半导体物理与器件第四版课后习题答案2
Chapter 2
Sketch
_______________________________________
Sketch
_______________________________________
Sketch
_______________________________________
From Problem , phase t x
ωλπ-=2
= constant Then
??
? ??+==?=-?πλωυωλπ2,02p dt dx dt dx From Problem , phase t x
ωλ
π+=2 = constant Then ??
? ??-==?=+?πλωυωλπ2,02p dt dx dt dx
_______________________________________
E
hc
hc
h E =
?==λλν Gold: 90.4=E eV ()()
19106.190.4-?= J So,
(
)(
)
()(
)
519
10
341054.210
6.190.410310625.6---?=???=λcm or
μλ254.0=m
Cesium: 90.1=E eV ()()
19106.190.1-?= J
So,
(
)(
)
()(
)
5
19
103410
54.6106.190.110310625.6---?=???=λcm or
μλ654.0=m
_______________________________________
(a) 9
34
10
55010625.6--??==λh p 2710205.1-?=kg-m/s
331
27
1032.11011.9102045.1?=??==--m p υm/s or 51032.1?=υcm/s (b) 9341044010625.6--??==λh p 2710506.1-?=kg-m/s 331
27
1065.11011.9105057.1?=??==--m p υm/s
or 51065.1?=υcm/s (c) Yes _______________________________________ (a) (i) ()()()
1931106.12.11011.922--??==mE p
2510915.5-?=kg-m/s
925
34
1012.110915.510625.6---?=??==p h λm or o
A 2.11=λ
(ii)()()()
1931106.1121011.92--??=p 241087.1-?=kg-m/s
1024
34
1054.3108704.110625.6---?=??=λm or o A 54.3=λ
(iii) ()()()
1931106.11201011.92--??=p 2410915.5-?=kg-m/s
1024
34
1012.110915.510625.6---?=??=λm or o
A 12.1=λ (b)
()()()
1927106.12.11067.12--??=p 23
10532.2-?=kg-m/s
1123
34
1062.210532.210625.6---?=??=λm or o
A 262.0=λ
_______________________________________
()03885.00259.02323=??
?
??==kT E avg eV
Now
avg avg mE p 2=
(
)()()19
31
106.103885.010
11.92--??=
or
2510064.1-?=avg p kg-m/s Now
925
34
10225.610064.110625.6---?=??==p h λm or
o
A 25.62=λ
_______________________________________
p
p p hc
h E λν==
Now
m
p E e
e 22= and
2
21???
? ??=
?=e
e e e h m E h
p λλ Set e p E E = and e p λλ10=
Then
2
2
102121???
? ??=
???? ??=
p e
p h m h
m hc
λλλ which yields
mc
h
p 2100=λ
100
221002
mc mc h hc hc E E p p =
?===λ ()()
100
1031011.922
8
31??=
-
151064.1-?=J 25.10=keV
_______________________________________
(a) 10
34
108510625.6--??==λh p
2610794.7-?= kg-m/s
4
31
2610
56.81011.910794.7?=??==--m p υm/s or 61056.8?=υcm/s
()()
243121056.81011.92
1
21??==-υm E
211033.3-?=J
or 219
21
1008.210
6.110334.3---?=??=E eV (b) ()()
23311081011.92
1
??=-E
2310915.2-?=J
or 419
23
1082.110
6.110915.2---?=??=E eV ()()
3311081011.9??==-υm p
2710288.7-?=kg-m/s
827
35
1009.910288.710625.6---?-??==p h λm or o
A 909=λ
_______________________________________
(a) ()()
10
8
3410
110310625.6--???===λνhc h E 15
1099.1-?=J Now
19
15
106.11099.1--??==??=e E V V e E
41024.1?=V V 4.12=kV (b)
()()
15311099.11011.922--??==mE p
231002.6-?=kg-m/s Then
1123
34
1010.11002.610625.6---?=??==p h λm or
o
A 11.0=λ
_______________________________________
6
34
1010054.1--?=
?=?x p 2810054.1-?=kg-m/s
_______________________________________
(a) (i) =??x p
2610
34
10783.810
1210054.1---?=??=?p kg-m/s (ii)p m p dp d p dp dE E ?????
?
??=??=?22 m
p
p p m p ?=
??=
22 Now mE p 2=
()()()
1931106.1161092--??= 2410147.2-?=kg-m/s so ()()
31
26
2410
910783.8101466.2---???=?E 1910095.2-?=J
or 31.1106.110095.219
19
=??=?--E eV
(b) (i) 2610783.8-?=?p kg-m/s (ii)()()()
1928106.1161052--??=p 231006.5-?=kg-m/s
()(
)
28
26
2310
510783.81006.5---???=?E 2110888.8-?=J
or 219
21
1055.510
6.110888.8---?=??=?E eV _______________________________________
322
34
10054.110
10054.1---?=?=?=?x p kg-m/s
1500
10054.132
-?=
?=??=m p m p υυ 36107-?=?υm/s
_______________________________________
(a) =??t E
()()
16
19
341023.8106.18.010054.1---?=??=?t s
(b) 10
34
105.110054.1--??=
?=?x p 251003.7-?=kg-m/s
_______________________________________
(a) If ()t x ,1ψ and ()t x ,2ψ are solutions
to
Schrodinger's wave equation, then
()()()()t t x j t x x V x t x m ?ψ?=ψ+?ψ??-,,,211
212
2 and
()()()()t t x j t x x V x t x m ?ψ?=ψ+?ψ??-,,,222
2
22
2 Adding the two equations, we obtain
()()[]t x t x x m ,,2212
2
2ψ+ψ???
- ()()()[]t x t x x V ,,21ψ+ψ+
()()[]t x t x t
j ,,21ψ+ψ??
=
which is Schrodinger's wave equation. So ()()t x t x ,,21ψ+ψ is also a solution.
(b) If ()()t x t x ,,21ψ?ψ were a solution to
Schrodinger's wave equation, then we could write []()[]21212
222ψ?ψ+ψ?ψ???
-x V x m
[]21ψ?ψ??
=t
j
which can be written as
???????ψ???ψ?+?ψ?ψ+?ψ?ψ-x x x x m 212
122
22
21222
()[]???????ψ?ψ+?ψ?ψ=ψ?ψ+t t j x V 12
2
121 Dividing by 21ψ?ψ, we find
?
?
?
????ψ??ψ?ψψ+?ψ??ψ+?ψ??ψ-x x x x m 2121212
1222
22
2112 ()??
?
????ψ?ψ+?ψ?ψ=+t t j x V 112211
Since 1ψ is a solution, then
()t j x V x m ?ψ??ψ?=+?ψ??ψ?-1
1212
12112
Subtracting these last two equations, we have
??
?
????ψ??ψ?ψψ+?ψ??ψ-x x x m 212122222212
t j ?ψ??ψ?=2
21
Since 2ψ is also a solution, we have
()t j x V x m ?ψ??
ψ?=+?ψ??ψ?-2
2222
22112 Subtracting these last two equations, we obtain
()02
221212=-?ψ???ψ??ψψ?-x V x
x m
This equation is not necessarily valid, which means that 21ψψ is, in general, not a solution
to Schrodinger's wave equation.
_______________________________________
12cos 23
1
2=???
???
+-dx x A π
()12sin 23
12=??
????++-ππx x A 121232=????????? ??--A
so 21
2=A
or 2
1
=A
_______________________________________
()1cos 22
/12
/12
=?+-dx x n A
π
()142sin 22
/12/12=??????++-ππn x n x A ??? ??==??
??????? ??--211414122A A
or 2=A
_______________________________________
Note that
10
*=ψ?ψ?
∞
dx
Function has been normalized.
(a) Now
dx a x a P o
a o o 2
4
exp 2?
???
?
???????? ??-=
dx a x a o
a o o
?
????
??-=
4
2exp 2
402exp 22
o a o o o a x a a ???
? ??-????
?
?-=
or
()??? ??--=???
?????-????
??--=21exp 1142exp 1o
o
a a P which yields 393.0=P (b)
dx a x a P o o
a a o o 2
2
4
exp 2?
???
?
???????? ??-=
dx a x a o o
a a o o
?
???
?
??-=
2
4
2exp 2
242exp 22
o o
a a o o o a x a a ???? ??-????
?
?-=
or
()()???
??
???? ??----=21exp 1exp 1P
which yields
239.0=P (c)
dx a x a P o
a o o 2
0exp 2?
????
?
??????? ??-= dx a x a o
a o o
?
????
??-=0
2exp 2
o a o o o
a x a a 02exp 22???
? ??-????
??-=
()()[]12exp 1---= which yields 865.0=P
_______________________________________
()dx x P 2
?
=ψ
(a)
dx x a a ??
?
????? ???
2cos 224
/0
π 4
/042sin 22a a a x x a ???
????
???
????? ????? ??+??
? ??=ππ
???
????
????
???? ????? ??+
??? ????? ??=a a a ππ42sin 2
42 ()()???
???+??? ??=π4182a a a or 409.0=P
(b) dx a x a P a a ???
????? ??=
?
π2
/4
/2cos 2 2
/4/42sin 22a a a a x x a ???
????
???
?
???? ????? ??+??
? ??=ππ
()???
???
?
???
?
???? ????? ??-
-??? ??+??
? ??=a a a a a ππππ42sin 84sin 42 ??
????--+=π4181041
2
or 0908.0=P
(c) dx a x a P a a ???
????? ??=
?
+-π22
/2
/cos 2 2
/2/42sin 2
2a a a a x x a +-???
????
???
?
???? ????? ??+??? ??=ππ
()()??
?
????
????
???? ??--
??? ??--??? ??+??? ??=a a a a a ππππ4sin 44sin 42 or 1=P
_______________________________________
(a) dx a x a P a ??
? ????? ??=
?
π2sin 224
/0
4
/0244sin 22a a a x x a ???
????
???
????? ????? ??-??
? ??=ππ
()??
?
???
?
????
???? ??-??? ??=a a a ππ8sin 82
or 25.0=P
(b) dx a x a P a a ??? ????? ??=
?
π2sin 222
/4
/ 2
/4/244sin 2
2a a a a x x a ???
????
???
?
???? ????? ??-??? ??=ππ
()()??
?
????
????
???? ??+
??? ??-??? ??-??? ??=a a a a a ππππ8sin 882sin 42 or 25.0=P
(c) dx a x a P a a ??? ????? ??=
?
+-π2sin 222
/2
/ 2
/2/244sin 22a a a a x x a +-???
????
???
?
???? ????? ??-??
? ??=ππ
()()???
????
????
???? ??-+
??
? ??--??? ??-??? ??=a a a a a ππππ82sin 482sin 42 or 1=P
_______________________________________
(a) (i) 48
12
1010
8108=??==k p ωυm/s or 610=p υcm/s
98
10854.710822-?=?==ππλk m
or o
A 54.78=λ (ii)()()
431101011.9-?==υm p
271011.9-?=kg-m/s
()()
24312101011.92
1
21-?==υm E
2310555.4-?=J
or 419
23
1085.210
6.110555.4---?=??=E eV (b) (i) 49
13
1010
5.1105.1-=?-?==k p ωυm/s or 610-=p υcm/s
991019.410
5.122-?=?==
π
πλk m or o
A 9.41=λ
(ii) 271011.9-?-=p kg-m/s
41085.2-?=E eV
_______________________________________
(a) ()()t kx j Ae t x ω+-=ψ, (b) ()()
2192
1
106.1025.0υm E =
?=- ()
2311011.92
1
υ-?= so
41037.9?=υm/s 61037.9?=cm/s
For electron traveling in x -direction, 61037.9?-=υcm/s
()()
4311037.91011.9?-?==-υm p
2610537.8-?-=kg-m/s
926
34
1076.710537.810625.6---?=??==p h λm
89
10097.810
76.722?=?=
=-π
λπk m 1- ()()
481037.910097.8??=?=υωk
or 1310586.7?=ωrad/s
_______________________________________
(a) ()()
4311051011.9??==-υm p
2610555.4-?=kg-m/s
8
26
3410
454.110555.410625.6---?=??==p h λm
88
1032.410
454.122?=?==-πλπk m 1- ()()
481051032.4??==υωk
131016.2?=rad/s (b) ()()
631101011.9-?=p
251011.9-?=kg-m/s
10
25
3410
27.71011.910625.6---?=??=λm 910
1064.810
272.72?=?=-π
k m 1-
()()
15691064.8101064.8?=?=ωrad/s
_______________________________________
()
()()
2
10
3122
34222
2210751011.9210054.12---???=
=ππn ma n E n
()21
2
100698.1-?=n E n
J
or
()
19
21
2106.1100698.1--??=n E n
or ()
3210686.6-?=n E n eV Then
311069.6-?=E eV
2
21067.2-?=E eV
231002.6-?=E eV
_______________________________________
(a) (
)
(
)(
)
2
10
3122
34222
2210101011.9210054.12---???==
ππn ma n E n (
)20
2
10
018.6-?=n J
or
()
()3761.010
6.110018.6219
20
2n n E n =??=--eV Then
376.01=E eV 504.12=E eV
385.33=E eV
(b) E
hc ?=
λ ()()
19106.1504.1385.3-?-=?E
191001.3-?=J
()()
19
8
341001.310310625.6--???=λ
710604.6-?=m or 4.660=λnm
_______________________________________
(a) 2
2
222ma n E n π =
()
(
)()2
23
22
3423
102.110
15210054.110
15----???=
?πn
()62
2
310538.21015--?=?n
or 2910688.7?=n (b) 151?+n E mJ (c) No
_______________________________________
For a neutron and 1=n :
(
)
(
)(
)
2
14
2722
342
221101066.1210054.12---??==ππma E
13103025.3-?=J
or
6
19
13110
06.2106.1103025.3?=??=--E eV For an electron in the same potential well: ()()()
2
1431
2
2
341
101011.9210054.1---??=
πE
10100177.6-?=J or
919
10
11076.310
6.110017
7.6?=??=--E eV _______________________________________
Schrodinger's time-independent wave equation
()()()()0222
2=-+??x x V E m
x x ψψ
We know that
()0=x ψ for 2a x ≥ and 2
a
x -≤
We have
()0=x V for 2
2a x a +<<-
so in this region
()()02222=+??x mE
x x ψψ
The solution is of the form ()kx B kx A x sin cos +=ψ where
2
2 mE
k =
Boundary conditions: ()0=x ψ at 2
,2a x a x +=-=
First mode solution:
()x k A x 111cos =ψ where
2
2
2112ma E a k ππ=?=
Second mode solution: ()x k B x 222sin =ψ where
2
2
222242ma
E a k ππ=?= Third mode solution: ()x k A x 333cos =ψ where
2
2
233293ma E a k ππ=
?= Fourth mode solution: ()x k B x 444sin =ψ where
2
2
2442164ma E a k ππ=
?= _______________________________________
The 3-D time-independent wave equation in
cartesian coordinates for ()0,,=z y x V is:
()()()2
22222,,,,,,z z y x y z y x x z y x ??+
??+??ψψψ ()0,,22=+z y x mE
ψ
Use separation of variables, so let ()()()()z Z y Y x X z y x =,,ψ
Substituting into the wave equation, we obtain
22222
2z
Z
XY y Y XZ x X YZ ??+??+?? 022=+XYZ mE
Dividing by XYZ and letting
222 mE
k =, we
find
(1)
01112222222=+???+???+???k z
Z
Z y Y Y x X X We may set
012
2
2222=+???-=???X k x
X k x X X x x Solution is of the form
()()()x k B x k A x X x x cos sin += Boundary conditions: ()000=?=B X and ()a
n k a x X x x π
=
?==0 where ....3,2,1=x n Similarly, let
2
221y k y Y Y -=??? and 2221z k z
Z Z -=???
Applying the boundary conditions, we find
a
n k y y π=, ....3,2,1=y n
a
n k z z π
=
, ...3,2,1=z n From Equation (1) above, we have
02222=+---k k k k z y x
or
2
222
22 mE
k k k k z y x =
=++ so that
()
2
222
222z y x n n n n n n ma
E E z y x ++=→π _______________________________________
(a)
()()()0,2,,22222=?+??+??y x mE
y y x x y x ψψψ
Solution is of the form:
()y k x k A y x y x sin sin ,?=ψ
We find
()y k x k Ak x y x y x x sin cos ,?=??ψ ()y k x k Ak x
y x y x x sin sin ,2
2
2?-=??ψ
()y k x k Ak y y x y x y cos sin ,?=??ψ
()
y k x k Ak y
y x y x y sin sin ,2
2
2?-=??ψ
Substituting into the original equation, we find:
(1) 0222
2=+--
mE k k y x
From the boundary conditions, 0sin =a k A x , where o
A a 40= So a
n k x x π
=
, ...,3,2,1=x n Also 0sin =b k A y , where o
A b 20= So b
n k y y π
=
, ...,3,2,1=y n Substituting into Eq. (1) above
???
? ??+=22222222b n a n m E y x n n y x ππ (b)Energy is quantized - similar to 1-D result. There can be more than one quantum state
per given energy - different than 1-D result.
_______________________________________
(a) Derivation of energy levels exactly the
same as in the text
(b) ()
212
22
222n n ma
E -=?π For 1,212==n n Then
2
2
223ma E π =?
(i) For o
A a 4= (
)
(
)()
2
1027
22
3410410
67.1210054.13---???=
?πE
2210155.6-?=J
or 319
22
1085.310
6.110155.6---?=??=?E eV
(ii) For 5.0=a cm
()
(
)()
2
227
22
34105.010
67.1210054.13---???=
?πE
3610939.3-?=J or
1719
36
1046.210
6.110939.3---?=??=?E eV _______________________________________
(a) For region II, 0>x
()()()0222222=-+??x V E m
x x O ψψ
General form of the solution is
()()()x jk B x jk A x 22222exp exp -+=ψ where
()O V E m
k -=2
22
Term with 2B represents incident wave and
term with 2A represents reflected wave. Region I, 0 ()()02122 12=+??x mE x x ψψ General form of the solution is ()()()x jk B x jk A x 11111exp exp -+=ψ where 2 12 mE k = Term involving 1B represents the transmitted wave and the term involving 1A represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that 01=A . Then ()()x jk B x 111exp -=ψ ()()()x jk B x jk A x 22222exp exp -+=ψ (b) Boundary conditions: (1) ()()0021===x x ψψ (2) 0 2 01==??= ??x x x x ψψ Applying the boundary conditions to the solutions, we find 221B A B += 112222B k B k A k -=- Combining these two equations, we find 212 122B k k k k A ????? ??+-= 212 212B k k k B ????? ??+= The reflection coefficient is 2 12 12*22*2 2???? ??+-==k k k k B B A A R The transmission coefficient is () 2 212 141k k k k T R T +=?-= _______________________________________ ()()x k A x 222exp -=ψ () ()x k A A x P 2*2 22 2exp -== ψ where ()2 22 E V m k o -= ()()( )34 19 3110054.1106.18.25.31011.92---??-?= 9210286.4?=k m 1- (a) For 101055-?==o A x m ()x k P 22exp -= ()()[] 109105102859.42exp -??-= 0138.0= (b) For 10101515-?==o A x m ()()[] 1091015102859.42exp -??-=P 61061.2-?= (c) For 10104040-?==o A x m ()()[] 1091040102859.42exp -??-=P 151029.1-?= _______________________________________ ()a k V E V E T o o 22exp 116-? ??? ??-???? ??? where ()2 22 E V m k o -= ()()( )34 19 3110054.1106.11.00.11011.92---??-?= or 2k 910860.4?=m 1- (a) For 10104-?=a m ()()[] 10 910 41085976.42exp 0.11.010.11.016-??-??? ??-??? ???T 0295.0= (b) For 101012-?=a m ()()[] 10 910121085976.42exp 0.11.010.11.016-??-??? ??-??? ???T 51024.1-?= (c) υe N J t =, where t N is the density of transmitted electrons. 1.0=E eV 20106.1-?=J () 23121011.92 1 21υυ-?==m 510874.1?=?υm/s 7 10874.1?=cm/s ()() 719310874.1106.1102.1??=?--t N 810002.4?=t N electrons/cm 3 Density of incident electrons, 108 10357.10295.010002.4?=?=i N cm 3- _______________________________________ ()a k V E V E T O O 22exp 116-???? ? ?-???? ??? (a) For ()o m m 067.0= () 2 22 E V m k O -= ()( ) ()() ( ) 2 /1234193110054.1106.12.08.01011.9067.02?? ??????????-?=--- or 9210027.1?=k m 1- Then ?? ? ??-??? ??=8.02.018.02.016T ()()[] 109101510027.12exp -??-? or 138.0=T (b) For ()o m m 08.1= 2k = ()( ) ()() ( ) 2/12 34193110054.1106.12.08.01011.908.12??? ?????????-?--- or 9210124.4?=k m 1- Then ??? ??-??? ??=8.02.018.02.016T ()()[] 109101510124.42exp -??-? or 51027.1-?=T _______________________________________ ()a k V E V E T o o 22exp 116-???? ? ?-???? ??? where ()2 22 E V m k o -= ()()() 34 1962710054.1106.1101121067.12---????-?= 1410274.7?=m 1- (a) ()()[] 14 141010274.72exp 121112116-?-??? ??-??? ???T []548.14exp 222.1-= 710875.5-?= (b) ()() 710875.510-?=T ()[] a 1410274.72exp 222.1?-= ()?? ? ? ??=?-6 14 10 875.5222 .1ln 10274.72a or 1410842.0-?=a m _______________________________________ Region I ()0 ()()()x jk B x jk A x 11111exp exp -+=ψ (incident) (reflected) where 2 12 mE k = Region II: ()()()x k B x k A x 22222exp exp -+=ψ where ()2 22 E V m k O -= Region III: ()()()x jk B x jk A x 13133exp exp -+=ψ (b) In Region III, the 3B term represents a reflected wave. However, once a particle is transmitted into Region III, there will not be a reflected wave so that 03=B . (c) Boundary conditions: At 0=x : ?=21ψψ 2211B A B A +=+ ?=dx d dx d 2 1ψψ 22221111B k A k B jk A jk -=- At a x =: ?=32ψψ ()()a k B a k A 2222exp exp -+ ()a jk A 13exp = ?=dx d dx d 32ψψ ()()a k B k a k A k 222222exp exp -- ()a jk A jk 131exp = The transmission coefficient is defined as * 11*3 3A A A A T = so from the boundary conditions, we want to solve for 3A in terms of 1A . Solving for 1A in terms of 3A , we find () {()()[]a k a k k k k k jA A 22212 22 131exp exp 4---+= ()()[]}a k a k k jk 2221exp exp 2-+- ()a jk 1exp ? We then find () () {()[a k k k k k A A A A 2212 22 21*33*11exp 4-= ()]2 2exp a k -- ()()[]} 2222 2 21exp exp 4a k a k k k -++ We have ()2 22 E V m k O -= If we assume that E V O >>, then a k 2 will be large so that ()()a k a k 22exp exp ->> We can then write ()() {()[]22 2 122221*33*11exp 4a k k k k k A A A A -= ()[]} 222 2 21exp 4a k k k + which becomes () () ()a k k k k k A A A A 2212 22 21*33*112exp 4+= Substituting the expressions for 1k and 2k , we find 2 2 2212 O mV k k = + and ()?? ??????????-=22 222122 mE E V m k k O ()()E E V m O -?? ? ??=2 22 ()()E V E V m O O ??? ? ??-??? ??=122 2 Then ()()??? ????????? ??-??? ?????? ??= E V E V m a k mV A A A A O O O 12162exp 22222 2 *33*11 ()a k V E V E A A O O 2*3 32exp 116-???? ? ?-???? ??= Finally, ()a k V E V E A A A A T O O 2*11* 332exp 116-???? ? ?-???? ??== _____________________________________ Region I: 0=V ()()?=+??0212212 x mE x x ψψ ()()()x jk B x jk A x 11111exp exp -+=ψ incident reflected where 2 12 mE k = Region II: 1V V = ()()()?=-+ ??0222 12 22x V E m x x ψψ ()()()x jk B x jk A x 22222exp exp -+=ψ transmitted reflected where () 2 122 V E m k -= Region III: 2V V = () () ()?=-+ ??0232 2232x V E m x x ψψ ()()x jk A x 333exp =ψ transmitted where () 2 232 V E m k -= There is no reflected wave in Region III. The transmission coefficient is defined as: * 11* 3 313*11*3313A A A A k k A A A A T ?=?=υυ From the boundary conditions, solve for 3A in terms of 1A . The boundary conditions are: At 0=x : ?=21ψψ 2211B A B A +=+ ???=??x x 2 1ψψ 22221111B k A k B k A k -=- At a x =: ?=32ψψ ()()a jk B a jk A 2222exp exp -+ ()a jk A 33exp = ???=??x x 3 2ψψ ()()a jk B k a jk A k 222222exp exp -- ()a jk A k 333exp = But ?=πn a k 22 ()()1exp exp 22=-=a jk a jk Then, eliminating 1B , 2A , 2B from the boundary condition equations, we find ()()2 313123121 1344k k k k k k k k k T += +?= _______________________________________ (a) Region I: Since E V O >, we can write ()()()0212212=--??x E V m x x O ψψ Region II: 0=V , so ()()0222222 =+??x mE x x ψψ Region III: 03=?∞→ψV The general solutions can be written, keeping in mind that 1ψ must remain finite for 0 ()()()x k B x k A x 22222cos sin +=ψ ()03=x ψ where ()2 12 E V m k O -= and 2 22 mE k = (b) Boundary conditions At 0=x : ?=21ψψ21B B = 22112 1A k B k x x =???=??ψψ At a x =: ?=32ψψ ()()0cos sin 2222=+a k B a k A or ()a k A B 222tan -= (c) 12122211B k k A A k B k ??? ? ??=?= and since 21B B =, then 2212B k k A ???? ??= From ()a k A B 222tan -=, we can write ()a k B k k B 22212tan ???? ??-= or ()a k k k 221tan 1???? ??-= This equation can be written as ??? ???????--=a mE E E V O 22tan 1 or ??? ??????-=-a mE E V E O 22tan This last equation is valid only for specific values of the total energy E . The energy levels are quantized. _______________________________________ ()2 22424n e m E o o n ∈-=π(J) ()2223 24n e m o o ∈-=π(eV) ( )() ( )[]( ) 22342123193110054.121085.84106.11011.9n ----????-=π or 2 58 .13n E n -= (eV) 58.1311-=?=E n eV 395.322-=?=E n eV 51.133-=?=E n eV 849.044-=?=E n eV _______________________________________ We have ??? ? ??-??? ? ???=o o a r a exp 112 /3100 πψ and * 10010024ψψπr P = ??? ? ??-???? ????=o o a r a r 2exp 1143 2ππ or ()??? ? ??-?=o o a r r a P 2exp 42 3 To find the maximum probability ()0=dr r dP ()() ???? ??-???????? ? ??-=o o o a r r a a 2exp 2423 ????????? ??-+o a r r 2exp 2 which gives o o a r a r =?+-=10 or o a r = is the radius that gives the greatest probability. _______________________________________ 100ψ is independent of θ and φ, so the wave equation in spherical coordinates reduces to ()()021222=-+ ??? ? ??????ψψr V E m r r r r o where ()r a m r e r V o o o 2 24 -=∈-=π For ??? ? ??-???? ???=o o a r a exp 112/3100πψ Then ??? ? ??-???? ??-??? ? ???=??o o o a r a a r exp 111 2 /3100π ψ so ??? ? ??-???? ???-=??o o a r r a r r exp 11 22 /51002 π ψ We then obtain 2 /5100211??? ? ???-=???? ??????o a r r r πψ ??? ? ???????? ??-???? ??-???? ??-?o o o a r a r a r r exp exp 22 Substituting into the wave equation, we have ??? ????????? ??--???? ??-???? ???-o o o o a r a r a r r a r exp exp 21122/52π ??? ???++r a m E m o o o 222 0exp 112 /3=???? ??-??? ? ??????? ???o o a r a π where ()22 2 241224o o o o a m e m E E -=∈-==π Then the above equation becomes ???????????--???????????? ??-???? ???o o o o a r r a r a r a 222 /321exp 11π 022222=????? ???? ??+-+r a m a m m o o o o o or ???????????? ??-???? ???o o a r a exp 112 /3π 0211222=?? ???????????? ??+-++-?r a a a r a o o o o is which gives 0 = 0 and shows that 100 indeed a solution to the wave equation. _______________________________________ All elements are from the Group I column of the periodic table. All have one valence electron in the outer shell. _______________________________________