Giai Bai Tap Vat Ly Luong Tu - Nguyen Tu

7/29/2019 Giai Bai Tap Vat Ly Luong Tu - Nguyen Tu

1/28

Chng 1: Tnh chat hat cua anh sang

1.1) Bc xa nhiet1) Tm nang suat phat xa toan phan cua VT nhiet o 400oC.

Giai:

Nang suat phat xa toan phan cua VT, nhiet o 400oC

Ap dung nh luat Stefan-Bolzmann (1.1)

4T.)T(I =

Vi ;428 K.m.W10x670,5 = 673273400T =+=

24812 m/W10x163,110x10x63,1)T(I ==

2) Tm bc song ng vi nang suat bc xa cc ai cua:a) C the ngi (370C)b) Day toc bong en ien 3000 Kc) Be mat cua mat tri (6000 K)d) Khi bom nguyen t no (107 K)

Gia s rang cac vat tren eu la VT.

Bc song ng vi nang suat bc xa cc ai:Giai:

Theo nh luat Wien (1.2):

K.m2898T.max =

Suy ra: ]m[T

2898max =

a) C the ngi: ]m[300

2898

max

=

b)Day toc bong en ien: ]m[3000

2898max =

c) Be mat cua mat tri: ]m[6000

2898max =

d)Khi bom nguyen t no: ]m[10x2898 7max =

1

3) Tm nang lng phat ra t mot dien tch 1cm2 tren be mat mot VT trong mot giay. Biet bcsong ng vi nang suat phat xa cc ai cua no bang 0,484m.

Giai:


2/28

Nang lng phat ra t dien tch 1cm2 tren be mat cua VT trong 1s chnh la nang suat phat

xa toan phan:

]s/cm/watt[T.I 224=

Ma theo nh luat Wien ta co

K.m2898T.max = 603,5987m484,0

K.m28982898T

max

=

=

=

22

24

48

s/cm/W8,7287cm10

)603,5987(x10x670,5I ==

4) Mot lo nung co nhiet o 1000K. Ca so quan sat co dien tch 250cm 2. Xac nh cong suat bcxa cua ca so o neu coi lo la vat en tuyet oi.(4.1)

Giai: Ta co cong suat bc xa:W5.1417)10x250()K1000)(K.m.W10x67.5(S.T.S.IW 444284 ====

5) Tm nhiet o cua mot lo neu mot lo nho cua no kch thc (2x3)cm 2, c moi giay phat ra 8,28calo. Coi lo nh mot vat en tuyet oi.(4.2)

Giai: Ta co

K1009T

10x018815.110x3x2x10x67.5

186.4x28.8

S.

WT

S.T.S.IW

12

48

4

4

=

==

=

==

Lu y: 1cal/s = 4.186W

6) Vat en tuyet oi co hnh dang mot qua cau ng knh d=10cm, mot nhiet o khong oi,Tm nhiet o cua no biet cong suat bc xa nhiet o nay la 12kcal/phut.(4.3)

Giai:

K828T

)10.(.10x67.5

)10x0697667.0)(phut/kcal12(

d

W

4

d4.

W

S.

WT

S.T.S.IW

218

3

22

4

4

=

=

=

=

=

==

2

Lu y: KW0697667.0KW10x186.460

10s/cal

60

10phut/kcal 3

33

=== 1

7) Tnh lng nang lng bc xa trong mot ngay em t mot ngoi nha gach trat va, co dien tchmat ngoai tong cong la 1000 m2. Biet nhiet o cua mat bc xa la 27oC va he so hap thu khi o bang


3/28

0,8. (4.5)

Giai:

Lng nang lng bc xa trong mot ngay em

W10x17.3)3600x24)(8.0)(1000()27273(x10x67.5

.S.T.t.WA

1048

4

=+=

==

8) Be mat kim loai nong chay co dien tch 10cm2 moi phut bc xa mot lng nang lng 4x104J.Nhiet o be mat la 2500K, tm:

a. Nang lng bc xa cua mat o neu coi no la vat en tuyet oi.b. Ty so gia cac nang suat bc xa toan phan cua mat o va cua vat en tuyet oi cung motnhiet o. (4.11)

Giai:

Cau a) Nang suat phat xa toan phan khi coi be mat kim loai la vat en tuyet oi la:

4T.I =

Nang lng do ca be mat S phat xa trong thi gian t = 1phut = 60s la:

J10x33.160x10x10x)2500(x10x67.5

t.S.T.

t.S.IA

5448

4

==

=

=

Cau b) Ty so 3.0J10x33.1

J10x45

4

===envatAxabcA

1.2) Hieu ng quang ien Compton1) Tm ong nang ln nhat theo n v eV cua quang electron neu cong thoat cua vat lieu la 2,33

eV va tan so cua bc xa la 3,19x1015 Hz ?

Giai:Theo (1.8)

0max WhK = vieV10x242,6J1

s.J10x626,6h

18

34

=

=

Ma

eV194,1310x9367795,131

eV10x242,6x10x14,21

J10x14,21

10x19,3x10x626,6h

1

1819

19

1534

==

=

=

=

Suy ra: eV86,1033,2194,13Kmax ==

3


4/28

2) Tm ong nang ln nhat theo n v eV cua quang electron neu ngng quang ien cua vat lieula 2800A0 va bc song cua bc xa la 1900A0 ?

Giai:

=

=

===

)10x2800

1

10x1900

1)(s/m(10x3).S.J(10x626,6

)

11

(hchhwhK

1010

834

000max

3) Tnh nang lng cua moi photon trong mot chum anh sang n sac cua a) song vo tuyen, cobc song = 1km; b) anh sang vang, co = 0,590m va c) cua tia X, co bc song = 1 A0. Tnh

nang lng ra n v eV.

Giai:Nang lng cua moi photon cua anh sang n sac:

=

c.hE

a) Song vo tuyen: km1=18

834 10x242,6x

1000

10x3x10x626,6E

=

b) Anh sang vang: m590,0 =S dung: s.eV10x14,4h 15=

6

815

10x590,0

10x3x10x14,4E

=

c) Tia X: 0A1=10

815

10

10x3x10x14,4E

=

4) Tnh bc song ' cua tia tan xa goc 90o oi vi tia X mem, = 0,2 nm va tia X cng, =0,02 nm.

Giai:

a) oi vi tia X mem: 0A2nm2,0 ==

Ap dung cong thc (1.15)

)cos1(

cm

h'

e

+=

Ta co: )90(cm

h)cos1(

cm

h 0c

ee

===

4


5/28

Bc song Compton 0834

34

c A0243,0s/m10x3kgx10x1,9

s.J10x626,6==

Vay: 000 A0243,2A0243,0A2' =+=

b) oi vi tia X cng: 0A2,0nm02,0 ==

Suy ra: 000 A2243,0A0243,0A2,0' =+=

5) Mot tia X bc song = 0,3A0 b tan xa di goc 300 do hieu ng Compton. Tnh bc song cua tia tan xa va ong nang cua electron.

Giai:

Tng t bc song tan xa:

)30cos1.(0243,0A3,0' 00 +=

ong nang electron:

)'

11.(hc'hhE e

==

6) Trong th nghiem Compton, photon trc tan xa co nang lng 50 keV. Tnh nang lng cua tiatan xa di goc 600.

Giai:

eV50000keV50c

hhE ====

eV10x5

10x3.s.eV10x14,4

E

hc4

815

==

Suy ra: )60cos1.(' 0c +=

'

s/m10x3.s.eV10x14,4

'

ch'E

815

=

=

7) Xac nh nang lng, ong lng, va khoi lng cua photon ng vi anh sang co bc song=0.6m. (v du 1)

Giai:

a) Nang lng J10x32.3)10x6.0

s/m10x3)(s.J10x626.6(

chE 19

6

834

==

=

b) ong lng s/m.Kg10x1.110x6.0

10x626.6h

c

Ep 27

6

34

==

==

c) Khoi lng: Kg10x68.310x3x10x6

10x626.6.ch

ch

cEm 36

87

34

22

=====

8) Gii han o ( ngng quang ien ) trong hien tng quang ien oi vi Cs la 0.653m. Xac5


6/28

nh van toc cc ai cua quang electron khi chieu Cesi bang anh sang tm co bc song 0.4 m. (v du

2).

Giai:

Ta co m10x4.0m4.0

m10x653.0m653.0

6

6

ng

==

==

Ta lai co:2

mv

)11

(hcc

hc

hwhKmax

2

00

0max =

=

==

Suy ra:

s/m10x5.6v

10).653.0

1

4.0

1(

Kg10x11.9

10x3x10x626.6x2)

11(

m

hc2v

5

max

6

31

834

0max

2

=

=

=

Lu y: Kg10x11.9m 31e =

9) Trong hien tng tan xa Compton, chum tia ti co bc song . Hay xac nh ong nang cuaelectron ban ra oi vi chum tan xa theo goc . Tnh ong lng cua electron o. ( v du 3)

Giai:

a) Xac nh ong nang cua electron ban ra

Ta co:+

==+==hchc

'hhK'hKhE

Ma )cos1(22

sin2 c2

e =

=

Suy ra:

)

2

sin2

2sin2

(hc

)

2

sin21

12

sin21

(hc

)

2sin21

11(

hc

2sin2

hchcK

2

c

2

c

2c

2c

2c2

c

+

=

+

+

=

+

=

+

=

Khi ong nang e at cc ai.=

c

cmax

2

2hcK

+

=

b) ong lng cua electron.

S dung nh luat bao toan ong lng

p'ppe

=+

Suy ra:

+

=+= cos'

h2)

'

h()

h(cos'pp2'ppp

22222

e

2

6


7/28

Biet2

sin2' 2c

+= ta tnh c ep

10)Tm cac ngng quang ien oi vi Liti, Natri, Kali, Cesi, biet cong thoat W cua electrontng ng vi cac kim loai o lan lt la: 2,4eV; 2,3eV; 2,0eV; va 1.9eV.(4.28)

Giai:

a) oi vi Li:0

0

0

0w

hcchev4.2W =

==

m517.010x6.1x4.2

)s/m0x3)(s.J10x626.6(19

834

0 ==

Lu y: 1 J10x6.1eV 19=

b) Na: m54.0

c) K: m62.0

d) Cs: m66.0

11)Tm van toc cc ai cua cac quang electron ban ra t be mat Cs va Pt khi chieu vao chung lanlt cac chum bc xa co bc song (4.30)

a. =1850Ao

b. =4227Ao

Cho biet cong thoat cua Cs la 1.9eV cua Pt la 4.09eV.

Giai:

d) oi vi Cs:

Ta co:

m

)Wc

h(2

v

Wh2

mv

WhK

02

0

2

0max

=

=

=

Suy ra

s/m10x3.1Kg10x11.9

)10x6.1x9.110x1850

10x3x10x626.6(2

v 631

19

10

834

=

=

e) oi vi Pt:s/m10x05.7v 5=

7


8/28

12)Khi chieu chum anh sang vao kim loai co hien tng quang ien xay ra, Neu dung mot hieu thekhang ien la 3V th cac quang electron b ban khoi kim loai b gi lai ca, khong bay sang anode c.

Biet tan so gii han o( tan so ngng ) cua kim loai la 6x1014s-1, hay tnh (4.32)

a. Cong thoat cua electron oi vi kim loai o.b. Tan so cua chum anh sang ti

Giai:

a) Cong thoat cua electron oi vi kim loai o:Tan so gii han o (tan so ngng) lien he vi cong thoat qua he thc

eV48.210x6.1

)s10x6)((s.J10x626.6(hW

19

11434

00 ===

Lu y: 1 J10x6.1eV19

=b) Tan so cua chum anh sang tiTa co cong dch chuyen electron qua hieu the khang ien bang ong nang cc ai cua

electron

0U

0max eUK =

Mac khac theo he thc Einstein: 0max WhK =

Suy ra: 000max WeUWKh +=+=

Vay: 1151434

19

0000 s10x32.110x6

10x626.6

10x6.1eVx3

h

eU

h

WeU

====

13)Hay xac nh hang so Planck, biet rang khi lan lt chieu bc xa tan so 1=2,2x1015s-1 va2=4,6x1015s-1 vao mot kim loai th cac quang electron ban ra eu b gi lai bi hieu ien the khang

ien U1 =6,5V va U2=16,5V (coi nh a biet ien tch electron va van toc anh sang). (4.33)

Giai:

Ta biet cong dch chuyen electron qua hieu the khang ien U ve tr so tuyet oi bang ong nang

cc ai cua electron.

eUKmax =

Mac khac theo he thc Einstein: 0max WhK =

Suy ra: 000max WeUWKh +=+=

Ap dung cong thc tren cho hai bc xa co tan so 21, tng ng vi cac hieu the khang ien

ta c:21

U,U

011 WeUh +=

022 WeUh +=

8


9/28

Suy ra: s.J10x66667.610)2.26.4(

)5.65.16(e)UU(eh

34

15

12

12 =

=

=

14)Dung nh luat bao toan ong lng va nang lng tng oi tnh, chng minh rang motelectron t do khong the hap thu hoan toan mot photon.(4.46)

Giai:

Chon he quy chieu gan lien vi electron trc khi hap thu photon. Neu e hap thu hoan toan 1

photon th ta co no phai thoa nh luat bao toan nang lng va xung lng. Ta co:

lbtnl:

2

2

2

e2

e

c

v1

cmhcm

=+ (1)

lbtxl

2

2

e

c

v1

vm

c

h0

=

+ (2)

Nhan (2) cho c, roi lay (1) tr (2) ta co:

2e

2

2

e

2

2

2

e cm

c

v1

cvm

c

v1

cm =

Chia 2 ve cho ta co:2e cm

1

c

v1

cv

c

v1

1

2

2

2

2=

hayc

v1

c

v11

c

v1

c

v1

2

2

2

2==

Hayc

v2

c

v1

c

v1

2

2

2

2

+= Suy ra 0)1c

v(

c

v0

c

v

c

v2

2

==

T o suy ra

a) Hoac a en vo ngha0v = 0=b) Hoac 1

c

v= : mau thuan vi thuyet tng oi

15)Chng minh rang mot electron t do khong the phat xa mot photon. (4.47)Giai:

Ly luan tng t, gia s e phat xa ra photon th ta co:9


10/28

lbtnl:

+

= h

c

v1

cmcm

2

2

2

e2

e

lbtxl

2

2

e

c

v1

vm

c

h0

+

=

T o chng minh no dan ti mau thuan.

16)Xac nh o tang bc song va goc tan xa trong hien tng Compton, biet bc song ban aucua photon la =0.03Ao va van toc cua electron bay ra la v= c=0,6 c. (4.50)

Giai:

ong nang cua electron ban ra bang nang lng e sau tan xa tr i nang lng ngh ban au:

)11

1(cmcm

c

v1

cmE

2

2

e

2

e

2

2

2

e

=

=

ong nang o ve tr so bang o giam nang lng cua photon sau khi tan xa:

'

hchc'hh =

Vay )11

1(cm

'

hchc

2

2

e

=

Suy ra: 0

2

e

A0135.0

)11

1(

h

cm1

1' =

=

Biet suy ra = '

Va goc tan xa c tnh bi:c

2

22sin

=

. Tc la ' 4063o=

17)Xac nh bc song cua bc xa Rontgen. Biet rang trong hien tng Compton cho bi bc xao, ong nang cc ai cua electron ban ra la 0,19MeV. (4.51)

Giai:

Theo bai 3, ta co ong nang cc ai cua electron ban ra la

c

cmax

22hcK

+

=

10


11/28

Suy ra 0max

2

e

e

A037.01K

cm21

cm

h=+=

18)Dung nh luat bao toan ong lng va cong thc Compton, tm he thc gia goc tan xa vagoc xac nh phng bay ra cua electron. (4.57)

Giai:

Goi 'p,p la ong lng cua photon trc va sau khi tan xa.

ep la ong lng cua electron ban ra ( ban au electron ng yen)

Theo lbtl

p'pp e =+

Goc gia cac vec t p va 'p la ; p va ep la Ve hnh

Theo hnh ve ta co:

=cos'pp

sin'ptg

Ma:

2sin2

h

'

h'p;

hp

2

c

+

=

=

=

Do o

+

=

+

+

=c

2

c

2c

1

2gcot

2sin2

coshh2

sin2

sinh

tg

11


12/28

19)Mot photon co bc song =0,11Ao bay en va cham vao electron va b tan xa theo goc 110o;con elctron bay ra theo goc 30o. Coi nh a biet khoi lng cua electron va van toc anh sang, tnh

hang so Planck. (4.58)

Giai:

Dung ket qua tren:

+

=

+

=

cm

h1

2gcot

1

2gcot

tg

e

c

Suy ra s.J10x4.6)1tg 2gcot

(cmh 34e =

=

12


13/28

Chng 2: Lng tnh song hat

2.1) Gia thuyet De Broglie1) Cac hat electron, proton, neutron va co cung ong nang, hat nao co bc song be hn.

Giai:

ong nang: = 2mv2

1T xung lng mT2p =

Bc song:mT2

h

p

h==

Vay hat co khoi lng ln (alpha) se co bc song be hn

2) Th nghiem nao cho thay moi electron eu co tnh chat song?Giai: Th nghiem Bibermann va Sushkin

3) Mot photon co nang lng Ep = 1,5 eV va mot electron co ong nang Ke = 1,5eV. Tnh bcsong cua chung, nhan xet.

Giai:

Bc song cua photon:

m828,0m10x28,85,1

10x3x10x14,4

E

ch 7

815

p ====

Bc song cua electron:

nm00201,110x00201,110x6,1x5,1kgx10x11,9x2

s.J10x626,6

T.m.2

h

p

h 91931

34

e ====

Vay electron co bc song ngan hn photon co cung nang lng.

4) Bc song cua 1 proton la = 0,113 pm (1 pm = 10-12 m).a) e gia toc cho proton co c van toc t v = 0, can dung hieu ien the bao nhieu?b) Tnh van toc cua proton.Giai:

a) Ta co: 0mv2

1V.q 2 =

Suy ra:

Volt9,64339

]c

s/m.kg[

10x6,1

)10x51121,3(x10x67,1x

2

1

q

vm

2

1V

22

19

2627

2

=

==

b)

Van toc cua proton

13

s/m51121,3m10x113,0x10x67,1

s.J10x626,6

m

hv

mv

h

p

h1227

34

==

===


14/28

5) Hat electron ban au ng yen c gia toc qua mot hieu ien the U. Xac nh bc songDeBroglie cua electron sau khi c gia toc trong 2 trng hp: ( v du 1, p77)

a. U=51Vb. U=510kV

Giai:

Cho U hoi coi nhu a biet em,e

Ta biet cong cua lc ien trng:

== maxKeU ong nang cua electron.

a) Trng hp U=51VV U khong ln nen van toc cua electron thu c khong ln lam ta co the s dung cac cong thc

trong c hoc phi tng oi ( c hoc Newton)

e

22

e

m2

p

2

vmeU ==

Suy ra: eUm2p e=

Bc song DeBroglie:

eUm2

h

p

h

e

==

Ta co Mev10x51.0eV51eUV51U 4===

Ma 2e42

e cm10eUcmMev51.0==

0

c

2

e

2

e

2

2

e

4

e

A72.12

10

cm

h

2

10

2cm

h10

cm10m2

h=====

ay: 0c A0243.0=

b) Trng hp U=510KV

Mev51.0KeV510eU ==

Ngha la ong nang cua electron=nang lng ngh cua electron.

Vay ta phai ap dung tng oi tnh.

ong nang cua electron bang

eUcm

c

v1

cm 2e

2

2

2

e =

14


15/28

eUcm

c

v1

cm 2e

2

2

2

e +=

2e

2

e

2e

2

e

2

2

cmeU

)cm2eU(eUcv

eUcm

cm

c

v1

+

+=

+=

Suy ra ong lng cua electron la:

2

e

2

e

2

e

2

ee

2

2

e

cm

cmeUx

cm2eU

)cm2eU(eUcm

c

v1

vmp

+

+

+=

=

)cm2eU(eUc

1 2e+=

Bc song De Broglie

)cm2eU(eU

hc

p

h

2

e+==

Ma 2e cmMev51.0eU ==

Suy ra: 0c2

e

2

e

2

e

A014.03

1

)cm2cm(cm

hc==

+=

6) Hoi phai cung cap cho hat electron them mot nang lng bang bao nhieu e cho bc songDeBroglie cua no giam t 100x10-12m en 50x10-12m? (5.6)

Giai:

Dung cong thc tng oi tnh (xem bai 1) ta c

)cm2eU(eU

hc

2

e+= ma eU=K

Suy ra

)cm2K(K

hc

2

e+=

Vi2

12

2

1

12

1

Km10x50

Km10x100

=

=

Suy ra: KeV45.0KKK 12 ==

2.3) He thc bat nh Heisenberg1) Mot vien an (m=50g) va mot electron (m=9,1.10-28 g) c xac nh co cung toc o v = 300

m/s, vi cung mot o bat nh 0,01%. Tm sai so kha d cua toa o cua chung, neu toa o c o ong

thi vi van toc trong cung mot th nghiem. Neu nhan xet.

15


16/28

Giai:

Ap dung he thc bat nh Heiseinberg

p.2x

2p.x

hh

Mav.m.2

xvmpmvp

== h

* oi vi vien an:

26

3

34

d

10x52,3300x%01.0x10x50..4

s.J10x626,6

v

vm.2

v/x

=

=

h

m

* oi vi electron:

m00193.0300x%01,0kgx10x11,94

s.J10x626,6

v

vm.2

v/x31

34

d

=

=

h

Suy ra sai so o toa o oi vi electron ln hn oi vi vien an rat nhieu c lan.2210x48,5

2) Mot hat nhan chuyen t trang thai kch thch ve trang thai c ban va phat ra 1 photon. Thi giansong trung bnh trang thai kch thch la 8,7 ps. Tm o bat nh trong nang lng cua photon.

Giai:

Ap dung:2

t.E h

Lay Suy ra:ps7,8t == eV0000378679.0s10x7,84

eV10x14.4

t4t.2E

12

15

=

=

=

hh

3) Thi gian song trung bnh cua 1 nguyen t 2 trang thai kch thch khac nhau la 12 ns va 23 ns.Tm o bat nh cua nang lng cua photon phat ra khi nguyen t chuyen t trang thai nay sang trang

thai kia.

Giai:

Ap dung:2

t.Eh

Pho ton phat ra chu s bat nh cua 2 trang thai kch thch

Trang thai kch thch 12ns cho o bat nh:

eV10x027468,0s10x124

eV10x14,4

t4

h

t.2E 6

9

15

11

1

=

=

=

h

Trang thai kch thch 23ns cho o bat nh:

eV10x0143312,0s10x234

eV10x14,4

t4

h

t.2E 6

9

15

22

2

=

=

=

h

16


17/28

Vay o bat nh tong cong: eV10x0418.0EEE 621=+=

4) Hat electron co ong nang T=15eV chuyen ong trong mot giot kim loai kch thc d=10-6m.Tnh o bat nh ve van toc ( theo %) cua hat o.(5.11)

Giai:

Ap dung he thc bat nh:

2x.p x

h

ay: vmpmvp;2

dx ===

Suy ra:22

d.v.m

h

mdv

h=

%01.0dmK2d.v.mv

v===

hh

5) ong nang cua electron trong nguyen t Hydro co gia tr vao c khoang 10eV. Dung he thcbat nh hay anh gia kch thc nho nhat cua nguyen t. (vi du 2, p.79)

Giai:

Gia s kch thc nguyen t bang l. Vay v tr cua electron theo phng x xac nh bi:

2

lxlx0

T he thc bat nh suy ra:x

xp

l2

p2

l

hh(*)

Ma o bat nh khong vt qua p:xp pp x


18/28

Suy ra:x2x2

pphh

=

Vay:x2

p minh

Nang lng cua dao ong t ieu hoa=ong nang + the nang2

2

22

222

2

kx2

1

mx8kx

2

1

m2

x4kx

2

1

m2

pE +=+=+=

hh

Gia tr cc tieu cua E ng vi 0dx

dE= , tc la

0kxmx4

03

0

2

=+ h

0kmx4

4

0

2

=+h

mk4

xmk4

x4

0

24

0

hh==

Vay nang lng nho nhat:

mk4

.k

2

1

m8

mk4

mk4

k

2

1

mk4m8

kx2

1

mx8E

2

2

02

0

2

min

hhh

h

h

h

+=+=

+=

2

2

1

m

k

4

1

m

k

4

1=+= hhh vi

m

k=

Vay: 2min2

1E = h PCM

7) Dung he thc bat nh xac nh o rong cua mc nang lng electron trong nguyen t Hydro trang thai ( 5.19)

a. C ban (n=1)b. Kch thch ng vi thi gian song s810

Giai:

a) trang thai c ban: Vay==t 0E b) trang thai kch thch: s10t 6==

Suy ra: eV102

E 7

h

18


19/28

Chng 3: Phng trnh Schrodinger

1) Chng minh rang neu 1 va 2 la hai nghiem cua phng trnh Schroedinger th = c11 +c22 cung la 1 nghiem cua phng trnh nay.

Giai:

Neu va la nghiem th no phai thoa (3.1)1 2

t

)t,z,y,x(i)t,z,y,x(.U)t,z,y,x(

m2

111

2

=+ hh

t

)t,z,y,x(i)t,z,y,x(.U)t,z,y,x(

m2

222

2

=+ hh

Gia s 2211 CC += thoa, ta co:

ti.U

m2

2

=+ hh

t

)CC(i)CC.(U)CC(

m2

221122112211

2

+

=+++ hh

t

)C(i

t

)C(i)C(U)C.(U)C(

m2)C(

m2

2211221122

2

11

2

+

=++ hh

hh

0t

)C(i)(U)(

m2C

t

)(i).(U)(

m2C 2222

2

21

11

2

1 =

++

+ h

hh

h

PCM.

2) Chng minh rang neu (x) la nghiem cua phng trnh Schroedinger dng (3.8), th(x).exp{-iEt/h} la nghiem cua phng trnh (3.2), trong o U khong phu thuoc vao t.

Giai

thoa (3.8) cho ta:

)x(.E)x(.Udx

)x(dm2

22

=+h

at the vao (3.2).)/iEtexp().x()t,x( h=

t

)/iEtexp()x((i)/iEtexp()x(.(U

x

)/iEtexp().x((

m2 2

22

=+

hhh

hh

)/t.E.iexp(.E).x(

)/iEtexp().iE.i

)(x()x(.(Ux

).x((

m2)/iEtexp(

2

22

h

hhh

hh

=

=

+

n gian 2 ve cho ta con lai (3.8). PCM)/iEtexp( h

3) Viet phng trnh Schrodinger oi vi hat vi mo:19


20/28

a. Chuyen ong trong trng the 22

1kxU =

Giai

0)kx

2

1E(

m2

dx

d 222

2

=+

h

b. Chuyen ong trong trng tnh ien Coulomb:r

ZekU o

2

= vio

ok4

1=

Giai

0)r

ZekE(

m2 2

02=++

h

c. Chuyen ong trong khong gian hai chieu di tac dung cua trng the 22

1krU =

Giai

0)yx(k2

1E

m2

dy

d

dx

d 2222

2

2

2

=

++

+

h

4) a)Tnh nang lng thap nhat c phep cua mot electron b gii han trong mot ho sau vo hanmot chieu co o rong bang ng knh hat nhan (khoang 1,4.10-14m).

b) Lap lai phep tnh tren cho mot neutron.

c) So sanh hai ket qua tren vi nang lng lien ket cua proton va neutron trong hat nhan

(khoang 10 MeV). Lieu ta co the ch i se thay electron trong hat nhan hay khong?.

Giai

Nang lng thap nhat c phep cua mot electron b gii han trong mot ho sau vo han mot

chieu co o rong bang ng knh hat nhan

Ap dung cong thc:

2

2

2

22

1a.m.8

ha.m.2

E == h

* Vi electron: me=9,11x10-31kg

J10x077,3)m10x4,1)(kg10x11,9)(8(

)s.J10x63,6(E 10

21431

234

1

==

MeV82,1920eV10x92082,1 9 ==

* Vi neutron: mn=1,68x10-27kg

MeV10eV10x04159,1)m10x4,1)(kg10x68,1)(8(

)s.J10x63,6(E 6

21427

234

1 ==

*Vay electron khong the ton tai trong nhan.

20


21/28

5) He so an hoi trong s dao ong cua mot phan t hai nguyen t vao khoang 103 joules/m2. Khoilng cua phan t la 4,1.10-26 kg.

a) c lng nang lng khong cua dao ong nay.

b) c lng hieu nang lng gia mc khong (n=0) va mc nang lng ng vi n=1.

c) T o hay tnh nang lng va tan so cua photon phat ra khi phan t chuyen t trang thai

ng vi n=1 ve trang thai ng vi n=0.

Giai

a) Nang lng dao ong khong

m/k4

h

2

m/k

2E 0

==

=hh

Chu y rang: 1J=6,6242x1018

eVh=6,626x10-34J.s

k=103J/m2

m=4,1x10-26kg

eV0514,0J10x23474,8kg10x1,4

m/J10

4

)s.J10x626,6(E 21

26

2334

0 ===

b) Nang lng dao ong vi n=1

eV154204,02

3E1 == h

Suy ra E1-E0 = 0,102802eV

c) eV102802,0EEE 01 ==

Tan so Hz10x48,2s.eV10x14,4

102802,0

h

EhE 13

15====

6)

Mot electron co ong nang K = 5,0 eV chuyen ong en mot hang rao co o cao U = 6 eV vao rong a = 0,70 nm. Tnh

a) bc song de Broglie cua electron,

b) he so truyen qua hang rao D cua electron,

c) he so truyen qua hang rao D cua electron neu o rong a giam con 0,30 nm.

d) lam lai cau b) cho trng hp o la proton thay v electron.

Giai

a) Bc song De Broglie

Trc khi electron ti b the, nang lng toan phan cua no ch la ong nang v the nang vung

nay bang khong.21


22/28

Suy ra 01931

34

A48823,5)J10x6,1)(eV0,5)(kg10x11,9)(2(

s.J10x626,6

mK2

h

p

h====

b) He so xuyen rao cua electron

)Ta2exp(D =

Vi2

0

2

h

)EU(m8T

=

Ta co:

eV/J10x60,1eV1

kg10x11,9m

10x626,6h

19

31

34

=

=

=

Vay19

234

19312

m10x11991,5

)s.J10x63,6(

)eV/J10x6,1)(eV5eV6)(kg10x11,9(8T

=

=

Suy ra: %077,000077,0)16788,7exp(10x7,0x10x1199,5x2exp(D 99 ====

c) He so xuyen rao D cua electron khi a giam con 0,3nm

Khi a=0,3nm th

%63,40463,0)07195,3exp()10x3,0x10x1199,5x2exp(D 99 ====

d) Trng hp cua proton: m=1,67x10-27kg

Khi o T=2,19211x1011.

Suy ra 2.T.a=306,895

Va D=5,21471x10-134 coi nh bang khong.

Vay he so truyen qua se vo cung nho oi vi hat nang.

22


23/28

Chng 4: Nguyen t

1) Cac electron trong nguyen t Hidro c kch thch e chuyen len trang thai ng vi so lngt chnh n = 4. Sau o cac electron se chuyen ve trang thai c ban (n =1). Co the quan sat thay bao

nhieu vach pho ?

Giai

T n=4 xuong n=1 co 03 vach thuoc day Lyman. Xem hnh di ay.

2) Tnh bc song dai nhat cua anh sang thuoc day Balmer.Giai

Bc song dai nhat ng vi s chuyen ve n=2 t n=3

m65,0m10x54545,6

)3

1

2

1(10x27,3

s/m10x3

)'n

1

n

1(R

c)

'n

1

n

1(R

c

7

22

15

28

22

22

==

=

==

=

3) Co bao nhieu trang thai cua electron trong nguyen t Hidro co cung so lng t chnh n = 3?Co bao nhieu trang thai cua electron trong nguyen t Hidro co cung nang lng 3,4 eV ?

Giaia) n=3 suy ra l=0, 1, 2

l=0 suy ra m=0

l=1 m=-1, 0, 1

l=2 m=-2, -1, 0, 1, 2

Vay cac trang thai la 3S0, 3P-1, 3P0, 3P1, 3D-2, 3D-1, 3D0, 3D1, 3D2

b) E = -3,4eV

Ta co:22

0

4

e1

)4(2

emE

h=

23


24/28

Suy ra: eV4,3n

EE

2

1n ==

Suy ra: 44,3

6,13

4,3

En 12 === . Vay n = 2. Co n2=4 trang thai.

4) Tnh goc nho nhat gia vect momen xung lng L va truc z, khi electron trang thai ng viso lng t qu ao l = 3.

Giai

Vi 3,2,1,0m3l == , hmL z =

Ve hnh

Da vao hnh ve thay : Goc nho nhat ng vi h3L z =

Theo hnh ta co:L

Lzcos =

Ma: hhh 12)13(3)1l(lL =+=+=

Suy ra: 23

12

3

L

Lz

cos === h

h

0

30=

5) Khi at nguyen t trong t trng ngoai, nang lng cua electron co gia tr phu thuoc so lngt nao?

Giai

Khi at nguyen t trong t trng ngoai, nang lng cua electron co gia tr phu thuoc vao so

lng t t m (ngoai so lng t n).

V t trng lam sinh ra nang lng phu:B.mBE B==

6) Tnh khoang cach gia hai vach pho ke tiep nhau trong hieu ng Zeemann, khi nguyen t c24


25/28

at trong t trng co B = 5 Tesla.

Giai

Khoang cach gia 02 vach pho ke tiep nhau trong hieu ng Zeemann khi nguyen t c at

trong t trng co B=5 Tesla

h

5xm.A10

h

B. 223B

==

s/10x995,6s/J10x626,6

Tesla5.Tesla/J10x27,9 10

34

24

==

7) Electron co hnh chieu cua momen t qu ao len phng z la z = 3B, electron o co the trang thai nao trong cac trang thai ng vi cac so lng t sau:

a) n=3, l=5; b) n=4, l=2; c) n=2, l=1; d) n =5, l = 4, e) n=2, l=3.

Giai

Ta co: 3m3 Bz ==

a) n=3 ; l=5 vo ly v l toi a la n-1

b) n=4 ; l=2 vo ly v l=2 th m toi a la l tc la 2

c) n=2 ; l=1 vo ly v m=3

e) n=2 ; l=3 vo ly v n=2 th l toi a la 1

d) n=5, l=4 hp ly

8) Nguyen t Hydro trang thai c ban hap thu photon nang lng 10,2 eV. Xac nh o bienthien momen xung lng quy ao cua electron, biet electron trang thai kch thch p. (6.13)l

Giai

Ban au nguyen t Hydro trang thai c ban. Nen electron mc l=0 ( trang thai s)

Suy ra 0)1l(lL =+= h

Khi kch thch trang thai p: l=1

Suy ra hh 2)1l(lL =+=

Vay: h2L =

9) oi vi electron hoa tr trong nguyen t Na, hoi nhng trang thai nang lng nao co the chuyenve trang thai ng vi n=3? Khi xet chu y en ca spin. (6.14)

Giai

oi vi electron hoa tr trong nguyen t Na, nhng trang thai ng vi n=3 co the la:

(khi cha ke en spin)D3,P3,S32,1,0l3n == Neu tnh en spin, tr mc c ban khong tach, cac mc khac tach 2:

25


26/28

2/52

2/3

2

2/3

2

2/1

2

2/1

2D3D3;P3P3;S3

Quy tac la chon:

;1l = 1;0j =

* Vay nhng trang thai electron co the chuyen ve la2/1

2 S3

va vi (n = 3, 4, 5, . . .)2/12 Pn 2/3

2 Pn

* Nhng trang thai electron co the chuyen ve la2/12 P3

vi (n = 4, 5, 6. . .)2/12Sn

vi (m = 3, 4, 5, . . .)2/32 Dm

* Nhng trang thai electron co the chuyen ve la2/32 P3

vi (n = 4, 5, 6. . .)2/12Sn

; vi (m = 3, 4, 5, . . .)2/32

Dm 2/52

Dm

* Nhng trang thai electron co the chuyen ve la2/32 D3

va vi (n = 4, 5, . . .)2/12

Pn 2/32

Pn

vi (m = 4, 5, . . .)2/52 Fm

* Nhng trang thai electron co the chuyen ve la2/52 D3

vi (n = 4, 5, . . .)2/32 Pn

; vi (m = 4, 5, . . .)2/52 Fm 2/7

2 Fm

10)Khao sat s tach vach quang pho mD-nP di tac dung cua t trng yeu (6.15) (bo)11)Trang thai cua nguyen t c ky hieu bi: , trong o X=S,P,D,F tuy theo so lng t

quy ao l; S la so lng t spin va j la so lng t momen toan phan cua ca vo electron. (6.16)

j

SX

12 +

Xac nh momen t cua nguyen t trang thai:

a. 31Fb. 2/32 D

Giai

Momen t cua nguyen t c tnh theo cong thc:

26


27/28

B)1J(Jg +=

Vi)1J(J2

)1L(L)1S(S)1J(J1g

+++++

+=

a) trang thai: :3F1Ta co: 3J,0S,3L ===

Suy ra: 112x2

120121g =

+++=

Vay: B12=

b) trang thai: :2/32 DTa co: 2/3J,2/1S,2L ===

Suy ra: 5/4g =

Vay: B155

2=

12)Nguyen t trang thai2

3;2 == sl co momen t bang 0. Tm momen toan phan cua nguyen t

o. (6.17)

Giai

?/J/ =r

Ta co: 0)1J(J2

)1L(L)1S(S)1J(J10g =

+++++

+=

Thay gia tr cua L, va S vao ta co:2

1J

4

3)1J(J ==+

Vay: hhr

2

3)1J(J/J/ =+=

13)Co bao nhieu electron s, electron p va electron d trong lp K,L,M ?Giai

s-electron p-electron d-electron

Lp K, n=1 2

Lp L, n=2 2 6

Lp M, n=3 2 6 10

Dien giai:Lp K co 2l(l+1)=2 e (s)0l1n ==

Lp L)p(e6)1l2(2'co1l

)s(e2)1l2(2'co0l2n

=+=

=+==

27


28/28

Lp M

)d(è10)1l2(2'co2l

)p(è6)1l2(2'co1l

)s(è2)1l2(2'co0l

3n

=+=

=+=

=+=

=

14)Lp ng vi n=3 cha ay electron, trong so o co bao nhieu electron (6.19)a. Cung 21=sm b. Cung 1=mc. Cung 2=md. Cung

21=sm va 0=m

e. Cung2

1=sm va 2=l

15)Trong nguyen t, cac lp K, L, M eu ay. Xac nh: (6.20)a. Tong so electron trong nguyen tb. So electron s, so electron p, va so electron d.c. So electron p co m=0.

Giai

a) Tong so electron: 2818823x22x21x2 222 =++=++=b) So electron s: 6 e gom 222 )s3()s(2)s1(

So electron p: 12e gom 66 )p3()p2(

So electron d: 10e gom 10)d3(

c) So electron p co m=0: co hai mc p ong gop vao bao gom mc v vay co 2electron p co m=0.

66 )p3()p2(

n=2, l=1, m= -1, 0, 1

n=3, l=1, m= -1, 0, 1

16)Viet cau hnh electron oi vi cac nguyen t sau ay trang thai c ban: (6.21)a. Bohrb. Carbonc. Natri

Giai

a. Bohr: Z = 5 122 )p2()s2()s1(b. Carbon: Z = 6 222 )p2()s2()s1(

1622

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Giai Bai Tap Vat Ly Luong Tu - Nguyen Tu