Download Homework # 6

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Transcript
Homework # 6
Chapter 6 Kittel
Phys 175A
Dr. Ray Kwok
SJSU
Prob. 1 – Kinetic Energy of electron gas
Adam Gray
Kinetic energy of electron gas.
Show that the kinetic energy of a threedimensional gas of N free electrons at 0 K is
3
U 0 = N  E f
5 .
In Kittel, we’re given the energy of a free electron
in this model to be:
 h2
E k = 
 2m

 K

2
To find the mean value of E over the volume of the
sphere in K space we use the definition:
E


1
 ∫ E
= 
 Vol ( C )  C
For our problem we have

 h 2 
1
∫∫∫ K 2 • K 2 sinθdkdθdφ

E = 
3 

 2m  (4 / 3)πK f 

 h 2 
1

 (4 / 5)πK 5f

E = 
3


 2m  (4 / 3)πK f 
(
)
Which results in the following mean energy per
electron of
 3  h 
K
E =  
2
2
f


 5  2 m 
Comparing to our initial energy equation (12), we
can rewrite this as
3
E =  E
5
f
So for N number of electrons, the total energy
would be
3
U 0 = N  E f
5
Prob. 2 – Pressure & Bulk modulus
Victor Chikhani
(a)
Derive a relation connecting the pressure (p) and
volume (V) of an electron gas at 0 K.
At 0 Kelvin entropy is constant (3rd Law of Thermo.)
When S=0 we can solve for the pressure (p) by taking the partial of
the internal energy (Uo) with respect to volume (V)
 ∂U o 
p = −

 ∂V 
3Nh  3π 2 N 
Uo =


10m  V 
2
Solution:
−1
2
3
 ∂U o 
3Nh 2  3π N  3  3π 2 N  2U o
p = −
=−

 − 2  =
 ∂V 
10m 3  V   V  3V
2
2
(b) Show that the bulk modulus (B) of an electron gas at 0 K is
 ∂p 
B = −V  
 ∂V 
Solution:
B=
5 p 10U o
=
3
9V
2U o
p=
3V
 2 ∂U o
 2 2U o
∂  2 
2 
B = −V 
+ Uo
−
+ Uo − 2 
  = −V 
 3V 3V
∂V  3V 
3V 
 3V ∂V
 4U o 6U o  10U o
B =
+
=
 9V
9V  9V
(c) Estimate for potassium (K), using Table 1, the value of the electron gas
contribution to B
Solution:
10U o 10 3Nε 10 3 1.40 ×10 22 
3.15 *1010 dyn
22 eV
B=
=
=
=

(2.12eV ) = 1.97 *10
3
3
9V
9 5V
9 5  cm
cm
cm 2

Answer agrees with value from Table 3, chapter 3.
Prob. 3 – Chemical potential in 2D
Jason Thorsen
Show that the chemical potential of a Fermi
Gas in two dimensions is given by:
for n electrons per unit area. Note: the density of
orbitals of a free electron gas in two dimensions is
independent of energy:
Solution:
We know the Fermi-Dirac distribution is given by
And we know the identity
We know that D(ε) is a constant so take it out of integral:
Plug in values and evaluate:
QED
Prob. 4 – Fermi gas in astrophysics
Michael Tuffley
(Wavevector is invariant in relativistic limit)
(Off by a factor of ~3)
Prob. 5 – Liquid H3
Daniel Wolpert
6.5 Liquid He3. The atom He3 has spin ½ and is a fermion. The density of
liquid He3 is 0.081g/cm3 near absolute zero. Calculate the Fermi energy εF and
the Fermi temperature TF.
(.081g/cm3/3.016(g/mole))(6.02x1023atoms/mole) = 1.625x1022 atoms/cm3
2 electrons per molecule = 3.25x1028 electrons/m3
εF = ħ2/2m ( 3π2N/V)2/3
= (1.055x10-34)2/(2(9.11x10-31)) * (29.6*3.25x1028)2/3 = 5.97x10-19 J
TF=εF/kb = 5.97x10-19 J/1.38x10-23 J/K
TF = 4.3x104 K
Prob. 6 – Frequency dependence of σ
John Anzaldo
Use the equation
for the electron drift velocity to show that the
conductivity at frequency w is:
where
Chapter 6 #6: Oscillation of Conductivity
Because E oscillates with w as
, we find that
, where
(From Eq. 42 on page 147)
Solving
gives
Plugging in
gives
and
.
Chapter 6 #6: Oscillation of Conductivity
Solving for
gives
Multiplying by
and
gives
Recall that J=σ
σE=nqv, where q=-e (p. 147148)
Chapter 6 #6: Oscillation of Conductivity
This gives the relation
Solving for
gives
which is what we set out to prove.
Prob. 7 – Dynamic Magnetoconductivity tensor
for free electron
Michael Tuffley
(a) Solve the drift velocity equation (51) to find the components
of the magnetoconductivity tensor.
QED
(b) Show that the dispersion relation for this wave in the medium is:
electromagnetic wave equation in a nonmagnetic isotropic medium
This gives rise to a set of four linear, homogeneous, algebraic equations. For
a nontrivial solution to exist, the determinant must vanish (Thornton 497.)1
QED
1Thornton,
Stephen T. Classical dynamics of particles and systems.
Belmont, CA: Brooks/Cole, 2004.
Prob. 8 – Cohesive energy of free electron
Fermi gas
Gregory Kaminsky
Cohesive energy of free electron Fermi gas. We define the
dimensionless length rs as r0/aH, where r0 is the radius of a sphere
that contains one electron, and aH is the Bohr radius h2/e2m.
a) Show that the average kinetic energy per
electron in a free electron Fermi gas at 0 K
is 2.21/rs2, where the energy is expressed
in rydbergs, with 1 Ry = me4/2h2.
Average kinetic energy: Integral of the
energy inside the sphere in k space, divided
by the volume of the sphere in k space.
kF
2
4
π
*
4
k
dk
∫
h 0
kF
2m
∫ 4π * k
0
h2 2
εF =
kF
2m
2
dk
3
= εF
5
Where kF is the
wavevector at
the Fermi
surface and εF
is the Fermi
energy.
Need to show that (3/5) εF = 2.21/rs2
3
3h 2 2
3h 2
3π 2 2 / 3 3 h 2
kF =
(
) = *
εF =
2
4
5
5 * 2m
5 * 2m
5
2
mr
0
π * r03
3
 9π 
 
 4
2/3
2.21 2.21a H2
2.21 * h 4
= 4 2 2
2 =
r0
rs
e m r0
Converting from
Rydbergs.
2.21 * h 4 me 4
h2
4 2 2 *
2 = 2.21 *
e m r0
2h
2mr02
Setting them equal to each other and canceling out
similar factors.
2/3
3  9π 
2.21 =  
5 4 
That’s true, you can check on the
calculator.
Q.E.D
b)Show that the coulomb energy of a point positive charge e
interacting with the uniform electron distribution of one electron
in the volume of radius r0 is –3e2/r0, or –3/rs in rydbergs.
Formula for the energy of electron due to point positive charge is just the
integral of the potential multiplied by the charge density within the sphere,
were interaction is occurring.
Charge density = δ = 4
3
e
e
Potential → V =
r
π *r3


 e  − e  − 3e 2 ro
−
e
3e 2


2
2

U = ∫∫∫ r sin θdrd θdφ ⋅ ρ ( r )
 = ∫ 4 πr dr 
 = 3 ∫ rdr = −
ro 0
2r o
 4 πr 3  r 
 r  0

o 
3

ro
c) Show that the coulomb self-energy of the electron
distribution in the sphere is 3e2/5r0, or 6/5rs in rydbergs.
r1
Energy required to bring a point charge close to a sphere is
r2
ρ(d 3 r1 )ρ(d 3 r2 )
r2
r1
Bring a ring close to a charged sphere is
r2
ρ(d 3r1 )ρ(4πr22 dr2 )
r2
Stack up all the shell up to r = ro
r1
ro
ro
ρ(d 3 r1 )ρ(4πr22 dr2 )
2 3
2
2
=
2
πρ
d
r
(
r
−
r
)
1
o
1
∫
r2
r2 = r1
This is the potential between a charged
sphere and the thick shelf enclosed it.
Total energy stored between the 2 charge-density is just the integral of it:
ro
3 2
5
r
r
r
16 2 2 5
2
2
2
2
2 2 o o
o
U = ∫ 2πρ (4πr1 dr1 )(ro − r1 ) = 8π ρ [
− ] = π ρ ro
3
5
15
0
With
ρ=
−e
4 3
πro
3
3e 2
U=
5ro
d) The Sum of (b) and (c) gives –1.80/rs for the total
coulomb energy per electron. Show that the
equilibrium value of rs is 2.45. Will such a material
be stable with respect to separated H atoms?
Factoring in the Kinetic energy of the electron
in a free Fermi gas at 0 K, basically adding in
the answer from part (a) to (b) and (c)
2.21 180
.
Energy = 2 −
rs
rs
To find the equilibrium
value of rs value I took the
derivative of Energy with
respect to rs
d
4.42 180
.
(E) = − 3 + 2
drs
rs
rs
4.42 180
.
Setting this equation to zero: d
(E) = − 3 + 2 = 0
drs
rs
rs
rs = 2.45
To check whether this is a stable
equilibrium, I took the second
derivative.
d2
13.3 2.6
2 (E) =
4 −
drs
rs
rs3
At E(rs = 2.45), this second
d
13.3
2.6
.
derivative is positive, so it is dr ( E ) = − 2.454 + 2.453 = 019
s
a stable equilibrium.
Thus since the second derivative is
positive, It is a stable equilibrium, and
such a metal will be stable with
respect to the separated H atoms at
zero Kelvin.
The end.
Prob. 9 – Static magnetoconductivity
Wanshan Li
tensor
Q: Let a static magnetic A:
field B lie along the z
a) From (52) p153, the
axis,
drift velocity is
a) find the static current
eτ
v x = − E x − ωcτv y [1]
density j in a static
m
electric field E.
eτ
v y = − E y − ωcτv x [2]
m
b) under strong
eτ
magnetic field (ωcτ >> 1
vz = − Ez
m
), find σ yx , σ xy
Problem 9: Current Density (2/3)
Plug [2] into [1], and
solve for vx
ωc eτ 2
eτ
−
Ex +
Ey
m
vx = m
1 + (ωcτ ) 2
plug [1] into [2], and
solve for vy.
− ωcτ 2 e
eτ
Ex −
Ey
m
m
vy =
1 + (ωcτ ) 2
From (42) p147, electric
current density
j x = n(−e)v x
Plug in vx, and define
electrical conductivity
2
ne τ
σo =
m
Then,
σ o ( E x − ωcτE y )
jx =
1 + (ωcτ ) 2
Problem 9: Hall conductivity (3/3)
Similarly for jy, jz.
In matrix form,
 jx 
 1
 

σo
j
=
ωτ
 y
2  c
ω
τ
+
1
(
)
c
j 
 0
 z

− ωcτ
1
0
 E x 
 
 E y 
1 + (ωcτ ) 2  E z 
0
0
b) Hall conductivity
σ yx =
σ oωcτ
1 + (ωcτ ) 2
When in high magnetic
field,
ωcτ >> 1
We have
σ yx =
σo
ωcτ
Plug in frequency
ωc =
eB
mc
and σ o , then
σ yx =
nec
= −σ xy
B
Prob. 10 – Maximum surface resistance
Nabel Alkhawlani
Consider a square sheet of side L, thickness d, and electrical resisitvity
ρ. the resistance measured between opposite edges of the sheet is
called the surface resistance Rsq = ρ L/Ld = ρ/d, which is
independent of the area L2 of the sheet. If we express ρ by 44, then
R sq =
m
n d e 2τ
Suppose now that the minimum value of the oscillation time is
d
determined by the scattering from the surfaces of the sheet, so that, τ ≈
v f
where vf is the fermi velocity. Thus the maximum surface
resistvity is Rsq ≈
in thickness that
mv f
nd 2 e 2
R sq ≈
show for a monatomic metal sheet one atom
h
e2
Sketch of the problem
λ~d
ρL ρL ρ
=
=
A Ld d
ne 2 τ 1
σ=
=
m
ρ
d
d3 = vol/atom
τ=
n = 1/d3 = # of atoms/vol
vf
R sq =
R sq =
m
mv f
mv f d
=
=
ne 2 τd ne 2 d 2
e2
With λ ~ d, p = mv = h / λ ~ h / d
R sq ≈
h
e2
= (6.6 x 10-34)/(1.6 x 10-19)2 = 26 kΩ
Document related concepts

Microplasma wikipedia, lookup

Polywell wikipedia, lookup