr
dr!
r!
Gravitational energy:
Exercise: Show EG = ! 35M 2GR
= ! BR
Start with EG = ! dmmGr
0
R
"
Express m!!and dm in terms of r.
m = # 4$r3
3dm = #dV = #4$r2dr
Gravitational energy:
EG = ! dmmGr
.
0
R
" m = # 4$r3
3dm = #dV = #4$r2dr
EG = !# 4$r3
3#4$r2drG
r=
0
R
" ! 163$ 2#2G
0
R
" r4dr = !163$ 2#2G R5
5
Use #2 = M 2
V 2 = 9M 2
16$ 2R6 EG = ! 35GM 2
R= ! 3
5GmN2 NN
2
R= ! BNN
2
R
Using: E = Ee + EG = CNe5/3
R2 ! BNN2
Rand dE
dR= 0
Find: the numerical value of R
Ee = CNe5/3
R2EG = ! BNN
2
R
Find the equilibrium radius by mimimizing the energy with respect to R.
Ee =340
34/3
4! 2( )2/3h2
meR2Ne5/3 = CNe
5/3
R2EG = " 3
5GM 2
R= " 3
5GmN
2 Nn2
R= " BNN
2
R
Use the following data to find B and C for the Sun.
mN = 1.67 !10"27 kg me = 9.1!1031 kg G = 6.67 !10"11 m3kg-1s"2
M! = 2 !1030 kg NN = 1.2 !1057 Ne = 0.6 !1057 !!!!h = 6.63!10"34 J-s
C = 1.36 !10"38 !!!!!!!!!!!!!!!!!B = 1.12 !10"64
Use the following data to find B and C for the Sun.
mN = 1.67 !10"27 kg me = 9.1!1031 kg G = 6.67 !10"11 m3kg-1s"2
M! = 2 !1030 kg N = 1.2 !1057 h = 6.63!10"34 J-s
C = 340
34/3
4! 2( )2/3h2
meNe5/3 = 2.8 "10#2
6.63"10#34( )29.1"10#31
0.6 "1057( )5/3 = 1.36 "10#38
B = 35Gmn
2Nn2 = 3
56.67 "10#11( ) 1.67 "10#27( )2 = 1.12 "10#64
Ee =340
34/3
4! 2( )2/3h2
meR2Ne5/3 = CNe
5/3
R2EG = " 3
5GM 2
R= " 3
5Gmn
2NN2
R= " BNN
2
R
E = Ee + EG = CNe5/3
R2 ! BNN2
RdEdR
= ! 2CNe5/3
R3 + BNN2
R2 = 0
R = 2CB
Ne5/3
N 2N
B = 1.12 "10!64 C = 1.36 "10!38
R =2 1.36 "10!38( ) 0.6 "1057( )5/3
1.12 "10!64 1.2 "1057( )2= 7.2 "103 km
Density of white dwarf
Find the density ! !M"V
Fermi Energy of electrons:
Find EF using EF = 53EeN e
Data: R = 7.2 "106 m M" = 2 "1030 kg Ne = NN / 2 = 0.6 "1057
Exercises!
R = 7.2 !106 km M! = 2 !1030 kg Ne = NN / 2 = 0.6 !1057
Density of white dwarf
" " 2 !1030 kg 43# 7.2 !106( )3 m3
= 1.28 !109 kg-m-3 = 1.28 !106 gm-cm-3
Fermi Energy of electrons:
EF = 53EeN e
Ee =CNe
5/3
R2 = 3.5!1042 J = 2.2 !1061 eV
EF = 53CNe
2/3
R2 = 53
1.36 !10$38( ) 0.6 !1057( )2/3
7.2 !106( )2= 3.1!10$4 J = .194 !106 eV
%non relativistic kinematics is becoming invalid.
Review!
Fermi energy:
EF = h2
8me3Ne!V
"#$
%&'
2/3
Electron "zero point" energy:
Ee =35NeEF Ee =
37/3
40 4! 2( )2/3h2
meR2 Ne5/3 = CNe
5/3
R2
Gravitational energy:
EG = ( 35GM 2
R= ( 3
5GmN
2 NN2
R= ( BNN
2
R
