Semiconductor physics and devices: basic principles [solutions manual]

188 Pages·2003·1.708 MB·English

by  Donald A. Neamen

Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

Chapter 1

F I
Problem Solutions G4πr3J
4 atoms per cell, so atom vol. =4H K

3
1.1
(a) fcc: 8 corner atoms × 1/8 = 1 atom Then F I
6 face atoms × ‰ = 3 atoms G4πr3J
4H K
Total of 4 atoms per unit cell
3
Ratio= ×100% ⇒ Ratio =74%
(b) bcc: 8 corner atoms × 1/8 = 1 atom 16 2r3
1 enclosed atom = 1 atom
(c) Body-centered cubic lattice
Total of 2 atoms per unit cell
4
d =4r =a 3 ⇒a = r
(c) Diamond: 8 corner atoms × 1/8 = 1 atom 3
6 face atoms × ‰ = 3 atoms F I
4 3
4 enclosed atoms = 4 atoms Unit cell vol. =a3 =H rK
Total of 8 atoms per unit cell 3
F I
G4πr3J
1.2 2 atoms per cell, so atom vol. =2H K
3
(a) 4 Ga atoms per unit cell
Then
4 F I
Density = b5.65x10−8g3 ⇒ 2HG4πr3KJ
3
Density of Ga =2.22x1022 cm−3 Ratio= F I ×100% ⇒ Ratio =68%
4r 3
H K
4 As atoms per unit cell, so that
3
Density of As =2.22x1022 cm−3
(d) Diamond lattice
(b)
8
8 Ge atoms per unit cell Body diagonal =d =8r = a 3 ⇒a = r
3
8
Density = b g ⇒ F I
5.65x10−8 3 Unit cell vol. =a3 =H 8r K3
3
Density of Ge =4.44x1022 cm−3 F I
G4πr3J
8 atoms per cell, so atom vol. 8H K
3
1.3
(a) Simple cubic lattice; a =2r Then
F I
Unit cell vol =a3 =(2r)3 =8r3 G4πr3J
F I 8H K
1 atom per cell, so atom vol. =(1)HG4πr3KJ Ratio= F 3I ×100% ⇒ Ratio =34%
3 H 8r K3
Then 3
F I
G4πr3J
H K
1.4
3
Ratio= ×100% ⇒ Ratio=52.4% From Problem 1.3, percent volume of fcc atoms
8r3 is 74%; Therefore after coffee is ground,
(b) Face-centered cubic lattice Volume =0.74cm3
d
d =4r =a 2 ⇒ a = =2 2r
2

c h
Unit cell vol =a3 = 2 2r 3 =16 2r3

3
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

Then mass density is
1.5 4.85x10−23
8 ρ= b g ⇒
(a) a =5.43 A° From 1.3d, a = r 2.8x10−8 3
3
( ) ρ=2.21gm/cm3
a 3 5.43 3
so that r = = =1.18 A°

8 8
1.8
Center of one silicon atom to center of nearest
neighbor =2r ⇒ 2.36 A° (a) a 3 =2(2.2)+2(1.8)=8 A°
so that
(b) Number density
a =4.62 A°
8
= b g ⇒ Density =5x1022 cm−3
5.43x10−8 3 Density of A = b 1 g ⇒ 1.01x1022 cm−3
(c) Mass density 4.62x10−8 3
b g
N(At.Wt.) 5x1022 (28.09) 1
= ρ= = ⇒ Density of B = b g ⇒ 1.01x1022 cm−3
N 6.02x1023 4.62x10−8
A
ρ=2.33grams/cm3 (b) Same as (a)
(c) Same material

1.6
1.9
(a) a =2r =2(1.02)=2.04 A° (a) Surface density
A
Now 1 1
= = b g ⇒
2r +2r =a 3 ⇒2r =2.04 3−2.04 a2 2 4.62x10−8 2 2
A B B
so that rB =0.747 A° 3.31x1014 cm−2
(b) A-type; 1 atom per unit cell
Same for A atoms and B atoms
1
Density = b g ⇒ (b) Same as (a)
2.04x10−8 3 (c) Same material

Density(A) =1.18x1023 cm−3 1.10
B-type: 1 atom per unit cell, so 1
(a) Vol density =
Density(B) =1.18x1023 cm−3 a3
o
1
Surface density =
1.7
a2 2
(b) o
(b) Same as (a)
a =1.8+1.0⇒ a =2.8 A°

(c) 1.11
1 2 Sketch
Na: Density = b g =2.28x1022 cm−3
2.8x10−8 3
1.12
Cl: Density (same as Na) =2.28x1022 cm−3 (a) F I
1 1 1
(d) H , , K ⇒ (313)
Na: At.Wt. = 22.99 1 3 1
Cl: At. Wt. = 35.45 (b)
F I
So, mass per unit cell
1 1 1
1 1 H , , K⇒ (121)
(22.99)+ (35.45) 4 2 4
= 2 2 =4.85x10−23
6.02x1023
4
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

1.13 2atoms
(a) Distance between nearest (100) planes is: = b g ⇒ 9.88x1014 cm−2
4.50x10−8 2
d =a =5.63 A°
(ii) (110) plane, surface density,
(b)Distance between nearest (110) planes is:
2atoms
1 a 5.63 = b g ⇒ 6.99x1014 cm−2
d = a 2 = = 2 4.50x10−8 2
2 2 2
or (iii) (111) plane, surface density,
F I
d =3.98 A° H3⋅1+3⋅1K
(c) Distance between nearest (111) planes is: = 6 2 = b 4 g
1 a 5.63 3 3 4.50x10−8 2
d = a 3 = = a2
3 3 3 2
or or 1.14x1015 cm−2
d =3.25 A°

1.15
1.14 (a)
(a) (100) plane of silicon (cid:150) similar to a fcc,
Simple cubic: a =4.50 A° surface density = b 2atomsg ⇒
(i) (100) plane, surface density, 5.43x10−8 2
1atom
= b g ⇒ 4.94x1014 cm−2 6.78x1014 cm−2
4.50x10−8 2
(b)
(ii) (110) plane, surface density, (110) plane, surface density,
1atom 4atoms
= b g ⇒ 3.49x1014 cm−2 = b g ⇒ 9.59x1014 cm−2
2 4.50x10−8 2 2 5.43x10−8 2
(iii) (111) plane, surface density, (c)
F I
H1K 1 (111) plane, surface density,
3 atoms
= 6 = 2 = 1 = b4atoms g ⇒ 7.83x1014 cm−2
1c h( ) 1 a 3 3a2 3 5.43x10−8 2
a 2 x ⋅a 2⋅
2 2 2
1.16
1
= b g ⇒ 2.85x1014 cm−2 d =4r =a 2
3 4.50x10−8 2
then
(b) ( )
4r 4 2.25
Body-centered cubic a = = =6.364 A°
(i) (100) plane, surface density, 2 2
(a)
Same as (a),(i); surface density 4.94x1014 cm−2
4atoms
(ii) (110) plane, surface density, Volume Density = b g ⇒
2atoms 6.364x10−8 3
= b g ⇒ 6.99x1014 cm−2
2 4.50x10−8 2 1.55x1022 cm−3
(iii) (111) plane, surface density, (b)
Distance between (110) planes,
Same as (a),(iii), surface density 2.85x1014 cm−2
1 a 6.364
(c) = a 2 = = ⇒
Face centered cubic 2 2 2
(i) (100) plane, surface density or
5
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

4.50 A°
1.20
b g
(c) 5x1016 (30.98)
Surface density (a) Fraction by weight ≈ b g ⇒
2atoms 2 5x1022 (28.06)
= = b g
2a2 2 6.364x10−8 2 1.10x10−6
or (b) Fraction by wbeightg
3.49x1014 cm−2 1018 (10.82)
≈ b g b g ⇒
5x1016 (30.98)+ 5x1022 (28.06)
1.17 7.71x10−6
Density of silicon atoms =5x1022 cm−3 and 4

valence electrons per atom, so
1.21
Density of valence electrons 2x1023 cm−3 1
Volume density = =2x1015 cm−3
d3
1.18 So
Density of GaAs atoms d =7.94x10−6 cm=794 A°
8atoms
= b g =4.44x1022 cm−3 We have a =5.43 A°
5.65x10−8 3 O
So
An average of 4 valence electrons per atom, d 794 d
= ⇒ =146
Density of valence electrons 1.77x1023 cm−3
a 5.43 a
O O

1.19

2x1016
(a) Percentage = x100% ⇒
5x1022
4x10−5%
1x1015
(b) Percentage = x100% ⇒
5x1022
2x10−6%

6
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 2
Solutions Manual Problem Solutions

Chapter 2

Problem Solutions p =5.4x10−25 kg−m/s

h 6.625x10−34
2.1 Computer plot λ= = ⇒
p 5.4x10−25
2.2 Computer plot or
λ=12.3 A°
2.3 Computer plot
(ii) K.E. =T =100eV =1.6x10−17 J
2.4 p = 2mT ⇒ p =5.4x10−24 kg−m/s
2πx
For problem 2.2; Phase = −ωt =constant h
λ λ= ⇒ λ=1.23 A°
Then p
F I
2π⋅dx −ω=0 or dx =v = +ωH λK (b) Proton: K.E. =T =1eV =1.6x10−19 J
λ dt dt p 2π b gb g
p = 2mT = 2 1.67x10−27 1.6x10−19
2πx
For problem 2.3; Phase = +ωt =constant or
λ
p =2.31x10−23 kg−m/s
Then
F I
2π dx dx λ h 6.625x10−34
⋅ +ω=0 or =v = −ωH K λ= = ⇒
λ dt dt p 2π p 2.31x10−23
or

2.5 λ= 0.287 A°
E =hν= hc ⇒λ= hc (c) Tungsten Atom: At. Wt. = 183.92
λ E For T =1eV =1.6x10−19 J
b g
Gold: E =4.90eV =(4.90) 1.6x10−19 J p = 2mT
So b gb g
b6.625x10−34gb3x1010g = 2(183.92) 1.66x10−27 1.6x10−19
λ= (4.90)b1.6x10−19g ⇒ 2.54x10−5 cm or
p =3.13x10−22 kg =m/s
or
λ= 0.254 µm h 6.625x10−34
b g λ= = ⇒
Cesium: E =1.90eV =(1.90) 1.6x10−19 J p 3.13x10−22
or
So
b gb g
6.625x10−34 3x1010 λ=0.0212 A°
λ= b g ⇒6.54x10−5 cm
(1.90) 1.6x10−19 (d) A 2000 kg traveling at 20 m/s:
p =mv =(2000)(20)⇒
or
or
λ= 0.654 µm
p =4x104 kg−m/s

2.6 h 6.625x10−34
λ= = ⇒
(a) Electron: (i) K.E. =T =1eV =1.6x10−19 J p 4x104
b gb g
or
p = 2mT = 2 9.11x10−31 1.6x10−19
λ=1.66x10−28 A°
or

9
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 2
Solutions Manual Problem Solutions

2.7 or
E = 3kT = 3(0.0259)⇒ E =1.822x10−22 J ⇒ E =1.14x10−3 eV
avg 2 2 Also
b gb g
or p=mv = 9.11x10−31 2x104 ⇒
E =0.01727eV
avg p =1.822x10−26 kg−m/s
Now
Now
p = 2mE
avg avg h 6.625x10−34
b g b g λ= = ⇒
= 2 9.11x10−31 (0.01727) 1.6x10−19 p 1.822x10−26
or λ=364 A°
p =7.1x10−26 kg−m/s
(b)
avg
Now h 6.625x10−34
p = = ⇒
h 6.625x10−34 λ 125x10−10
λ= = ⇒
p 7.1x10−26 p =5.3x10−26 kg−m/s
or Also
λ=93.3 A° p 5.3x10−26
v = = =5.82x104 m/s
m 9.11x10−31
2.8 or
E =hν = hc v =5.82x106 cm/s
p p λ
Now
p
Now 1 1b gb g
F I E = mv2 = 9.11x10−31 5.82x104 2
p2 h 1 G h J2 2 2
E = e and p = ⇒ E = H K
or
e 2m e λ e 2m λ
e e E =1.54x10−21 J ⇒ E =9.64x10−3 eV
Set E = E and λ =10λ
p e p e

Then
F I F I 2.10
hc = 1 HG h KJ2 = 1 HG10hKJ2 hc b6.625x10−34gb3x108g
λp 2m λe 2m λp (a) E =hν= λ = 1x10−10
which yields or
λ = 100h E =1.99x10−15 J
p 2mc Now
b g
hc hc 2mc2 E =e⋅V ⇒1.99x10−15 = 1.6x10−19 V
E = E = = ⋅2mc=
p λ 100h 100 so
p
b gb g V =12.4x103V =12.4 kV
2 9.11x10−31 3x108 2
= ⇒ b gb g
100 (b) p = 2mE = 2 9.11x10−31 1.99x10−15
So
=6.02x10−23 kg−m/s
E =1.64x10−15 J =10.3keV
Then
h 6.625x10−34
2.9 λ= = ⇒ λ=0.11 A°
p 6.02x10−23
1 1b gb g
(a) E = mv2 = 9.11x10−31 2x104 2
2 2

10
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 2
Solutions Manual Problem Solutions

2.11
(a) ∆p = h = 1.054x10−34 ⇒ (b) ∆t = 1.0b54x10−34g ⇒
∆x 10−6 (1) 1.6x10−19
∆p =1.054x10−28 kg−m/s or
(b) ∆t =6.6x10−16 s
F I
hc p
E = =hcH K = pc
λ h 2.16
So (a) If Ψ(x,t) and Ψ (x,t) are solutions to
b gb g 1 2
∆E =c(∆p)= 3x108 1.054x10−28 ⇒ Schrodinger(cid:146)s wave equation, then

or
∆E =3.16x10−20 J ⇒ ∆E =0.198eV −h2 ⋅∂2Ψ1(x,t)+V(x)Ψ(x,t)= jh∂Ψ1(x,t)
2m ∂x2 1 ∂t

and
2.12
(a) ∆p = h = 1.054x10−34 ⇒ −2hm2 ⋅∂2Ψ∂2x(2x,t)+V(x)Ψ2(x,t)= jh∂Ψ2∂(tx,t)
∆x 12x10−10
Adding the two equations, we obtain
∆p =8.78x10−26 kg−m/s − 2 ∂2
h ⋅ Ψ(x,t)+Ψ (x,t)
(b) b g 2m ∂x2 1 2
1 (∆p)2 1 8.78x10−26 2 +V(x) Ψ(x,t)+Ψ (x,t)
∆E = ⋅ = ⋅ ⇒ 1 2
2 m 2 5x10−29 ∂
∆E =7.71x10−23 J ⇒ ∆E =4.82x10−4 eV = jh∂t Ψ1(x,t)+Ψ2(x,t)
which is Schrodinger(cid:146)s wave equation. So

Ψ(x,t)+Ψ (x,t) is also a solution.
2.13
1 2
(a) Same as 2.12 (a), ∆p =8.78x10−26 kg−m/s (b)
If Ψ ⋅Ψ were a solution to Schrodinger(cid:146)s wave
(b) 1 2
b g equation, then we could write
1 (∆p)2 1 8.78x10−26 2 − 2 ∂2 a f a f
∆E = ⋅ = ⋅ ⇒ h Ψ ⋅Ψ +V(x) Ψ ⋅Ψ
2 m 2 5x10−26 2m ∂x2 1 2 1 2
∆E =7.71x10−26 J ⇒ ∆E =4.82x10−7 eV ∂ a f
= jh Ψ ⋅Ψ
∂t 1 2
2.14 which can be written as
L O
∆p = h = 1.054x10−34 =1.054x10−32 −h2 NMΨ ∂2Ψ2 +Ψ ∂2Ψ1 +2∂Ψ1 ⋅∂Ψ2QP
∆x 10−2 2m 1 ∂x2 2 ∂x2 ∂x ∂x
L O
∆p 1.054x10−32 ∂Ψ ∂Ψ
p =mv⇒ ∆v = = ⇒ +V(x)Ψ ⋅Ψ = jhNMΨ 2 +Ψ 1QP
m 1500 1 2 1 ∂t 2 ∂t
or
Dividing by Ψ ⋅Ψ we find
∆v =7x10−36 m/s L 1 2 O
−h2 NM 1 ⋅∂2Ψ2 + 1 ⋅∂2Ψ1 + 1 ∂Ψ1 ∂Ψ2QP
2m Ψ ∂x2 Ψ ∂x2 ΨΨ ∂x ∂x
2.15 2 1 L 1 2 O
(a) ∆p = h = 1.054x10−34 ⇒ +V(x)= jhNM 1 ∂Ψ2 + 1 ∂Ψ1QP
∆x 10−10 Ψ ∂t Ψ ∂x
2 1
∆p =1.054x10−24 kg−m/s

11
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 2
Solutions Manual Problem Solutions

Since Ψ is a solution, then 2.19
1 z
∞
−h2 ⋅ 1 ⋅∂2Ψ1 +V(x)= jh⋅ 1 ⋅∂Ψ1 Note that Ψ⋅Ψ*dx =1
2m Ψ ∂x2 Ψ ∂t 0
1 1
Function has been normalized
Subtracting these last two equations, we are left
(a) Now
w i−thh 2 LNM 1 ∂2Ψ2 + 2 ∂Ψ1 ∂Ψ2OQP P = aoz4LNM 2 expFHG−xIKJOQP2dx
2m Ψ ∂x2 ΨΨ ∂x ∂x a a
2 1 2 0 o F Io
z
= jh 1 ∂Ψ2 = 2 ao 4expHG−2xKJdx
Ψ ∂t a a
2 o 0 o
S i n−c2ehm 2ΨΨ21 is∂ ∂a2xlΨs2o2 a+ sVo(luxt)io=n, jwheΨ1 ma∂y∂Ψ tw2 r ite o r = a2o FH−2aoIKexpFHG−a2oxIKJ0ao 4
2 2 L F I O F I
S u b−trha2ct⋅ing2 the⋅se∂ Ψla1s⋅t ∂twΨo2 e−qVua(txio)n=s,0 w e obtain P = −1NMexpHG−42aaoKJ −1QP =1−expH−21K
o
2m ΨΨ ∂x ∂x which yields
1 2
This equation is not necessarily valid, which P =0.393
means that ΨΨ is, in general, not a solution to (b)
1 2 F F II
Schrodinger(cid:146)s wave equation. aoz2G 2 G−xJJ2
P = H expH KK dx

a a
2.17 ao 4 o F Io
Ψ(x,t)= A sin(πx) exp(−jωt) 2 aoz2 G−2xJ
= expH Kdx
z z
+1 +1 a a
Ψ(x,t)2dx =1= A2 sin2(πx)dx o Fao 4 I oF I
−1 −1 = 2 H−aoKexpHG−2xKJao 2
or
L O a 2 a
A2 ⋅NM1 x− 1 sin(2πx)QP+1 =1 or o o ao 4
L F IO
which yie2lds 4π −1 P = −1NMexp(−1)−expH−1KQP
2
A2 =1 or A= +1,−1,+ j,− j
which yields
P =0.239

(c)
2.18 F F II
Ψ(x,t)= A sin(nπx) exp(−jωt) azoG 2 G−xJJ2
z z P = H expH KK dx
+1 +1 a a
Ψ(x,t)2dx =1= A2 sin2(nπx)dx 0 o o
or 0 0 = 2 azoexpFHG−2xIKJdx = 2 FH−aoIKexpFHG−2xIKJao
L O a a a 2 a
A2⋅NM1 x− 1 sin(2nπx)QP+1 =1 or o 0 o o o 0
2 4nπ 0 P = −1exp(−2)−1
which yields
which yields
A2 =2 or P =0.865

A= + 2 ,− 2,+ j 2,− j 2

12
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 2
Solutions Manual Problem Solutions

2.20
(a) kx−ωt =constant 2.22
b g
Thend x dx ω E = h2n2π2 = b 1.054x10gb−34 2π2n2 g
k dt −ω=0⇒ dt =vp = + k 2ma2 2 9.11x10−31 100x10−10 2
or so
1.5x1013 E =6.018x10−22n2 (J)
v = =104 m/s
p 1.5x109 or
v =106 cm/s E =3.76x10−3n2 (eV)
p Then
(b)
n=1⇒ E =3.76x10−3 eV
2π 2π 2π 1
k = λ ⇒λ= k = 1.5x109 n =2⇒ E2 =1.50x10−2 eV
or n=3⇒ E =3.38x10−2 eV
λ=41.9 A° 3

Also 2.23
p = h = 6.625x10−34 ⇒ (a) E = h2n2π2
λ 41.9x10−10 2ma2
or b g
1.054x10−34 2π2n2
p =1.58x10−25 kg−m/s = b gb g
Now 2 9.11x10−31 12x10−10 2
b gb g
hc 6.625x10−34 3x108 =4.81x10−20n2 (J)
E =hν= =
λ 41.9x10−10 So
or E =4.18x10−20 J ⇒ E =0.261eV
1 1
E =4.74x10−17 J ⇒ E =2.96x102 eV
E =1.67x10−19 J ⇒ E =1.04eV
2 2

(b)
2.21
b g hc hc
ψ(x)= Aexp −j kx+ωt E −E =hν= ⇒λ=
2 1 λ ∆E
where or
b gb g
2mE 6.625x10−34 3x108
k = λ= ⇒
h 1.67x10−19 −4.18x10−20
b g b g
2 9.11x10−31 (0.015) 1.6x10−19 λ=1.59x10−6 m
= or
1.054x10−34 λ=1.59µm
or
k =6.27x108 m−1
2.24
Now
b g (a) For the infinite potential well
ω= E = (0.015) 1.6x10−19 E = h2n2π2 ⇒n2 = 2ma2E
h 1.054x10−34 2ma2 h2π2
or so
b gb g b g
ω=2.28x1013 rad /s 2 10−5 10−2 2 10−2
n2 = b g =1.82x1056
1.054x10−34 2π2

or
13
Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 2
Solutions Manual Problem Solutions

n =1.35x1028 2mE
where K =
(b)
2
h
2π2
∆E = h (n+1)2 −n2 Boundary conditions:
2ma2 ψ(x)=0 at x = +a , x = −a
2π2 2 2
= h (2n+1)
So, first mode:
2ma2
ψ (x)= AcosKx
or
1
b g b g
∆E = 1.054x10b−34 2πgb2(2)g1.35x1028 where K = π so E = π2h2
2 10−5 10−2 2 a 1 2ma2
Second mode:
∆E =1.48x10−30 J ψ (x)= BsinKx
2
lEanrgeregr yth iann t h1e0 (mn+J.1 ) state is 1.48x10−30 Joules where K = 2π so E = 4π2h2
(c) a 2 2ma2
Quantum effects would not be observable. Third mode:
ψ (x)= AcosKx
3
2.25
For a neutron and n b= 1: g where K = 3π so E = 9π2h2
2π2 1.054x10−34 π2 a 3 2ma2
E = h = b gb g
1 2ma2 2 1.66x10−27 10−14 2 F o uψrth( xm)o=deB: sinKx
or 4
E =2.06x106 eV where K = 4π so E = 16π2h2
1
a 4 2ma2
For an electron in the same potential well:
b g
1.054x10−34 2π2
E = b gb g 2.27
1 2 9.11x10−31 10−14 2 The 3-D wave equation in cartesian coordinates,
for V(x,y,z) = 0
or
∂2ψ(x,y,z) ∂2ψ(x,y,z) ∂2ψ(x,y,z)
E =3.76x109 eV + +
1 ∂x2 ∂y2 ∂z2

2mE
2.26 + ψ(x,y,z)=0
Schrodinger(cid:146)s wave equation 2
h
∂2ψ(x) 2m Use separation of variables, so let
+ (E −V(x))ψ(x)=0 ψ(x,y,z)= X(x)Y(y)Z(z)
∂x2 h2
Substituting into the wave equation, we get
We know that

a −a
ψ(x)=0 for x ≥ and x ≤ ∂2X ∂2Y ∂2Z 2mE
2 2 YZ + XZ + XY + XYZ = 0
−a +a ∂x2 ∂y2 ∂z2 h2
V(x)=0 for ≤ x ≤
2mE
2 2 Dividing by XYZ and letting k2 = , we
so in this region 2
h
∂2ψ(x) 2mE obtain
∂x2 + h2 ψ(x)=0 (1) 1 ⋅∂2X + 1⋅∂2Y + 1 ⋅∂2Z +k2 =0
Solution is of the form X ∂x2 Y ∂y2 Z ∂z2
ψ(x)= AcosKx+BsinKx We may set
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