半导体物理与器件第四版课后习题答案2

Chapter 2

Sketch

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Sketch

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Sketch

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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

_______________________________________

h E =

?==λλν Gold: 90.4=E eV ()()

19106.190.4-?= J So,

(

)(

)

()(

)

519

341054.210

6.190.410310625.6---?=???=λcm or

μλ254.0=m

Cesium: 90.1=E eV ()()

19106.190.1-?= J

So,

(

)(

)

()(

)

103410

54.6106.190.110310625.6---?=???=λcm or

μλ654.0=m

_______________________________________

(a) 9

55010625.6--??==λh p 2710205.1-?=kg-m/s

331

1032.11011.9102045.1?=??==--m p υm/s or 51032.1?=υcm/s (b) 9341044010625.6--??==λh p 2710506.1-?=kg-m/s 331

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

1012.110915.510625.6---?=??==p h λm or o

A 2.11=λ

(ii)()()()

1931106.1121011.92--??=p 241087.1-?=kg-m/s

1024

1054.3108704.110625.6---?=??=λm or o A 54.3=λ

(iii) ()()()

1931106.11201011.92--??=p 2410915.5-?=kg-m/s

1024

1012.110915.510625.6---?=??=λm or o

A 12.1=λ (b)

()()()

1927106.12.11067.12--??=p 23

10532.2-?=kg-m/s

1123

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

106.103885.010

11.92--??=

2510064.1-?=avg p kg-m/s Now

925

10225.610064.110625.6---?=??==p h λm or

A 25.62=λ

_______________________________________

p p hc

h E λν==

Now

p E e

e 22= and

21???

? ??=

?=e

e e e h m E h

p λλ Set e p E E = and e p λλ10=

Then

102121???

? ??=

???? ??=

p e

p h m h

m hc

λλλ which yields

p 2100=λ

100

221002

mc mc h hc hc E E p p =

?===λ ()()

100

1031011.922

31??=

151064.1-?=J 25.10=keV

_______________________________________

(a) 10

108510625.6--??==λh p

2610794.7-?= kg-m/s

2610

56.81011.910794.7?=??==--m p υm/s or 61056.8?=υcm/s

()()

243121056.81011.92

21??==-υm E

211033.3-?=J

or 219

1008.210

6.110334.3---?=??=E eV (b) ()()

23311081011.92

??=-E

2310915.2-?=J

or 419

1082.110

6.110915.2---?=??=E eV ()()

3311081011.9??==-υm p

2710288.7-?=kg-m/s

827

1009.910288.710625.6---?-??==p h λm or o

A 909=λ

_______________________________________

(a) ()()

3410

110310625.6--???===λνhc h E 15

1099.1-?=J Now

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

1010.11002.610625.6---?=??==p h λm or

A 11.0=λ

_______________________________________

1010054.1--?=

?=?x p 2810054.1-?=kg-m/s

_______________________________________

(a) (i) =??x p

2610

10783.810

1210054.1---?=??=?p kg-m/s (ii)p m p dp d p dp dE E ?????

??=??=?22 m

p p m p ?=

??=

22 Now mE p 2=

()()()

1931106.1161092--??= 2410147.2-?=kg-m/s so ()()

2410

910783.8101466.2---???=?E 1910095.2-?=J

or 31.1106.110095.219

=??=?--E eV

(b) (i) 2610783.8-?=?p kg-m/s (ii)()()()

1928106.1161052--??=p 231006.5-?=kg-m/s

()(

)

2310

510783.81006.5---???=?E 2110888.8-?=J

or 219

1055.510

6.110888.8---?=??=?E eV _______________________________________

322

10054.110

10054.1---?=?=?=?x p kg-m/s

1500

10054.132

-?=

?=??=m p m p υυ 36107-?=?υm/s

_______________________________________

(a) =??t E

()()

341023.8106.18.010054.1---?=??=?t s

(b) 10

105.110054.1--??=

?=?x p 251003.7-?=kg-m/s

_______________________________________

(a) If ()t x ,1ψ and ()t x ,2ψ are solutions

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 Adding the two equations, we obtain

()()[]t x t x x m ,,2212

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??

which can be written as

???????ψ???ψ?+?ψ?ψ+?ψ?ψ-x x x x m 212

122

21222

()[]???????ψ?ψ+?ψ?ψ=ψ?ψ+t t j x V 12

121 Dividing by 21ψ?ψ, we find

????ψ??ψ?ψψ+?ψ??ψ+?ψ??ψ-x x x x m 2121212

1222

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

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

2=???

???

+-dx x A π

()12sin 23

12=??

????++-ππx x A 121232=????????? ??--A

so 21

2=A

or 2

_______________________________________

()1cos 22

/12

=?+-dx x n A

()142sin 22

/12/12=??????++-ππn x n x A ??? ??==??

??????? ??--211414122A A

or 2=A

_______________________________________

Note that

*=ψ?ψ?

∞

Function has been normalized.

(a) Now

dx a x a P o

a o o 2

exp 2?

???

???????? ??-=

dx a x a o

a o o

????

??-=

2exp 2

402exp 22

o a o o o a x a a ???

? ??-????

?-=

()??? ??--=???

?????-????

??--=21exp 1142exp 1o

a a P which yields 393.0=P (b)

dx a x a P o o

a a o o 2

exp 2?

???

???????? ??-=

dx a x a o o

a a o o

???

??-=

2exp 2

242exp 22

o o

a a o o o a x a a ???? ??-????

?-=

()()???

???? ??----=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

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

/2cos 2 2

/4/42sin 22a a a a x x a ???

????

???

???? ????? ??+??

? ??=ππ

()???

???

???? ????? ??-

-??? ??+??

? ??=a a a a a ππππ42sin 84sin 42 ??

????--+=π4181041

or 0908.0=P

????? ??=

+-π22

/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

/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

/ 2

/4/244sin 2

2a a a a x x a ???

????

???

???? ????? ??-??? ??=ππ

()()??

????

???? ??+

??? ??-??? ??-??? ??=a a a a a ππππ8sin 882sin 42 or 25.0=P

+-π2sin 222

/ 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

1010

8108=??==k p ωυm/s or 610=p υcm/s

10854.710822-?=?==ππλk m

or o

A 54.78=λ (ii)()()

431101011.9-?==υm p

271011.9-?=kg-m/s

()()

24312101011.92

21-?==υm E

2310555.4-?=J

or 419

1085.210

6.110555.4---?=??=E eV (b) (i) 49

1010

5.1105.1-=?-?==k p ωυm/s or 610-=p υcm/s

991019.410

5.122-?=?==

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

106.1025.0υm E =

?=- ()

2311011.92

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

1076.710537.810625.6---?=??==p h λm

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

3410

454.110555.410625.6---?=??==p h λm

1032.410

454.122?=?==-πλπk m 1- ()()

481051032.4??==υωk

131016.2?=rad/s (b) ()()

631101011.9-?=p

251011.9-?=kg-m/s

3410

27.71011.910625.6---?=??=λm 910

1064.810

272.72?=?=-π

k m 1-

()()

15691064.8101064.8?=?=ωrad/s

_______________________________________

()

()()

3122

34222

2210751011.9210054.12---???=

=ππn ma n E n

()21

100698.1-?=n E n

()

2106.1100698.1--??=n E n

or ()

3210686.6-?=n E n eV Then

311069.6-?=E eV

21067.2-?=E eV

231002.6-?=E eV

_______________________________________

(a) (

)

(

)(

)

3122

34222

2210101011.9210054.12---???==

ππn ma n E n (

)20

018.6-?=n J

()

()3761.010

6.110018.6219

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

()()

341001.310310625.6--???=λ

710604.6-?=m or 4.660=λnm

_______________________________________

(a) 2

222ma n E n π =

()

(

)()2

3423

102.110

15210054.110

15----???=

?πn

()62

310538.21015--?=?n

or 2910688.7?=n (b) 151?+n E mJ (c) No

_______________________________________

For a neutron and 1=n :

(

)

(

)(

)

2722

342

221101066.1210054.12---??==ππma E

13103025.3-?=J

13110

06.2106.1103025.3?=??=--E eV For an electron in the same potential well: ()()()

1431

341

101011.9210054.1---??=

πE

10100177.6-?=J or

919

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

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 mE

k =

Boundary conditions: ()0=x ψ at 2

,2a x a x +=-=

First mode solution:

()x k A x 111cos =ψ where

2112ma E a k ππ=?=

Second mode solution: ()x k B x 222sin =ψ where

222242ma

E a k ππ=?= Third mode solution: ()x k A x 333cos =ψ where

233293ma E a k ππ=

?= Fourth mode solution: ()x k B x 444sin =ψ where

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

Use separation of variables, so let ()()()()z Z y Y x X z y x =,,ψ

Substituting into the wave equation, we obtain

22222

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 y Y Y x X X We may set

012

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

221y k y Y Y -=??? and 2221z k z

Z Z -=???

Applying the boundary conditions, we find

n k y y π=, ....3,2,1=y n

n k z z π

, ...3,2,1=z n From Equation (1) above, we have

02222=+---k k k k z y x

222

22 mE

k k k k z y x =

=++ so that

()

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?-=??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?-=??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

222n n ma

E -=?π For 1,212==n n Then

223ma E π =?

(i) For o

A a 4= (

)

(

)()

1027

3410410

67.1210054.13---???=

?πE

2210155.6-?=J

or 319

1085.310

6.110155.6---?=??=?E eV

(ii) For 5.0=a cm

()

(

)()

227

34105.010

67.1210054.13---???=

?πE

3610939.3-?=J or

1719

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

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

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

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

12*22*2

2???? ??+-==k k k k B B A A R The transmission coefficient is

()

212

141k k k k T R T +=?-= _______________________________________

()()x k A x 222exp -=ψ ()

()x k A A x P 2*2

2exp -==

where ()2

22 E V m k o -=

()()(

)34

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

3110054.1106.11.00.11011.92---??-?=

or 2k 910860.4?=m 1-

(a) For 10104-?=a m

()()[]

910

41085976.42exp 0.11.010.11.016-??-??? ??-??? ???T 0295.0=

(b) For 101012-?=a m

()()[]

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

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= ()

E V m k O -= ()(

)

()()

(

)

/1234193110054.1106.12.08.01011.9067.02??

??????????-?=---

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 -=

()()()

1962710054.1106.1101121067.12---????-?=

1410274.7?=m 1- (a) ()()[]

141010274.72exp 121112116-?-??? ??-??? ???T []548.14exp 222.1-=

710875.5-?= (b)

()()

710875.510-?=T

()[]

a 1410274.72exp 222.1?-=

()??

??=?-6

875.5222

.1ln 10274.72a or 1410842.0-?=a m

_______________________________________

Region I ()0, 0=V (a) Region I:

()()()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 .

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

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

21*33*11exp 4-=

()]2

2exp a k --

()()[]}

2222

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

122221*33*11exp 4a k k k k k A A A A -= ()[]}

222

21exp 4a k k k + which becomes

()

()a k k k k k A A A A 2212

21*33*112exp 4+=

Substituting the expressions for 1k and

2k , we find

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

*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

12 mE

k =

Region II: 1V V =

()()()?=-+

??0222

22x V E m x

x ψψ

()()()x jk B x jk A x 22222exp exp -+=ψ

transmitted reflected where ()

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

()

232 V E m k -=

There is no reflected wave in Region III. The transmission coefficient is defined as:

11*

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

()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

.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

/3100π

ψ so

???

??-???? ???-=??o o a r r a r r

exp 11

/51002

ψ We then obtain

/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

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.

_______________________________________