hoa dai cuong.pdf

7/22/2019 Hoa dai cuong.pdf

1/275http://hhud.tvu.edu.vn

CHNG 1

CC KHI NIM V NH LUT CBN CA HA HC

A. MC TIU, NHIM V

1. MC TIUHc xong chng 1 sinh vin bit v hiu:

- Cc khi nim cbn: Cht, nguyn t, nguyn t, phn t, khi lngnguyn t, khi lng phn t, khi lng mol, ng lng...

- Hn v.- Mt snh lut cbn ca ho hc.- Mt sphng php xc nh khi lng phn tv khi lng nguyn

t.

2. NHIM V

Tm hiu vcc khi nim cbn ca ho hc, hn vSI, cc nh lutc bn ca ho hc, cc phng php xc nh khi lng phn t v khilng nguyn t t hiu c v c kh nng vn dng c cc kinthc ca chng vo thc hnh v luyn tp.

3. VPHNG PHP

Kt hp cht ch gia s hng dn ca gio vin vi s t hc, tnghin cuca sinh vin. Cn ht sc coi trng khu luyn tp v thc hnh nm vng c cc vn ca chng ny.

4. TI LIU THAM KHO

- Ho hc i cng 1: Trn Thnh Hu, nh xut bn i hc sphm- Ho hc i cng: Nguyn c Chuy, nh xut bn gio dc

- Ho hc i cng: o nh Thc, nh xut bn i hc Quc gia HNi

- Bi tp ho hc i cng: o nh Thc, nh xut bn gio dc- Bi tp ho i cng : Dng Vn m, nh xut bn Gio dc- Ho hc i cng : L Mu Quyn, nh xut bn Gio dc- C s l thuyt ho hc- Phn bi tp: L Mu Quyn, nh xut bn

khoa hc v kthut


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Trng i hc cng nghip hni gio trnh ho i cng1

B. NI DUNG

1. MT SKHI NIM CBN CA HA HC

1.1. Cht:Cht l tp hp cc tiu phn c thnh phn, cu to, tnh cht xc nhv c thtn ti c lp trong nhng iu kin nht nh.

V d1: C6H6, O2, H2O,...Cht m phn tc cu to bi mt loi nguyn tc gi l n

cht.V d2: Ag, O2, O3, ...Cht m phn tc cu to bi hai loi nguyn ttr ln c gi l

hp cht.V d3: NaCl, H2O, CaCO3, C2H5OH,...Tcc khi nim vn cht, hp cht va c cp trn kt hp

vi cc kin thc c, ta c ssau (hnh 1.1)

HNH 1.1. Shthng phn loi cc cht

Tp hp ca cc phn tcng loi c gi l nguyn cht, nhkh H2nguyn cht; nc (H2O) nguyn cht;....

Tp hp gm cc phn tkhc loi c gi l hn hp, khng kh l hnhp gm rt nhiu kh khc nhau trong N2v O2chim t l ln nht (mtcch gn ng ngi ta coi khng kh gm 4/5 nit, 1/5 oxi vthtch)

Cc khi nim ny c minh ha hnh 1.2.

Tp hp vt cht c thl hng thhoc hdth. Khng kh l hng th,hp kim inox l hng th, mt cc nc c cnc lng v nc l hdth.


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HNH 1.2.Minh ha cc khi nim n cht, hp cht, hn hp

1.2.Nguyn t, nguyn tho hc, phn t

1.2.1. Nguyn t: l ht nhnht ca nguyn t ho hc m khng th phnchia c vmt ho hc.

V d5: nguyn tH, O, Na, Cl...Nguyn t l loi ht rt Nhv rt nh. Tu thuc vo mi nguyn t

ho hc m khi lng ca mt nguyn t10-23 10-21g, cn ng knh camt nguyn tvo khong 10-8cm.

hnh dung v th tch ca mt nguyn t, c thhnh dung nh sau:Nu coi mi nguyn tu c dng hnh cu c ng knh 10-8 cm th qubng bn c ng knh 4 cm c thcha c khong 1024nguyn t.Nguyn tca cc nguyn tho hc khc nhau th c khi lng v kch thckhc nhau.

Chng ta tha nhn nguyn t c cu to bi 3 loi ht c bn l:electron (e), proton (p) v ntron (n), bng 1.2 cho chng ta bit c im c

bn ca ba loi ht .


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BNG 1.1. Bn knh cng ho tr(A0) v khi lng nguyn tca mt snguyn tha hc

Nguyn t R (A0) M(1,673.10-23)

H 0,30 1,008

O 0,66 15,994

S 1,04 32,064

Cl 0,99 35,453

Br 1,14 79,904

I 1,33 126,904

BNG 1.2. Khi lng, in tch ca electron, proton, ntron

KHI LNG IN TCH

Kg vC (u) Culong Quy c

Electron 9,109.10-31 5,55.10-4 -1,6021.10-19 -1

Proton 1,672.10-27 1,007 + 1,6021.10-19 +1

Ntron 1,675.10-27 1,009 0,0 0,0

bng 1.2 im cn ch l: in tch ca cc ht cbn.Ntronl ht khng mang in, tc l ht trung ho in, c k hiu l

0n. Mi ht proton mang in tch dng l +1,6021.10-19Culong. in tch ny

chnh l in tch cbn, thng c k hiu l e0. Trsny c quy cchn lm n vnn: mi ht proton mang mt n vin tch dng, c khiu l 1p. Mi ht electron mang mt n vin tch c trstuyt i bng trsin tch ca mt ht proton nhng ngc du. V thmi electron mang mtn vin tch m, k hiu l e.

Cng cn ch , khi lng ca electron rt nh so vi khi lng caproton, ntron.

Tsliu ca bng 1.2, ta c tlcc khi lng nhsau:

,518351 =e

p

m

m(ln) 8,0 1838=

e

n

m

m(ln)

V vy trong cc php tnh thng thng, ta coi me 0.Cng tbng trn ta thy

pn

m m10

> . Trong cc php tnh thng thng ta chp

nhn sgn ng:1

10 pn mm (vC)


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1,6021.10-19l in tnh nhnht nn c gi l n vin tnh nguyntv c k hiu l e0.

M hnh nguyn tc tha nhn rng ri hin nay l: Nguyn t chnh dng ca mt khi cu. Tm ca nguyn tl ht nhn tch in dng. V

ca nguyn tgm cc electron chuyn ng quanh ht nhn. Sn vintch dng ca ht nhn bng sn vin tch m ca v. Nguyn ttrungha vin

V d6: Ht nhn nguyn tnatri (Na) c 11 n vin tch dng (khiu Z = 11), vnguyn tNa c 11e, tc l c 11 n vin tch m. Vynguyn tNa trung ha vin, c vit l Na0hay Na. Nu v mt l do no vnguyn tNa cn 10e, ta c ion dng hay cation natri, c vit l Na+.

Tng t, ht nhn nguyn tClo (Cl) c 17 in tch dng (Z = 17), vnguyn tCl c 17e nhng nu cnguyn tCl c thm mt e tr thnh 18e,nguyn tCl khi ny khng cn trung ha vin na, ta c ion m, hay anionclo, Cl-

Biu din qu trnh bin i nguyn t trung ha in thnh ion nu trnnhsau:

Na - e Na+Cl + e Cl-

1.2.2. Nguyn tha hc: Tp hp cc loi nguyn tm ht nhn c cng sn vin tnh dng (Z) l mt nguyn tha hc.

V d7: Nguyn toxi c sn vin tch dng ca ht nhn bng 8.Trong thc tc 3 nguyn toxi vi khi lng khc nhau l 16, 17, 18 nhngu c sn vin tch dng ca ht nhn bng 18, l cc nguyn tkhcnhau - cc ng v ca nguyn t oxi. 3 nguyn t ny c vit nh sau:

hay 8O16 8O

17 8O18

Nhvy sn vin tch dng ca ht nhn (Z) l yu tquyt nhca mt nguyn tha hc. Tr sZ thay i d ch1 n vng ngha vivic chuyn tnguyn tha hc ny sang nguyn tha hc khc.

V d8: Hai nguyn tc cng khi lng nguyn t l 40 (vC), mtnguyn tc Z = 19, nguyn tkia c Z = 20. l 2 nguyn tca hai nguyn

tho hc, mt l 19K40(ng vthng gp ca K) v mt l 20Ca40.

Cn phn bit cc khi nim nguyn t, nguyn t, n chtV d9: K hiu O dng ch1 nguyn tca nguyn toxi. cng

l k hiu ca nguyn toxi.K hiu O2- (gisxut hin trong qu trnh in phn nhm oxit nng

chy) ch1 ion oxi. ion ny c to ra tnguyn t ca nguyn toxi. l 2dng th hnh ca oxi.

K hiu O2ch1 phn tn cht oxi.

K hiu O3 ch1 phn tn cht ozon. O2v O3K hiu H2O ch1 phn tnc. H2O l mt hp cht v trong thnh phn phntc 2 nguyn tl hiro v oxi.


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Xt tng tvi cc trng hp khc, c thkhi qut nhsau:Nguyn tha hc l khi nim rng, dng chcc ht v cng nh:

nguyn t, ion c cng sn vin tch dng ca ht nhn.Nguyn tl mt khi nim chmt dng tn ti cthca nguyn tho

hc. V th, khi ni n nguyn t c th th cng c ngha l ni n mtnguyn tho hc.n cht l khi nim chmt dng tn ti cthca nguyn tha hc.

Khi ni n mt n cht cng c ngha l ni n mt nguyn tha hc.1.2.3. Phn t: l phn tnhnht ca mt cht c thtn ti c lp nhngvn ginguyn tnh cht ca cht .

V d10: H2l phn tH2, chy c v dng lm nhin liu.2H2+ O2 2H2O H < 0 (tonhit)

CO2l phn tcacbon ioxit, khng chy c.Phn tc to nn tcc ht nhhn (nguyn thay ion)Phn tc to ra tcc nguyn tca cng mt nguyn tha hc l

phn tn cht.V d11: Cl2. O2, O3, P4, S8,...Cc n cht khc nhau ca mt nguyn tha hc c gi l cc dng

th hnh ca nguyn t.V d 12: Nguyn t oxi c 2 dng th hnh thng gp l oxi (O2) v

ozon (O3). Nguyn t cacbon c 2 dng th hnh ph bin l than ch v kimcng, ...

Phn tc to ra thai loi nguyn tca 2 nguyn tha hc trlnl phn thp cht.

V d13: HCl, HClO,...Phn tc thc to ra t1 nguyn tth l phn tn nguyn t.V d14: kh him

Mt sc im vphn t:- Vkhi lng phn t: C nhng phn trt nh(H2), c nhng phn t

nng nhng glucoz C6H12O6 (180 vC) v cng c nhng phn t siunng nhpolime (c khi lng phn ttrung bnh chng chc vn vC). Chitit c trnh by phn sau.

- Vin tch: th phn t trung ha vin. V thcn phi phn bitphn tvi gc tdo: K hiu SO3chphn tanhirit sunfuric; k hiu chgc tdo c to thnh tc thi (thi gian tn ti v cng ngn) trong phnng.

- Vcu to ha hc:y l mt vn ln, phn ny chxt mt scim vhin tng ng phn.

ng phn l hin tng cc cht c cng cng thc phn t, nhng ccu to khc nhau nn c tnh cht khc nhau, cc cht l cc ng phn

Xut pht tc im vcu trc, ta c ng phn cu tov ng phn

khng gian.


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V d15: t cng thc C2H6O ta c 2 ng phn cu to l ru etylicCH3CH2OH v imetyl ete CH3OCH3; tcng thc abC = Cab ta c 2 loi ng

phn khng gian l cis v trans.Trong ng phn cu to c ng phn mch cacbon; v tr (nhm chc,

lin kt bi,...), ng phn nhm nh chc. Trong ng phn khng gian cng phn hnh hc, ng phn quang hc v vn vcu dng.Ch : Khi xt phn tcn quan tm ti hnh dng hay hnh hc phn t.

Thc nghim xc nh cgc lin ktv di lin kt. Cc yu thnh hcthng gn lin vi di v tnh cht ca phn t. Mt shnh dng phn tthng gp nh: ng thng (cc nguyn t trong phn tc phn b trnmt ng thng); c gc (cc nguyn tthng l 3 hay 4 nguyn tlin ktvi nhau to ra gc khc gc 1800); lp th(khi khng gian nhthp tam gic,tdin u, bt din u,...), minh ha hnh 1.3.

HNH 1.3.Mt shnh dng phn ta) CO; b) CO2, phn tthngc) H2O, phn tgcd) NH3, phn tthp tam gice) CH4, phn ttdin

1.3. Khi lng nguyn t, khi lng phn t, khi lng mol1.3.1. Khi lng nguyn t: l khi lng ca mt nguyn t, khi lngnguyn t c xc nh bng tng khi lng ca tt c cc ht to thnhnguyn t.

Cn phn bit khi lng nguyn t tng i v khi lng nguyn ttuyt i.

a) Khi lng nguyn ttuyt i: l khi lng thc ca mt nguyn ttrong khng gian c tnh bng kilogam.

V d 16: mS= 5,3.10-23g = 5,3.10-26kg

mFe= 9,274.10-23g = 9,274.10-26kg

mC= 19,9206.10-24g = 19,9206.10-27kg


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Khi lng ny cc knhb, khng thun tin cho vic cn, o, ong,m c nn gp kh khn khi phi tnh ton trong cc bi ton ha hc.

thun tin cho vic tnh ton ngi ta dng mt hkhc, gi l khilng nguyn ttng i.

b) Khi lng nguyn ttng i(nguyn tkhi): AChn

12

1 ln khi lng ca mt nguyn tcacbon ng v12 (C12) lm

mt n vkhi lng v c gi l n vcacbon (vC hay u), ta c:

kgkgm

dvC C 2727

12 10.,6605112

.10,926019

121

===

12106605.,1 2712 == kg

mA CC

56106605.,1

.10,274927

26

==

kg

kgAFe

32106605.,1

.10,3527

26

==

kg

kgAS

Nh vy, khi lng nguyn t tng i l mt tr s khng c thnguyn. Nhng trong thc t ta vn hay dng mt cch ngn gn Fe= 56vC

hay 56u (dng n vcacbon: vC) v coiC

m1212

1 l 1vC (1u).

T cc v d trn cn nh: khi lng nguyn t tng i (klt) = khilng nguyn ttuyt i (klt) : sAvgar (N)

KLT= klt: N1.3.2. Khi lng phn t:ly khi lng ca 1 phn tchia cho 1 n vkhilng th c khi lng phn ttng i ca phn t.

Hoc: ly tng khi lng nguyn t tng i ca tt ccc nguyn tto nn phn t. Thng c k hiu l: M

V d17: 442.16122

=+=COM (hay: 44 vc), thng vit l CO2= 441.3.3. Mol-Khi lng mola)mol:Mol l lng cht cha 6,023.1023ht vi m.

Tkhi nim ny khi dng mol cn phi chr loi ht vi m.V d18: 1mol nguyn tH, 1mol phn tH2, 1mol ion H

+,...b) Khi lng mol: Khi lng mol nguyn tca mt nguyn t(A) l

khi lng ca 1 mol nguyn tca nguyn t. n vg/molV d19: khi lng mol nguyn tca hiro bng 1,008 g/mol ( hay AH

= 1,008 g/mol)Khi lng mol phn tca mt cht (M) l khi lng ca 1 mol phn

tcht .V d 20: Khi lng ca 1 mol phn t nc bng 18,015 g/mol (hayg molg molM OH /18/015,182 = ).


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Cch tnh smol:)( XX

XX AM

mn =

y: Xm l khi lng ca X. NuXl nguyn tth dng XA l khilng mol nguyn tca X. Nu Xl mt cht th XM l khi lng mol phn

tca X.1.4. ng lng1.4.1. ng lng ca mt nguyn t:L sphn khi lng ca nguyn t c thkt hp hoc thay th1,008 phn khi lng ca hiro hoc tm phnkhi lng ca oxi.

K hiu ng lng l: Theo nh ngha trn, ta c:

1008,1 =H ; 00,= 8O 23=Na ; 20=Ca

Tnh ngha trn ddng xc nh c: ng lng ca C trong CO2l 3 cn trong CO l 6.

C thtnh ng lng ca mt nguyn ttheo cng thc:

i

ii n

A=

Trong : iA l khi lng nguyn t ca nguyn t; in l ha tr canguyn t.1.4.2.ng lng ca mt hp cht:L sphn khi lng ca hp cht tc dng va vi mt ng lng ca cht khc.

V d21: Bit 9=Al . Tphn ng: 2Al + 6HCl 2AlCl3+ 3H2, ddngtnh c 5,36=HCl

Bit 40=NaOH . Tphn ng: 2NaOH + H2SO4Na2SO4+ 2H2O, tnhc 49

42= SOH

1.4.3. Cch tnh ng lng:

a) Trng hp chung:n

MA )(=

Trong : )(MA l khi lng nguyn thay khi lng phn t.

n l se trao ib) Cc trng hp cth:

+ i vi nguyn tho hc:H

A= vi H l ha trca nguyn tha

hc.

+ i vi hp cht:n

M= vi n l selectron trao i.

Nu hp cht l:- Oxit: Th n l tng ha trca oxi c trong oxit.- Axit: Th n l snguyn thiro c trong axit c thay th.- Baz: Th n l ha trca kim loi c trong baz.- Mui: Th n l tng ha trca kim loi c trong mui.


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V d22: ng lng ca Fe2O3l: 7,262.3222

16032

==++= OFe

ng lng ca H2SO4trong phn ng:

2NaOH + H2SO4Na2SO4+ 2H2O l 49

2

9842

== SOH NaOH + H2SO4NaHSO4+ H2O l 98

1

9842

== SOH ng lng ca Ca(OH)2l: 37

2

742)(

== OHCa ng lng ca Ca3(PO4)2l: 7,51

2.3222

310243 )(

==++= POCa

ng lng ca Fe, trong phn ng:Fe + 2H+ Fe2++ H2l 282:56 ==Fe

Cn trong phn ng: 2Fe + 3Cl22FeCl3l 67,183:56 ==Fe Ch :Theo nh ngha vng lng th ng lng l mt i lng

khng c n v.1.4.4. ng lng gam: ng lng gam ca mt cht (n cht hay hpcht) l lng cht c tnh bng gam v c trsng bng ng lngca cht .

Nu k hiu ng lng gam l lg, ta c: lgAl = 9g v 9=Al 1.4.5. S ng lng gam(s lg):c xc nh bng s gam chia chong lng gam.

Cng thc tnh:Slg = sgam : lg

Cho mHCl= 18,25g SlgHCl= 18,25g/36,5g = 0,5Nhvy sng lng gam cng l i lng khng c n v.

2. HN VMt trong cc vn ca ha hc, l bi ton ha hc. Bi ton ha

hc c t ra da trn yu cu ca thc t i sng, thng qua cc thcnghim ha hc. V thcc kt qu thng qua cc con sphi c ngha xcnh. Cho nn cc con sny buc phi c n v. Chc p sbng sngth cha m cn cn phi c n vng.

Mt lng vt cht lun c biu thbng trsc km theo n v.Lng vt cht = Trs. n v

Hin nay, c hai xu hng: Dng hn vquc t(hSI) v dng n vtheo thi quen.Trong qu trnh hi nhp vi quc t, chng ta nn dng hnvquc t(hSI)2.1. Hn vquc t(hSI)

i hi vo lng quc thp ti Pari vo thng 10 nm 1960 thngqua cc quy c vn vo v cc khi nim tng ng.

Trong chng ny ch xt hn vcth.2.1.1. HSI cs

Gm by i lng c chn lm c s cng vi n v ca mi ilng km theo, c a ra bng 1.3


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BNG1.3.By i lng cbn ca ho lng quc t(hSI)

I LNG N VO

K hiuTn gi K hiu Tn gi Quc t Vit Nam

Chiu di l mt m m

Khi lng m kilogam kg kg

Thi gian t giy s s (giy)

Cng dng in I Ampe A A

Nhit T Kelvin K K

Lng cht n Mol mol mol

Cng nh sng I Cadela (nn) cd cd

Ngoi ra cn c hai n vbsung thng dng l

Gc phng radian rad Rad

Gc khi sterdian sr sr

2.1.2.n vdn xut tn vSI csCc n vdn xut thn vSI csc xc nh ph hp vi ccnh lut vt l cng nhquan hgia cc i lng lin quan.

V d23: n vca lc F, theo nh lut th2 ca Niutn: F = m.a l lcgy ra mt gia tc l 1 m/s2cho vt c khi lng tnh 1kg. Vy lc F sc nvl kg.m.s-2, c k hiu l Niutn, ngha l: 1N = 1 kg.m.s-2

BNG 1.4.Mt sn vdn xut tn vSI csa)n vc tn ring

I LNG N V K HIU KHI NIM

Lc Niutn (Newton) N kg.m.s-2

p sut Patcan (Pascal) Pa N.m2 (hay kg.m-1.s-2)

Nng lng Jun (Joule) J kg.m2.s-2

Cng sut Oat (Watt) W J.s-1(hay kg.m2.s-3)

in tch Culong (Coulomb) C A.s

in th Von (Volt) V J.C

-1

(hay J.A

-1

.s

-1)

Tn s Hec (Hertz) Hz s-1


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b) Cc i lng khng c tn ring

I LNG N V K HIU

Din tch mt vung m2

Thtch mt khi m3

Vn tc mt/giy m.s-1

Gia tc mt/giy2 m.s-2

Khi lng ring kilogam/met khi kg.m-3

Cng in trng von/mt V.m-1

2.2.n vphi SITthi quen hng ngy trong cuc sng m mt sn vkhng thuc h

SI vn thng xuyn c dng. Tuy nhin khi dng cc n vny cn phitm mi lin hqua li gia chng.Mt sn vphi SI thng dng

BNG1.5. Mt sn vphi SI

N V THA SI VN VSI CSHAY

DN XUTI LNG

Tn K hiuChiu di Angstrom A0 10-10m

Thtch lit l 10-3m3

Nhit bch phn 0C t (0C) =T - 273,15

Thi gianpht

gi

min

h

1 min = 60s

1h = 3600s

p sut

atmotphe

bar

mm thy ngn

atm

bar

mmHg

1 atm = 1,013.105Pa

1 bar = 105Pa (1atm)

(1mmHg =1/760 atm)

Nng lng

ec

calo

oat gi

kiloat gi

electron Von

erg

cal

Wh

kWh

eV

10-7J

4,184J

3600J

3600kJ

1,602.10-19J

in tchn vtnh in

cgs

ues cgs

Gc phng o (/180)radMomen lng cc in bai (Debye) D 1/2,9979.10-29c.m


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2.3. Cc hng svt l

BNG 1.6. Cc hng svt l thng dng

HNG S K HIU GI TR

Hng sAvgar (Avogadro) NA 6,0223.1023/mol

n vkhi lng nguyn t u 1g/N = 1,6605.10-24g

Khi lng electron me 9,1095.10-28g

5,4858.10-4u

Khi lng proton mp 1,67258.10

-24

g1,00724u

Khi lng ntron mn 1,6748.10-24g

1,00862u

in tch nguyn t e0 1,6021.10-19C

4,8.10-10ues cgs

Hng sFaraday F 96487,0 C/mol 96500 C/mol

Hng sPlng (Planck) h 6,6256.10

-34

JsVn tvs nh sng (trong chnkhng)

c 2,99725.1018m/s = 300.000km/s

Thtch mol phn tcc chtkh

V0 22,41 l/mol = 0,02241 m3/mol

Hng skh R 8,3144J/mol.K

8,2054.10-21atm/mol.K

1,98 cal/mol.K

Hng sBnxman(Bolzman) k 1,38054.10-23J/KHng sRitbe (Rydberg) RH 109677,57 c.m

-1

Manhton Bo (Bohr) B 9,2732.10-24J/T

Bn knh Bo (Bohr) a0 5,29167.10-19cm = 0,529 A0

2.4. Hn vnguyn tTrong ha hc lng t (nhng ni dung vcu to ca vt cht) chng

ta dng hn vnguyn t(vn hay au). Trong hny quy c cc lng sauy bng n v:


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Hng sPlng rt gn 0,12

=

hh (vi =h 6,625.10-34J.s)

Bn knh Bo thnht a00,529 1,0Khi lng ca mt electron, me=9,109.10

-31kg 1

in tch cbn e0= 1,6021.10-19

culong 1Tch 40= 1 (quy c ny c thdng chung vi bn quy c trn hocdng ring mnh n).

Tcc quy c trn ta c n vca nng lng tng ng s l vnhay au hay l hactri (hartree).

V d 24: Gii phng trnh sring cho h 1 electron 1 ht nhn (H,He+, Li2+,...) tm c biu thc tnh nng lng l:

2

0

22

40

2

)4(

1

2 =

hn

emZEn trong

2

h=h

0 l hng sin mi trong chn khnga)Hy tnh nng lng ng vi n = 1 cho: H, He+, Li2+

b) Hy tm mi lin hgia hai n vnng lng l vn vi eVTrli:Tsliu cc bng trn, ta c: me= 9,1095.10

-28g = 9,1095.10-31kgh = 6,6256.10-34J.se0= 1,6021.10

-19C40= 1,112650056.10-10J-1.C2.m-1

a) Thay cc sliu vo biu thc tnh nng lng Enta c:E1= -13,6xZ

2 (eV) hay E1= 22

1Z vn (hay hactri)

Vi H: Z = 1 E1= - 13,6 (eV) hay E1= - 0,5 (vn)Vi He+: Z = 2 E1= - 54,4 (eV) hay E1= - 2,0 (vn)Vi Li: Z = 3 E1= - 122,4 (eV) hay E1= - 4,5 (vn)

b) Tcc kt qutrn ta c mi lin h: 1(vn) = 27,2(eV)

3. MT SNH LUT CBN CA HA HC

3.1. nh lut bo ton khi lng (Lmnxp nh bc hc Nga v Loavazinh bc hc Php)Bo ton vt cht l quy lut chung ca tnhin, trong cuc sng hng

ngy hiu mt cch n gin l quy lut tng khng i. Ha hc l khoa hcvcc cht v sbin i gia cc cht, nn quy lut vbo ton vt cht cthhin rt y . Trong ha hc l nh lut bo ton khi lng. nh lutny c nhiu cch pht biu khc nhau:

Tng khi lng cc cht tham gia phn ng bng tng khi lng cccht thu c sau phn ng.

Mt cch tng qut : C s bo ton vt cht trong cc phn ng hahc.

Xt phn ng dng tng qut:


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A1+ A2+ + An B1+ B2 + + BnTh lun c:

'n21n11 BBBAAAm...mmm...mm +++=+++

Hay: ==

=n'

1'iA

n

1iA 'ii

mm

Xt vphng din l thuyt th khi lng cc cht thc skhng cbo ton, v phn ng ha hc lun lun km theo qu trnh gii phng hay hpthnng lng di dng nhit.

Nm 1905, Anhstanh ch ra rng: Khi lng ca mt vt v nnglng ca n lin hvi nhau bi h thc E = mc2. Trong c l vn tc nhsng bng 3.108m/s.

Nu gi E l nng lng km theo ca phn ng ha hc th sthay ikhi lng trong phn ng ha hc sl: E = m.c2

m = E/c2, do E rt nh, c rt ln nn m l v cng b (khng ngk). Do khng pht hin thy s thay i khi lng ca phn ng ha hc(m), nn nh lut bo ton khi lng ca Lmnxp vn c tha nhn.

ng dng ca nh lut:Gii nhanh bi ton ha hc bng phng phpbo ton khi lng. Ch khi gii bi, khng tnh khi lng ca phn khngtham gia phn ng, cng nh phn cht c sn, v d nh nc c sn trongdung dch.

V d1: Hn hp X gm Fe, FeO, Fe2O3. Cho lung CO i qua ng ngm(g) hn hp X nung nng. Sau khi kt thc th nghim thu c 64,0g cht rnA trong ng sv 12,32 lt kh B (27,3oC; 1atm) c tkhi so vi H2l 20,4.

Tm m?Trli:Phng trnh phn ng chung:

X + CO = A + CO2m 64,0

)5(0,,4).22,3273 27.(1

.273.1,3212molnB =+

=

Gi x l smol CO2ta c: 44x+28(0,5x)/0,5=20,4.2=40,4 x=0,4 (mol).Theo nh lut bo ton khi lng ta c:

mX+ mCO= mA+ mCO2mX= mA+ mCO2 - mCO= 64 + 0,4.44 - 0,4.28 = 70,4(g)

V d2: Cho hn hp axit hu cA, B tc dng vi ru a chc C thuc hn hp nhiu este, trong c este E. t chy ht 1,88 g E cn mtlng va l 1,904 lt oxi (ktc) thu c CO2v hi H2O vi tlthtchtng ng l 4/3. Xc nh cng thc phn tca E bit tkhi ca E so vikhng kh nhhn 6,5.

Trli:

Phng trnh phn ng chungE + O2CO2+ H2OTheo nh lut bo ton, ta c:


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mE+ mO2= mCO2+ mH2O=1,88+(1,904/22,4).32=4,6 (g)Gi a l smol CO2th 3a/4 l smol H2O.Vy: 44a + (3a/4).18 = 4,6 a = 0,08 (mol)mC = 0,08.12 = 0,96 (g)

mH = (3/4).0,08.2 = 0,12 (g)mO = 1,88 - 0,96-0,12 = 0,8 (g)

x : y : z = 5:12:816

,80:

1

,120:

12

,960=

Do cng thc n gin nht ca E l C8H12O5v cng thc phn tl(C8H12O5)n.

Theo : mE< 29x6,5 = 188,5 n=1Vy cng thc phn tca E l: C8H12O5

V d3: C mt chn A cha dung dch Na2CO3c khi lng l g1, cnchn B cha dung dch HCl c khi lng g2. t gt= g1+ g2Trn dung dch trong chn A vi chn B, lc u cho phn ng xy ra

hon ton ri cn chai cc v ha cht c khi lng tng cng l gs.a) C thc nhng trng hp no vmi quan hgia hai trsgtv gs?

Ti sao?b) C thxy ra trng hp gt< gskhng? ti sao?Trli:a) Xy ra mt trong hai trng hp sau:Trng hp 1: gt> gs v xy ra phn ng

Na2CO3+ 2HCl 2NaCl + H2O + CO2Lng CO2thot ra khi dung dch lm gim khi lng cc cht cn li

trong cc sau phn ng.Trng hp 2: gt= gsv xy ra phn ng

Na2CO3+ HCl NaCl + NaHCO3Khng c cht no thot ra khi chn, nn khi lng trc v sau phn

ng bng nhau.b) Khng thxy ra trng hp gt< gsv iu ny tri vi nh lut bo

ton khi lng.

3.2. nh lut thnh phn khng i(Prt-nh bc hc Php)Mt hp cht ha hc d c iu chbng cch no cng u c thnh

phn khng i.Gii thch: Nu chp nhn quan im v cu to nguyn t v cu to

phn tth thnh phn ca mt cht bt kchnh l thnh phn ca mt phn tcht . Trong mt phn tca mt cht xc nh, th snguyn tca nguynt l xc nh, khng i. V vy thnh phn khi lng ca nguyn t cngkhng i.

Vai tr ca nh lut: cho php phn bit mt cht ha hc vi mt hnhp ch: Thnh phn ca mt cht khng thay i cn thnh phn ca hnhp thay i theo phng php iu ch.


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Hn chca nh lut: chng khi cht c cu trc phn thay cu trctinh thhon chnh, khng ng i vi hp cht khng nh thc.V d4: Tni dung ca nh lut dthy H2O c tlvsnguyn t trong

phn tH : O = 2 : 1 Nc c iu chtheo mt scch sau:

OH2COO2CH

OHSONaSOHSONa2OH2OH2

22t

24

2424242

2t

22

0

0

+++++

3.3. nh lut tlbinh lut ny c nh bc hc Prut a ra vo nm 1806.

Nu hai nguyn tha hc to vi nhau mt shp cht th cc lng camt nguyn tkt hp vi cng mt lng ca nguyn tkia tlvi nhau nh

cc snguyn nh.V d5: Xt hp cht gia C v O l CO v CO2. Dthy lng O kthp vi cng lng C lp thnh tsl 1: 2.

V d6: Xt cc oxit ca nitbng sau1.7.BNG 1.7.Mt skt quthc nghim thu c khi xc nh thnh phn

nguyn ttrong cc oxit ca nit

Thnh phn % khilng

Tllng oxiTn oxit

N O

Sphn khi lngoxi ng vi mt

phn khi lngnit

initoxit 63,7 36,3 0,57 1

Nitoxit 46,7 53,3 1,14 2

inittrioxit 36,8 63,2 1,71 3

Nitioxit 30,4 69,2 2,28 4

initpentoxit 25,9 74,1 2,85 5

Tcc kt quthc ngim thu c bng 3.1, ta ly:

: 5: 4: 3: 2157,0,852:

,570,282:

,570,711:

57,,141:

,570,570 =

o

Nhvy t lkhi lng oxi ng vi 1 phn khi lng nit trong ccoxit trn l 1:2:3:4:5

Cng c thxc nh c cc sny bng cch thng qua thnh phn %vkhi lng gia nitv oxi, ddng xc nh c cng thc ca cc oxittng ng:

initoxit: N2O, Nitoxit: NO, inittrioxit: N2O3, Nitioxit: NO2vinitpentoxit: N2O5

Tcc cng thc ny,nu ly cng mt lng nit tng ng nhnhau(gis2 mol nit) th smol oxi tng ngN2O NO N2O3 NO2 N2O5


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1

2 :

2

2 :

3

2 :

4

2 :

5

2

Smol tng ng ca oxi sl 1:2:3:4:5 ng nhin tlny cng l tlvsnguyn toxi trong cc oxit tng ng khi kt hp vi hai nguyn t

nit. V d7: Vn dng nh lut thnh phn khng i v nh lut tlbicho SO2v SO3.

+ iu chSO2: S + O2SO2Na2SO3+ H2SO4 Na2SO4+ H2O + SO2CuSO3 CuO + SO2

+ iu chSO3: SO2+ O22SO3Fe2(SO4)3 Fe2O3 + 3SO3

+ Ddng xc nh c cc lng oxi kt hp cng vi mt lng lu

hunh lp thnh ts2 : 3 ( hoc cc lng lu hunh kt hp vi mt lngoxi lp thnh ts3 : 2).

3.4. nh lut Avgaro:(chp dng cho cht kh)3.4.1.nh lut Avgar

Nh khoa hc ngi Italia - Avgar a ra mt gi thuyt vchtkh, vsau c gi l nh lut v mang tn ng: nh lut Avgar.

Trong cng iu kin vnhit v p sut, nhng th tch bng nhauca cc cht kh u cha cng mt sphn t.

Hqu: 00C, 1atm (760 mmHg) 1mol kh bt kchim thtch l 22,4lt.

nhng iu kin nhnhau vnhit v p sut 1 mol kh bt kuc thtch bng nhau.

V vy: Thtch m 1 mol kh chim chc gi l thtch mol phn tca kh.

Thtch mol phn tca mi cht kh 00C, 1atm l 22,4 lt.nh lut ny chp dng cho cht kh, kchn hp cc kh. Cc kh

u c chung c im: Khong cch gia cc kh rt ln, kch thc ca cckh li rt nhv vy c thbqua kch thc ca cc phn tkh khi cc

phn tkh c coi nhnhng cht im.Tc im ny, thy rng: cng iu kin vnhit ,p sut tc dng

ln cc kh l nhnhau th trong nhng thtch bng nhau ca cc kh schacng mt s lng nhnhau vcc phn tkh. T ddng thy c s

phn tkh tlthun vi smol kh. Cho nn khi lm cc bi ton vcht khthng dng mi lin hsau:

cng iu kin vnhit v p sut, nhng th tch bng nhau cami cht kh u cha cng mt smol kh.

V d8: Np y vo mt bnh kn 0,5 mol kh H2ri cn ton bbnh.

Sau khi tho ht H2ra, v lm th nghim nhtrn vi kh CO2. Hai ln cn thykhi lng khc nhau l m gam. Tm m.


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Nu cng lm th nghim nhtrn vi 0,5 mol kh H2, nhng mun hai lncn m m = 0 th cn dng bao nhiu mol CO2? Bit cc th nghim u ctin hnh cng iu kin vnhit v p sut.

Trli:

Theo nh lut Avgar, th nghim trc mi kh u c 0,5 mol.Theo cng thc: 1 mol CO2nhiu hn 1 mol H2l 42 gamVy 0,5 mol CO2nhiu hn 0,5 mol H2lm = 0,5.42 = 21(g)

trong th nghim sau, m = 0 th )(122

gmm HCO ==

Vy )0227(,044:12

molnCO ==

Cn ch rng: iu kin tiu chun (ktc), t = 00C hay T = 273,15 K;P = 1atm, mt mol kh bt ku chim thtch l 22,4 dm3(hay22,4 lt)3.4.2. Phng trnh trng thi ca kh l tng

Khi khong cch gia cc phn tkh l rt ln, kch thc ca cc phntl khng ng k(bqua) th lc tng tc gia cc phn tkh cng khngng k(bqua) khi kh c gi l kh l tng. Phng trnh lin hgianhit T, p sut P v thtch V ca kh c gi l phng trnh trng thica kh l tng:

RtPV = vi 1 mol khnRTPV = vi n mol kh

Hoc RTM

mPV = RT

M

dRT

VM

mP == ; trong dl khi lng ring

ca kh.Trong trng hp c mt hn hp kh l tng chim mt th tch V nhit T, th p sut ton phn c xc nh bi nh lut alton:

=i

iTP P, vi iPl p sut ring phn ca kh i

=i

iT nV

RTP , vi in l smol ca kh i

R l hng skh, trsR phthuc vo n vo p sut, thtch cn Tphi biu ththeo nhit Kenvin.

Hng skh R c sdng rt rng ri trong tnh ton, v vy cn lu

cch biu thn vca n sao cho thng nht. Tphng trnh trng thi,ta c:

Tmol

PVR

).(1=

Tphng trnh ny chcn thay cc gi trbng sca P, V, T th tmc gi trca hng skh R.

Theo hthng n vhp php ca nc ta, khi lc o bng Niutn trn1m2(N.m-2) cn nng lng o bng Jun (J) cn thtch o bng mt khi (m3),th:

P = 1,01324.105Pa = 1,01324.105N.m-2= 1,01324.105kg/m.s2

V = 0,022415m3

Thay vo, ta c:


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1_1112

23

25

.314 .,8...

,3148,15273.1

,0224150..

.10,013241

.1KmolJKmol

s

mkg

Kmol

msm

kg

Tmol

PVR ====

( v 1J = N.m; 1N = 1kg.m/s2

1J = kg.m2

/s2

)Tuy nhin, hin nay trong nhiu ti liu cn dng nhiu hthng n vkhc nhau, nn cng cn phi bit thm.

Nu biu din R bng n vcalo th da vo mi lin h1cal =4,184J1J =1/4,184 = 0,239cal. Thay vo trn ta c:

R = 8,314.0,239cal.mol-1.K-1= 1,987cal.mol-1.K-1.Nu p sut o bng dyn/cm2v thtch tnh bng cm3(theo hCGS:

di o bng centimt (cm), khi lng o bng gam (g), thi gian o bng giy(s). Khi lc tnh bng dyn, p sut tnh bng dyn/cm2, nng lng tnh bng

ec, 1ec =1dyn.cm = 10

-7

J, cn thtch o bng C.m

3

).Thay cc sliu vo biu thc (P =1,01324.106dyn/cm2; V = 22413cm3; T= 273,15K), ta c:

117117

32

6

...10,3148....10,3148,15273.1

22413..10,013241 === KmolecKmolcmdyn

Kmol

cmcm

dyn

R

Trong trng hp p sut tnh bng atmotphe vt l v th tch o bnglt, th:

11...,08205015273,.1

,41522.1 == KmollatmKmol

latmR

Nhvy tuthuc vo n vo ca p sut v thtch m hng skh Rc cc gi trkhc nhau.

(Cc gi trca R c xc nh iu kin tiu chun)V d9: tm khi lng ring ca kh flo 1atm v 25oC.Trli:Trc tin tm thtch ca 1mol flo 1atm v 25oC.

P

RTV =

RTPM

VMd .==

T = 273,15 + 25 = 298,15 K

Vy: ).55(,1...,0820

1.00 .,38 111

1

= lgKmollatm

atmmolgd

3.5. nh lut ng lngNh bc hc alton ngi Anh pht biu nh lut vo nm 1792, c ni

dung nhsau:

Cc nguyn t ha hc kt hp vi nhau hay thay th cho nhau theonhng phn khi lng tlvi ng lng ca chng.Xt phn ng: A + B AB


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Ta lun c:B

B

A

A

B

A

B

A mmhaym

m

=

= hay slgA = slgB

nh lut c pht biu bng cch khc: Cc cht tc dng vi nhautheo cng mt sng lng gam.

Tc l: Trong phn ng ha hc mt ng lng gam ca cht ny chkt hp hoc thay thmt ng lng gam ca cht khc.V d 10: Oxi ho cn thn 0,253 g Mg thu c 0,420 g MgO. Tm

ng lng ca Mg?Trli:p dng nh lut bo ton khi lng cc cht, ta c:

MgOOMg mmm =+ 2 Thay s: 420,0,2530

2=+ Om gmO 167,02 =

p dng nh lut ng lng: OMg

O

Mg

m

m

=

12,128,1670.2530

=

=

= OO

Ng

Mg m

m

V khi lng ca oxi c trong oxit bng khi lng ca oxi tham gia

phn ng

V d 11: Phn tch st oxit, thy t l Fe l 70% v khi lng. Tmng lng ca Fe ?

Trli:

p dng nh lut ng lng, ta c:O

Fe

O

Fe

m

m=

67,18870100

70 === OO

FeFe m

m

Nhn xt: Tv d10 thy rng nh lut ng lng gip cho vic giinhanh bi ton ha hc m khng cn phi cn bng phng trnh phn ng hahc.

4. MT SPHNG PHP XC NH KHI LNG MOL PHN TCA CHT KH V CHT LNG DBAY HI

Xc nh khi lng mol phn tca mt hp cht l mt vic lm cn thit

khi lp cng thc phn tca mt hp cht. C hai phng php thng dngkhi xc nh khi lng phn tca cht kh v cht lng dbay hi.4.1. Da vo tkhi hi4.1.1. Khi lng ring ca mt cht kh

Khi lng (tnh theo gam) ca mt n vthtch kh (tnh theo lt haydm3) ti mt nhit xc nh v p sut xc nh, c gi l khi lng ringca kh .

K hiu: DXl khi lng ring ca kh X.

Tnh ngha, ta c:X

X

X V

mD =


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Trong mXl khi lng ca kh X chim thtch l VXti nhit vp sut ang xt. Nu xt 1 mol kh ti iu kin tiu chun, ta c:

4,22X

X

MD = Mx= 22,4.DX

Vi: MXl khi lng mol phn tca kh X.4.1.2. Tkhi hiTkhi hi ca kh A so vi kh B l tskhi lng ring ca kh A so

vi kh B cng iu kin vnhit v p sut.Hay: Tkhi hi ca kh A so vi kh B l tskhi lng ca V lt kh

A so vi V lt kh B cng iu kin vnhit v p sut.K hiu tkhi hi ca kh A so vi kh B l: dA/B, theo nh ngha ta c:

B

ABA D

Dd =/

Kt hp vi biu thc trn, ta c: BBAAB

ABA MdMMMd .// ==

Ch : khi xc nh tkhi hi ca kh A vi kh B th phi a vcng

th tch nhnhau, th khi t s:B

A

B

A

B

A M

M

m

md == mi l tkhi. Cn khi khc

nhau vthtch th tskhi lngB

A

B

A

M

M

m

m khng phi l tkhi.

V d1: Lng hi ca mt cht A nng hn lng kh nitcng iu

kin vnhit v p sut l 2 ln. Hy xc nh:a) Khi lng phn tca cht A.

b) Khi lng ring ca A ti 1atm v 250CTrli:a) ta c 28

2=NM , theo phng trnh BBAA MdM ./=

Tm c: 56282.22/

=== NNAA MdM b) C MA= 56 (g/mol)

Tphng trnh:T

PV

T

VP

o

oo = )(,452415273,.1

,15.298,4.221l

PT

TVPV

o

oo ==

Vy: )/(29,2,4524

56 lgV

MD AA ==

4.2. Da vo phng trnh trng thi ca kh l tng

Tphng trnh trng thi ca kh l tng: RTM

mnRTPV ==

RTPV

mM =

Tbiu thc ny ,chng ta tnh c khi lng mol phn tca kh cn

kho st.


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V d2: C 3 gam mt cht lng c ho hi hon ton, lng hi nychim mt thtch l 1,232 lt 250C, 1 atm. Xc nh khi lng mol phn tca cht .

Trli:

Tphng trnh : PV = nRT , n =m/M )( /,559

,2321.1

,15298....,0820.3

.

.. 11molg

latm

KKmollatmg

VP

TRmM ==

5. MT SPHNG PHP XC NH KHI LNG NGUYN TCA CC NGUYN T

5.1. Phng php Canizaro

Canizaro lm nh sau: ly mt s hp cht ca nguyn t cn kho st

em i xc nh khi lng phn tca cc hp cht , sau phn tch hahc xc nh sn vkhi lng ca nguyn tkho st c trong tng phnt. Gi trnhnht trong cc gi trkhi lng tm c ca cc hp cht trnsl khi lng nguyn tca nguyn t.

V d1: Xc nh khi lng nguyn tca cacbon, bng 1.8.

Csl lun ca phng php Canizaro: trong cc phn thp cht khcnhau ca cng mt nguyn tphi c mt phn tca mt hp cht chchamt nguyn tca nguyn t. V vy, nu ly c cng nhiu hp cht khcnhau ca mt nguyn tem i xc nh nguyn tkhi ca nguyn t th sc nhiu khnng c c mt hp cht m mt phn thp cht chchamt nguyn tca nguyn tcn xc nh nguyn tkhi.

BNG 1.8.Nguyn tkhi ca cacbon

Tn hp cht Phn tkhi % khi lngSn vnguyn t

cacbon c trong 1 phn t

hp chtCacbondioxit 44 27,27 (44.27,27)/100=12

Cacbonoxit 28 42,86 12

Axetylen 26 92,31 24

Benzen 78 92,31 72

Axeton 58 62,67 36

Tkt quthu c bng 1.8 cho thy nguyn tkhi ca cacbon phil 12.


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Tcc kt quthc nghim thu c bng 1.8, ddng xc nh ccng thc phn tca :

Cacbon ioxit: CO2, cacbon oxit: CO, Axetilen: C2H2, benzen: C6H6vaxeton l C3H6O. Hp cht chcha mt nguyn tcacbon trong th nghim trn

l CO2v CO.T c s l lun trn thy rng chnh xc ca phng php khngc cao v kt quxc nh nguyn tkhi phthuc vo khi lng cc chtem i kho st. V thnu ly c cng nhiu cht khc nhau th kt quthuc scng cao.

Tuy nhin, phng php ny c hn ch ch khng xc nh cnguyn tkhi ca cc kim loi v a scc kim loi khng to c cc hpcht thkh hay dbay hi.

5.2. Phng php uy Lng- Pti5.2.1. Phng phpuy Lng- Pti

a)Nhit dung nguyn t:nhit dung ca mt nguyn tl nhit lng cnthit nng nhit ca mt mol nguyn tca nguyn tln 1.

b) Nhit dung ring (t nhit): Nhit dung ring l nhit lng nngnhit ca mt gam cht rn ln 10.

Thc nghim xc nh c gn ng nhit dung nguyn tca ncht rn l: 265 J/mol.K 6,3 cal/mol.K

Nu k hiu nhit dung ring l C, th:

C.A 6,3 CA3,6

Bit c C th xc nh c A, y l phng php gn ng v nhit

dung nguyn tchp dng c cho mi n cht rn.5.2.2. Phng phpuy Long - Pti kt hp vi ng lng

Cc bc tin hnhBc 1: Da vo phng php uy Lng- Pti tm khi lng nguyn t

gn ng )(A :C

A3,6

Bc 2: Da vo nh lut ng lng tm ng lng ng:B

A

B

A

m

m

=

Bc 3: Xc nh ho trgn ngHbng biu thc:

=A

H

Bc 4: TH suy ra ho trchnh xc (H) bng cch: chly phn nguyncaHv ho trl snguyn.

Bc 5: Bit c ho trchnh xc H, th tm c khi lng nguyn tchnh xc (A) bng biu thc: A = H.

V d2: Mt kim loi c tnhit l 0,22. Khi oxi ho 0,162 g kim loi ththu c 0,306 g oxit. Xc nh nguyn tlng chnh xc ca kim loi:

Trli:Khi lng nguyn tgn ng ca kim loi l: 64,28

22,0

,36==A


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ng lng chnh xc ca kim loi l:

OO m

m=

98

,162306 0,0

,1620=

== O

Om

m

Ho trgn ng ca kim loi l: 18,39

,6428==

=

AH

Vy ho trchnh xc ca kim loi l: H = 3Do khi lng nguyn tchnh xc l: A = H. = 3.9 = 27 (Al)

5.3. Phng php khi phL mt trong nhng phng php tt nht xc nh khi lng nguyn

t.Sn gin ca my khi ph gm c 3 phn (hnh 1.4). B phn

ngun in vi hiu in thU to dng ion dng ca cc kim loi cn xcnh khi lng nguyn t(A), tip n l ng cong vi p sut kh trong ngrt thp. ng cong c t trong mt ttrng vi cng ttrng l B tch thnh cc dng ion ca cc ng v. Cui cng l bphn c gn knh nhthu vtr ca cng dng ion dng.

HNH 1.4. Scc bphn chnh ca khi phAston

Nguyn tc lm vic ca my khi ph: Da vo mi quan hca bn knhr ca quo chuyn ng ca ion dng c in tch q vi khi lng A caion dng :

2)(2

Bru

qA =

Tkt qu xc nh c A v t l ca ng v tng ng, t xcnh c khi lng nguyn ttng i ca ng vkho st.

Chng hn khi tm khi lng nguyn tca cacbon, php o khi phcho bit trong tnhin cacbon gm hai ng vl 12C v 13C vi tltng ngl 98,982% v 1,108%. T cc d kin thc nghim ny, xc nh c khilng nguyn tca cacbon trong tnhin l:

01108 1212,100 ,108.1892 1398,.12 =+


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C. HTHNG BI TPI. Bi tp tlun(c li gii v khng c li gii)

I.1. Bi tp c li giiBi 1 Cho bit skhi, sproton, sntron v selectron ca cc nguyn

t Na2311 v S3216

Li giiNa2311 S

3216

Skhi A 23 32Sprotron 11 16Sntron N 12 16Selectron 11 16Bi 21. Trong phng th nghim ngi ta thu c 27g nc. Hy cho bit:a) C bao nhiu mol nc?

b) C bao nhiu nguyn thiro?2. Tnh khi lng nguyn t tuyt i ca oxi,bit rng nguyn t ca

nguyn t ny c khi lng nguyn t tng i bng 15,9994. Cho N =6,022.1023mol-1Li gii1. Smol nc l:Sphn tnc c trong 1,5 mol nc l:1,5mol.6,022.1023phn t.mol-1= 9,0345.1023phn t

Snguyn thiro l:9,0345.1023.2 = 1,8069.1024nguyn t

2. Khi lng nguyn ttuyt i ca oxi bng:15,9994/6,022.1023= 26,564.10-24g

Bi 3 Mt nguyn t ca nguyn t X c bn knh l 1,44 Ao, c khilng ring thc l19,36 g/cm3. Nguyn tny chchim 74% thtch ca tinhth, phn cn li l rng.

a) Xc nh khi lng ring trung bnh ca ton nguyn t, t suy rakhi lng mol nguyn t.

b) Bit nguyn tang xt c 118 ntron v khi lng mol nguyn tbmg tng khi lng proton v ntron. Tnh sproton.

Li giia) Khi lng ring trung bnh ca ton nguyn tl:

dd10074= 16,2636,19

74100

74100 === dd g/cm3


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Vi gi thuyt nguyn t c dng hnh cu, bn knh bng 1,44 Ao, th

tch nguyn tc tnh bi cng thc 33

4rV = , do khi lng nguyn tl:

gm 2338 10.,70416 32,2610 ).441,1416 (,3

3

4 ==

Khi lng mol nguyn tl:12323 .796,196.10,70410 32022.,6. === molgmNM hay 1.197 = molgM

b) Theo bi, ta c: 79197118 ==+=+=+= ZZNZmmM np Bi 4a) Trong scc ht nhn nguyn tch )(20782 Pb c tsN/Z l cc i v

heli )(42He c N/Z l cc tiu. Hy thit lp t s N/Z cho cc nguyn t vi.822


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bay hi v ui khng kh ra khi bnh. Lm lnh bnh ti nhit phng, cnli c khi lng 25,817g. p sut kh quyn o trn p kl 742 mmHg.

Tnh khi lng mol ca benzen. Xc nh cng thc phn tca benzen.Li gii

Da vo phng trnh trng thi ca kh l tng, tnh smol benzen:

)10 (83.7,082 373,0

,24720760

7423 mol

RT

PVn =

==

Khi lng benzen = 25,817 - 25,201 = 0,616 (g)

Khi lng mol ca benzen l: )( .,2781088.,7

,6160 13

=== molgM

mM

Giscng thc phn tca benzen l: (CH)n, th (12,02 +1).n = 78,2Suy ra: n = 6. Vy cng thc ca ben zen l C6H6.Bi 6 Mt bnh c dung tch 2 lt cha 3g CO2 v 0,1g He 170C. Tm p

sut ring phn ca CO2v ca He, t xc nh p sut chung ca hn hpkh.

Li giiXc nh smol mi hn hp kh:

)0682(0,,0144

32

molnCO ==

)025(0,,0034

,10

molnHe ==

p sut ring phn ca mi kh trong hn hp:

)812(,02

0821 290,0,06820.2

2atm

V

RTnP COCO ===

)30(,02

0821 290,0,0250.atm

V

RTnP HeHe ===

p sut chung ca hn hp kh: )11(,1,30812 0,02 atmPPP HeCO =+=+= Bi 7 Mt bnh cu dung tch 1lt cha 2,69g PCl5 c lm bay hi

hon ton 250oC. p sut o c nhit ny l 1 atm. nhit nyPCl5c thbphn ly theo phn ng: PCl5(k) PCl3(k)+ Cl2(k)

Hy cho bit iu kin th nghim trn p sut ring phn ca cc khPCl5, PCl3, v Cl2l bao nhiu?

Li gii

Tnh smol PCl5khi cha xy ra phn ng phn ly)0129(,0

208

,6925

molM

mnPCl ===


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p sut ca PCl5khi cha phn ng )( 05PCl

p

)553(,01

082 523,0,012905

5

0 atmV

RTnP PClPCl ===

p sut ny nhhn p sut thc to c l 1 atm.

Vy, khi phn ng xy ra ta c: )(1235 atmPPPP ClPClPCl =++= Tphng trnh phn ng phn ly PCl5

PCl5(k) PCl3(k) + Cl2(k)

Ban u )553(,005

atmpPCl = 0 0

Phn ng - x +x +x

Ti thi im o xPP PClPCl =0

55 x x

Ta c: 1,5530550

=+=+=++= xxPxxxPP PClPCl Vy: x = 1 - 0,553 = 0,447Do : )(447,

23atmoxPP

ClPCl===

)106(,0,447553 0,0,55305

atmxPPCl === Bi 8 Cho 3,2 g ng tc dng vi oxi thu c 3,6 g mt oxit ca ng.

Xc nh ng lng ca ng v cng thc ca ng oxit to thnh.Li giiKhi lng oxi tham gia phn ng l: 3,6 - 3,2 = 0,4(g)Da vo nh lut ng lng, ta c:

O

Cu

O

Cu

m

m=

648,40

.23=== O

O

CuCu m

m

Ha trca Cu l: 164

64==

=

AH

Vy cng thc ca ng oxit to thnh l: Cu2O

Bi9 Mun trung ha 10 g mt dung dch axit nng 10% cn dng 10gdung dch KOH 12,47%. Tnh ng lng ca axit.

Li gii

Theo nh lut ng lng, ta c:KOH

axit

KOH

axit

m

m

= KOHKOH

axitaxit m

m=

Tcc dkin ca bi: gmaxit 1= ; gmKOH 247,1= ; 561=56 =KOH

Thay cc gi tri ny vo cng thc tnh c: 45=axit Bi 10 Ha tan hon 1,44 g mt kim loi M ha trII Vo 150 ml dung

dch H2SO40,5M c dung dch A. trung ha dung dch A cn 30 ml dung

dch NaOH 1M. Xc nh tn ca kim loi tham gia phn ng. Li giiPhng trnh phn ng: M + H2SO4MSO4+ H2 (1)


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Phn ng ca axit dc trong dung dch A vi NaOH:H2SO4+ 2NaOH Na2SO4+ 2H2O (2)

Smol H2SO4em dng: )075(,0,5.0,15042

moln SOH ==

Smol NaOH tham gia phn ng (2): )03(.0.1.030 molnNaOH ==

Smol H2SO4d:2SO4H

n d-= )015(,021 molnNaOH =

Smol H2SO4 tham gia phn ng (1): )06(,0,012075 0,01)(42 moln SOH ==

Theo (1): )06(,01)(42 molnn SOHM == , tng ng vi 1,41 g.

T tnh c khi lng nguyn tkim loi:AM= 1,41:0,06 = 24 g/mol (Mg)

I.2. Bi tp khngc li gii

Bi 1 in vo bng sau:

K hiu nguyn t Mg2412 Ag10647 Ba

13756

Skhi - - -

Sin tch ht nhn - - -

Sproton - - -

Selectron - - -

Sntron - - -

Bi 2 Nguyn tca mt nguyn tk hiu l X, c tng scc loi htl 193, trong sproton l 56.

a) Hy xc nh skhi ca X.b) Tnh khi lng nguyn tv khi lng ht nhn ca nguyn tX

va tm c. Cho bit tskhi lng nguyn tv khi lng ht nhn tcho nhn xt.

Bi 3 Da vo nh ngha hy xc nh khi lng nguyn t ra kg cho

mt n v khi lng nguyn t (1u). T kt qu tnh c hy suy ra khilng nguyn ttuyt i ca oxi, bit oxi c khi lng nguyn ttng i l15,99749u.

Bi 4a) Tnh khi lng mol nguyn tca Mg v ca P nu bit khi lng

nguyn ttuyt i (KLT) ca chng l 40,358.10-27 kg v 51,417.10-27 kg.b) Xc nhkhi lng tuyt i ca N v Al nu bit khi lng tng

i (klt) ca chng l 14,007u v 26,982u.Bi 5: Tnh ng lng ca axit v baztrong cc phn ng sau:

a) Ca(OH)2+ H3PO4 CaHPO4+ 2H2Ob) 3NaOH + H3PO4 Na3PO4+ 3H2Oc) Ba(OH)2+ 2H3PO4Ba(H2PO4)2+ 2H2O


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Bi 6 Cn bng phng trnh v tnh ng lng ca cc cht oxi ha vcht khtrong cc phn ng sau:

a) KMnO4+ H2C2O4 + H2SO4K2SO4+ MnSO4+ CO2+ H2Ob) KMnO4+ Na2CO3+ H2SO4MnO2+ Na2SO4+ KOH

c) K2Cr2O7+ FeSO4+ H2SO4Cr2(SO4)3+ Fe2(SO4)3+ K2SO4+ H2Od) CrCl3+ Br2+ KOH K2CrO4+ KBr + KCl + H2OBi 7 Mt kim loi M tc dng va vi 4,032 lt kh Cl2iu kin

tiu chun, thu c 16,02g MCl3theo phng trnh phn ng:2M + 3Cl22MCl3

a) Xc nh khi lng nguyn tca kim loi M.b) Tnh khi lng ring ca M, Suy ra tlphn trm ca thtch thc

sso vi thtch ca tinh th, bit rng M c bn knh 0,431 Ar = v khi lngring thc l 2,7g/cm3.

Bi 8 Cquan nghin cu v trNASA cho rng c thdng BaO2cung cp O2cho tu vtrdo phn ng:2BaO22BaO + O2

a) Hy cho bit cung cp oxi cho con tu c thtch 10.000 lt 250Cv 0,20 atm th khi lng BaO2l bao nhiu?

b) Hi sau bao lu th lng oxi sc dng ht, nu 200C on phihnh tiu th1 lt oxi trong 1 pht.

Bi 9 Mt bnh cu c th tch V cha mt kh l tng p sut 650mmHg. Ngi ta rt mt lng kh c thtch 1,52 cm3p sut 1atm. Kh cn

li trong bnh gy ra mt p sut 600 mmHg. Xc nh thtch V ca bnh vigithit l mi php o u tin hnh cng mt nhit .Bi 10 Ly 2,70g hn hp A gm canxi cacbua v nhm cacbua ho tan

trong dung dch HCl 2M th thu c mt lng kh c tkhi hi so vi hirobng 10.

a) Xc hn thnh phn phn trm ca hn hp A.b) Tnh thtch kh thu c 270C v 836 mmHg.c) Tnh p sut ring phn ca tng kh trong hn hp kh thu c iu

kin cho.II. Bi tp trc nghim

Bi 1 Trong 280 g st c bao nhiu nguyn tst? Khi lng ca mtnguyn tst l bao nhiu gam?

p na) 3.1024 nguyn t;9,33.10-23 g b) 3.1024 nguyn t; 8,54.10-23 gc) 3,5.1023 nguyn t; 9,68.10-23 g d) 3,5.1023 nguyn t; 8,97.10-23 gBi 2 C bao nhiu mol phn tN2 trong 280 g N2? ktc, lng nit

trn chim thtch l bao nhiu lt?p na) 9,94 mol; 224 b) 10 mol; 224

c) 10 mol; 225 d) 9,95 mol; 225Bi 3 trung ho 7 g axit cn 8 g KOH. Xc nh ng lng ca axit.


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p na) 48,80 b) 48,85c) 49,00 d) 49,00Bi 4Xc nh ng lng ca lu hunh trong cc oxit cha 40% v

50% lu hunh vkhi lng.p na) 5,34 v 8,00 b) 5,30 v 8,20c) 5,37 v 7,95 d) 5,33 v 8,00Bi 5 Mt bnh cu cha 7 kg oxi di p sut 35 atm. Khi oxi c rt ra

khi bnh th p sut gim xung ti 12 atm. Hi c bao nhiu kg oxi c lyra?

p na) 4,54 kg b) 4,75 kgc) 4,85 kg d) 4,60 kgBi 6 Mt bnh cu c y kn cha 200g oxi di p sut 3,5 atm v

70C. Tnh khi lng cacbon ioxit trong cng thtch , cng nhit nhng di p sut 2,8 atm.

p na) 215 g b) 218 gc) 220 g d) 222 gBi 7 Xc nh thnh phn thtch ca hn hp oxi v ozon, bit tkhi

hi ca hn hp ny i vi hiro bng 17,6.

p n:a) 1:2:

32=OVO V b) 1:4: 32 =OVO V

c) 1:3:32

=OVO V d) 1:1: 32 =OVO V

Bi 8 Mt hn hp kh nitv hiro c tkhi hi so vi hiro l 3,6.Sau khi un nng vi bt st 5500C th tkhi ca hn hp so vi hiro tngln bng 4,5.

1. Xc nh thnh phn theo tlvsmol ca hn hp kh trc v sauphn ng.

2. Tnh phn trm thtch ca nitv hiro tham gia phn ng.p n

1. a) 4:1:22

=N nHn ; 2:5:1:: 322 =NHHN nnn

b) 3:1:22

=N nHn ; 2:3:1:: 322 =NHHN nnn

c) 3:2:22

=N nHn ; 2:3:2:: 322 =NHHN nnn

d) 2:1:22

=N nHn ; 1:3:1:: 322 =NHHN nnn

2. a) 50% N2 v 36,5% H2 b) 50% N2 v 37,5% H2c) 45,5% N2 v 37,5% H2 d) 45,5% N2 v 36,5% H2


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Bi 9 Magie c khi lng mol l 24,31g/mol, c khi lng ring l1,738 g/cm320oC. Tnh:

1. Khi lng nguyn ttrung bnh ca mt nguyn tMg2. Thtch ca mt mol nguyn tMg

3. Thtch trung bnh ca mt nguyn tMg4. Bn (knh tnh ra picomet) gn ng ca nguyn tMg, bit nguyn tMg c dng hnh cu

p n1. a) 4,030.10-23g b) 4,030.10-22g

c) 4,037.10-23g d) 4,037.10-23g2. a) 13,88 cm3/mol b) 14,00 cm3/g

c)12,87 cm3/g d) 13,99 cm3/g3. a) 2,340.10-23 cm3 b) 2,340.10-22 cm3

c) 2,323.10-23 cm3 d) 2,323. 10-22cm34. a) 178 pm b) 177 pm

c) 176 pm d) 179 pmBi 10 Khi in phn nc, thu c 3,6g mt hn hp kh. Lng kh

ny chim mt thtch l 6 lt 170C. Tnh p sut ring ca oxi v hiro.p n

a) atmPH 7920,02 = , atmPO 3920,02 = b) atmPH 7827,02 = , atmPO 3863,02 =

c) atmPH 7927,02 = , atmPO 3963,02 = d) atmPH 7929,02 = , atmPO 3969,02 =

Bi 11 Mt bnh kn c dung tch 1 lt cha 2,34 g CO v 1,56 g CO2300C. Xc nh p sut ring phn ca mi kh v p sut ton phn.

p na) 2,18 atm, 0,96 atm v 3,14 atm b) 2,08 atm, 0,88 atm v 2,96 atmc) 2,08 atm, 0,981 atm v 3,061 atm d) 2,08 atm, 0,88 atm v 3,061 atm


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CHNG 2

CU TO NGUYN TV HTHNG TUN HON CCNGUYN THA HC

A. MC TIU NHIM

1. MC TIUHc xong chng 2 sinh vin bit v hiu:- Thuyt lng tPlng.- Lng tnh sng ht ca nh sng.- Thuyt Bri.- Nguyn l bt nh Haixenbec.- Tin vhm sng- Bi ton h1 electron, 1 ht nhn- Cc khi nim: Mt xc sut, obitan nguyn t(AO), spin electrron,

obitan spin (ASO),...- Hnh dng cc obitan s, p, d- Bn slng tv mi quan hgia bn slng t- Cu hnh electron ca nguyn t. Cc c s vit ng cu hnh

electron ca nguyn t.- Cc vn vht nhn nguyn t.- nh lut tun hon v h thng tun hon, cc tnh cht bin i tun

hon.

2. NHIM V

Tm hiu vcc mt svn tin chc lng t, chc lng tvcu to nguyn t, t hiu c v c khnng vn dng c cc kinthc ca chng vo thc hnh v luyn tp.

3. PHNG PHP

Thng xuyn s dng cc phng php quy np v loi suy vo vic

nghin cu cc kin thc ca chng ny di shng dn ca gio vin. Cntng cng rn luyn thng qua cc bi tp luyn tp v thc hnh hiu suhn vcc vn ca chng, t nng cao c nng lc tduy ca ngihc.

B. NI DUNG

Mu:Nguyn tl phn tnhnht ca mt nguyn tha hc m vnmang cc tnh cht ha hc ca nguyn t, vmt in tch th nguyn t

trung ha vin. Vthnh phn th nguyn tgm c ht nhn mang in tchdng do cc ht proton v ntron to thnh, cn vmang in tch m do cc


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ht electron to nn. chng ny trc tin chng ta tm hiu vlp vca htnhn trc sau mi nghin cu ht nhn sau.

1. MT S VN TIN C HC LNG T V CU TO CA

VT CHT1.1. Lng tnh sng ht ca nh sng. Thuyt lng tca M. Planck1.1.1. Thuyt sng nh sng

- nh sng l sng in t, lan truyn trong chn khng vi c = 3.108m/s= 300.000km/s, c c trng bng:

+ Bc sng (khong cch lan truyn trong mt dao ng)

+ Tn sdao ng :

c

= (2.1) (sdao ng trong 1 giy: s-1)

+ Ssng v: c == 1 (2.2) (sdao ng c thc hin khi sng truync mt on l , n v: Cm3)

c im ca nh sng: Khi truyn tmi trng vt cht ny sang mitrng vt cht khc, vn tc nh sng gim nhng tn skhng thay i.

u im: Gii thch c nhng hin tng c lin quan n s truynsng nh: giao thoa, nhiu x, hin tng quang in.

Nhc im: Khng gii thch c cc dkin thc nghim vshp thv pht sng khi i qua mi trng vt cht.1.1.2. Thuyt lng tca M. Planck(Nh bc hc c 1858 1947, Nobelvt l 1918)

Vo nhng nm u ca thk20 vi hng lot cc pht minh mi ra i.Trong cc pht minh ny nu dng cc nh lut vchc v nng lng ca

Niutn th khng gii thch c. Nm 1900 nh Vt l hc ngi c lM.Planck a ra thuyt lng tmang tn ng: "Mt dao ng tvi tn schc thpht ra hay hp thnng lng theo tng n vnguyn vn, tnglng gin on, c gi l lng t nng lng. Lng tnng lng tlthun vi tn sca dao ng". Ni cch khc, nh sng c tnh cht ht vn mang nng lng:

h= (2.3)

h l hstl, hin nay gi l hng sPlng34.10,6256 =h J.s

l nng lng ca photon l tn snh sng (s-1) ngha quan trng ca thuyt lng tPlng l pht hin ra tnh cht

gin on hay tnh cht lng tho ca nng lng trong cc hvi m. T,Plng cho bit nh sng ngoi tnh cht sng cn c tnh cht ht. Tnh cht htc da ra di dng githuyt.


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V d 1: Hy cho bit v sao hng s Plng h c th nguyn l [nnglng].[thi gian]?

Trli: Ththc h= , ta c

=h (2.4)

M T1= (T l khong thi gian dao ng t thc hin c 1 daong - chu k dao ng)Vy Th = . T biu thc ny chng t h phi c th nguyn l [nng

lng].[thi gian]T thy rng, Tnh cht gin on ca nng lng c gi l tnh cht

lng tha ca nng lng . S lng tha ny l tnh cht chung ca cci lng vt l.

Theo thuyt lng tca Plng, mt dao ng tkhi dao ng vi tn sphi c nng lng blng tha l E vi cc gi trgin on l snguyn

ln lng ttc dng, tc l:0, 1, 2,..., n

Hay: 0 , h , h2 , ..., nh Vy c thbiu din mt cch tng qut nng lng E theo cng thc

nhE = ; vi n = 0; 1; 2;...V d2: Mt bc xin tc tn s= 3,7717.1014s-1. Tm nng lng

do bc x gy ra vi n = 1; 2; 3. Tnh cho 1 vi ht v 1 mol vi ht.Trli:

Vi 1 vi ht, ta c:E = nhVi 1 mol vi ht, ta c:Emol= N.E = 6.023.10

23.nhThay s, ta c:

n = 1: E1= 1.6,625.10-34J.s.3,7717.1014s-12,49875.10-19 J

Emol1,505.105J/mol

n = 2: E24,9975.10-19J

Emol3,01.103J/moln = 3: E37,49625.10-19J

Emol 4,515.105J/mol

1.1.3. Lng tnh sng, ht ca nh sng

Theo thuyt tng i ca Anhxtanh nmg lng v vn tc c ca nhsng c lin hvi nhau qua biu thc:

2mc= (2.5)Mt khc, theo hthc ca Plng: h=


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Kt hp hai hthc trn, ta c:

2mch = hayh

mc2= ,

c=

Nn :mCh= (2.6)

Hthc (2.6) thhin tnh cht sng - ht ca nh sng hay tnh cht nhnguyn sng - ht ca nh sng: Tnh cht sng c th hin ch khi lantruyn trong khng gian vi vn tc c th c bc sng , cn tnh cht ht thhin chmi ht photon mang mt khi lng l m.

Nhvy, khi thuyt tng i ca Anhxtanh ra i chng minh chogithuyt ca Plng vtnh cht ht ca nh sng l hon ton ng n.

Tcc biu thc trn t c:

chmc == 2 c.m =h/

Tch sc.m chnh l xung lng (p) ca photon ang xt, tc l:

p= m.c = h/ p = h/ (2.7)Xung lng p biu thcho tnh cht ca ht photon, ( hay )biu thcho

tnh cht sng ca photon. V thphng trnh (2.7) cng biu thcho sthngnht vtnh cht sng v ht ca nh sng.

V d3: Tnh nng lng v xung lng p ca photon vng nh sngxanh (=6,66.1014s-1)ca by sc cu vng.

Trli:Ta c 11434 10.,666..10,6256 == ssJh

J19.10,412254

Ta ch

p = vc

= nnc

hp

=

Do : 118

1142234

..,470751.103

.10,666.....10,6256

= smkgms

sssmkgp

Lu : 1J = kg.m2.s-2

V d4: Hy cho bit kin sau y ng hay sai? ti sao?"Photon l mt loi ht cbn nn n phi c khi lng m cnh"Trli:

Ththc

hmcp == . , ta c

1

=c

hm

m l khi lng ca photon, h v c l cc hng s bit, nn:

1

..103....10,6256 8

1234

=

smssmkgm )(1..10,212 42 kgm

=


38/275



Nhvy, khi lng m ca photon c trsphthuc vo trsca bcsng . Do khng thni nhtrn c.

Xt cho v d3, tnh c m = 4,911.10-36kg.

1.2. Hthc Bri. Nguyn l bt nh Haixenbec1.2.1. HthcBriXut pht t c im ca electron trong nguyn t, nm 1924 bng

phng php loi suy Bri suy rng tnh cht sng-ht ca nh sngkhng phi l tnh cht ring bit ca nh sng, m l tnh cht chung cho miht vt cht bt k nh: electron, proton, rtron,... T ng a ra githuyt:

Schuyn ng ca mi ht vt cht c khi lng m v vn tc v ulin kt vi mt sng c bc sng c xc nh theo hthc:

vm

h

.

= (2.8)Ni mt cch n gin:Mi dng vt cht vi m u c tnh cht sng v

tnh cht ht.Trong biu thc (2.8) iukin ht vi m hay vt thvm c tnh cht

sng l: Vn tc vphi l v cng lnV nguyn tc, h thc Bri c nghim ng cho mi ht vi m

cng nhvt thvm. Tuy nhin cc vt thvm khc ht vi m chccvt th v m c khi lng rt ln cn vn tc chuyn ng li rt nh nn

bc sng ca sng lin kt tnh theo hthc trn c gi trv cng nh, do

tnh cht sng khng cn c ngha.V d5: p dng hthc Bri, hy tnh bc sng cho cc trnghp di y ri rt ra kt lun cn thit.

a) Mt electron c khi lng me= 9,1.10-31kg, chuyn ng vi vn tc

v = 106m.s-1.b) Mt chic xe vn ti khi lng bng 1 tn, chuyn ng vi vn tc v

= 100 km.h-1.Li gii:

a) 010631

2234

,287.10,287

.10..10,19

....10,62561 Am

smkg

ssmkg

mc

h====

Nhn xt: Vi di sng lin kt tnh c, th sng lin kt Bri cmt ngha quan trng v lun i km vi hvi m. V kch thc ca nguyntrt nhc10-10m = 1A0.

b) 02853

2234

.10,3852

360010.10

....10,6256

.A

smkg

ssmkg

cm

h

=== Nhn xt: Xe vn ti thuc hvm, sng lin kt cc k nh, nn khng

c ngha.V d6: Hy xc nh bc sng cho hai trng hp sau:

a) Khi electron chuyn ng vi nng lngl 1eVb) Khi proton chuyn ng c nng lng l 1eV


39/275



Li giia) xc nh bc sng ca electron, ta phi xc nh c vn tc v

ca n. Khi electron chuyn ng n mang mt ng nng l:

2

2mv

T = mT

v

2

=

Vy:01019

1931

2234

122610.26,1210.,22676110.,610 1.2,92

...10.,6256

2Amm

ssmkg

mT

h

mv

h===

===

b, Khi proton chuyn ng ta c thp dng cng thc:

hchE == Ehc=

Thay s, ta c: 010619

834

1241210.1241210.,2412110.,61

10.10 3625.,6AmmE ===

=

1.2.2. Nguyn l bt nh bt nh HaixenbecTcc kt qutrn chng ta thy: Cc vt thvm do c khi lng rt

ln (c thcoi l mt v cng ln so vi ht vi m), nhng vn tc chuyn ngca chng li rt nh, v vy da vo chc cin ca Niutn hon ton xc

nh c chnh xc quo chuyn ng ca cc vt thvm. Tc lxc nhc chnh xc cta ln vn tc (xung lng) ca vt thvm. Nguynnhn chnh y l do vn tc chuyn ng ca vt thvm rt nh, nn vtthvm khng c tnh cht sng.

i vi cc ht vi m, v dnhelectron lun tn ti mt sng vt chtlin i i km (do vn tc ca ht vi m l v cng ln). Nn trong thgiichuyn ng ca ht vi m khng cn tn ti khi nim quo hnh hc thngthng. Nhvy khi nim quo hon ton khng cn c ngha trong vicm tchuyn ng ca electron hay mt ht vi m bt k.

T, Haixenbec a ra nguyn l c ni dung nhsau:"Khng thxcnh c ng thi chnh xc cta ln vn tc (hay xung lng) ca ht vim, do khng th xc nh c chnh xc quo chuyn ng ca ht".

Ni dung ny c pht biu vo nm 1927.

Hthc ca nguyn l: )9.2(.m

hvq

Cng c tha ra hthc ca nguyn l di dng:

h pq. (2.10) vim

h=h

Trong : q: bt nh vta .

v: bt nh vvn tc (hay bt nh vxung lng)


40/275



h = const, doTbiu thc (2.9) nhn thy, vi ht vi m chn, tl

m

:v

constq

=

Ththc ny ta thy, nu q cng nhtc l ta q ca ht c xcnh cng chnh th vn tc v ca n cng thiu chnh xc, tc l v cng ln vngc li.

Nhvy, i vi ht vi m, khi bit c chnh xc vn tc chuyn ngca n, th khng thxc nh c chnh xc to(q), ng i ca n (tcl khng xc nh c chnh xc quo chuyn ng ca n). T khinim quo trnn khng cn c ngha khi nghin cu cc ht vi m. Thayvo , ch c th ni n xc sut c mt ca n mt v tr no trongkhng gian.

V d7: Da vo nguyn l Haixenbec, hy thtnh bt nh vvtrq ri cho nhn xt vi cc trng hp sau:

a) Githit ve= 3.106m/s ; me= 9,1.10

-31kgb) Mt vin n sng sn vi m = 1g chuyn ng vi vn tc 30m/s. Gi

thuyt rng sai stng i v vn tc cho c2 trng hp l v/v = 10-5.Li gii

a)Vi e:mv

h

mv

hq

.10.. 5=

V:

= 510

v

v

089316

2234

10 22.22.10,22.10,19...103

....10,62561 Ammkgsm

ssmkg====

b)Vi vin n:01927

315

2234

10.,21210.,21210..10.30

...10.,6256Am

kgsm

ssmkgq

===

Gi tr ny qu b khng mt thit b no o c. Vy nguyn l btnh Haixenbec vi hvm khng c ngha.

V d 8: Mt ht c ng knh c 1 micron, khi lng 10-10g. Htchuyn ng Braonvi vn tc khong 10-4cm/s. Gi thuyt php o ta

t mc chnh xc vo khong 1% kch thc ht. C thxem ht l ht vim (nhelectron) c hay khng? Hy gii thch.Li gii

Nu mt ht xc nh c chnh xc cta ln vn tc (xung lng)ca ht mt trng thi chuyn ng th khng thcoi ht l ht vi m.

Xt cth:K hiu ng knh ca ht l x, ta c: x = 1m = 10-6m

Nn x = 1%.x = 10-2.10-6m = 10-8m.Theo bi: m = 10-10g = 10-13kg

cm sm skgm

ssmkgmx

hmq

hvx1113

138

2234.10,6256.10,6256

10.10....10,6256

..

===

=

=


41/275



scmvscmvx411 10.10,6256 ==

Vn tc chuyn ng ca ht ny c sai srt nhso vi chnh gi trcavn tc . iu ny c ngha l vn tc (hay xung lng) ca ht cng c

xc nh hon ton chnh xc nhta . V vy ht ny l ht vm chkhngphi l ht vi m (vi ht)

V d 9: Gi s php o ta x ca electron c chnh xc vokhong 10-3, ng knh nguyn t (khong 10-8cm). C th xc nh cchnh xc vn tc chuyn ng vxca electron hay khng?

Li giiTheo bi: x = 10-3.10-8cm = 10-11cm

Ta c: smkgm

ssmkg

mx

hvx

103113

2234

.10,6256

.10,19.10

....10,6256

.

==

=

So snh gi tr xv va tm c vi vn tc nh sng trong chn khng l3.108 m/s, nhn thy xv > c. Vy khng th xc nh c vn tc vx caelectron khi bit chnh xc ta ca n. Nhvy khi nim quo chuynng ca electron l khng c ngha.

2. MT SNGUYN L V TIN CA CHC LNG T2.1. Tin vhm sng. Nguyn l chng cht trng thi2.1.1. Tin vhm sng

Mi trng thi ca mt ht vi m (electron) c m tbng 1 hm sxc nh phthuc vo ta q v thi gian t , k hiu l ),( tq c gi l hmsng hay hm trng thi.

Trong trng hp ht vi m trng thi dng - l trng thi m nnglng ca hkhng phi l mt hm sca thi gian t - hm sng hay hmtrng thi ca ht vi m chl hm sca ta q , khi hm trng thi cvit l )(q . Ty, chng ta chxt hm sng trng thi dng

Vnguyn tc: Mi thng tin vht vi m u thu c thm )(q * ngha ca hm sng )(q :+ Hm sng )(q khng c ngha vt l trc tip nhng bnh phng m

un ca hm sng 2)(q cho bit mt xc sut tm thy ht ti ta tng

ng no trong khng gian.+ Xc sut tm thy ht trong mt phn t th tch dV bao quanh mt

im no trong khng gian l dVd 2=

Nhvy mt xc sut c xc nh bi hthc:dV

d= 2

+ Nu ly tch phn trong ton b khng gian th xc sut tm thy ht

trong ton bkhng gian bng 1 (theo l thuyt xc sut).1

2 =giankhng

dV (2.11)


42/275



Biu thc (2.11) c gi l iu kin chun ho ca hm sng. Hmsng khi thomn iu kin ny c gi l hm chun ha.

Ch : Hm sng )(q phi thomn mt siu kin: n tr, lin tc,hu hn v khvi.

2.1.2. Nguyn l chng cht trng thiNguyn l ny bao gm cc ni dung sau:1. Nu mt ht vi m vo trng thi c m tbi hm sng )(q th

trng thi cng c thc m t bi hm sng )(qc , vi c l tha s(hng s).

T ngi ta ni: trng thi ca mt ht vi m c thc m tbicc hm sng m cc hm ny chkhc nhau mt hng s.

2. Nu mt ht vi m vo trng thi c m tbi hm sng )(1 q hoc vo trng thi c m tbi hm sng )(2 q th ht vi m cng c

thc m tbi hm sng )(q , vi:)()()( 2211 qcqcq += (2.12)

c1, c2 l cc hng sc gi l hs thp hm sng m cc hs thp c ngha l tl(hay trng s) ng gp ca cc hm sng tng ngvo hm sng )(q . Biu thc (2.12) c pht biu l: hm sng )(q l thp tuyn tnh cc hm )(1 q , )(2 q .

Nu ht vi m ang xt c thvo nhiu hn hai trng thi, chng hn ntrng thi m mi trng thi c m tbng mt hm sng fi, tc l: f1, f2,...,fi,..., fnth ht vi m ny cng c thvo trng thi:

=

=++++= ni

iinnii fcfcfcfcfcf1

2211 ... (2.13)

V d1: Slai hoa sp l slai ha trong c sthp tuyn tnh mthm obitan nguyn t (AO) s vi mt hm AO-p (c th ly AO-px, AO-py,hoc AO-pz), kt quc hai hm sng lai ha:

)(2

1

2

1

2

1..1 pspspcsc ps +=++ ==

)(2

12

12

1..2 pspspcsc ps = ==

Ti sao cs= cp? Ti sao csv cpu bng2

1

Li giiHai hm sng AO-s v AO-p hon ton bnh ng trong vic m t

chuyn ng ca electron trong nguyn t. Do khi electron c xt vo

trng thi c m tbi mt trong hai hm thp tuyn tnh trn, th tnghm AO-s v hm AO-p u ng gp mt cch bnh ng nhau vo hm sngthp tuyn tnh. Vy buc phi c cs= Cp


43/275



Theo ngha vt l ca hm sng th sng gp ca cc hm s v p vohm sng thp tuyn tnh l ng gp vxc sut, tc l:

2

2

2

1

=sc v

2

2

2

1

=pc

Ta d thy mi lng trn u bng 1/2. Kt quny tng ng trongmi hm lai ha sp th tng hm s, hm p ng gp mt na, tc 50%.

2.2. Tin vphng trnh SringNm 1926, Sringl ngi u tin a ra mt phng trnh vsau

mang tn ng. y l mt trong nhng phng trnh u tin v quan trng cachc lng t.

Hm sng m t trng thichuyn ng ca ht vi m phi thomnphng trnh:

)()( qEqH = (2.14)Phng trnh (1) l phng trnh Sring khng ph thuc thi gian(phng trnh Sringtrng thi dng)

Trong : Um

hH += 2

2

2

8 c gi l ton tHamintn.

T phng trnh Sringc thvit di dng

0)(8

2

22 =+ UE

h

m (2.15)

Phng trnh (2.14) v (2.15) l phng trnh Sringtrng thi dng.

Trong :m l khi lng ca ht vi mE l nng lng ton phnU l thnng ca ht.h l hng sPlanck

2

2

2

2

2

22

zyx

+

+

== c gi l ton tLaplacePhng trnh (2.14) v (2.15) l phng trnh vi phn o hm ring cp

2.Mi thng tin vhht vi m u sc bit, nu gii c phng trnh

ny,c ngha l tm c cc gi trca hm v nng lng E tng ng.Gii phng trnh Sring s thu c v s nghim 1, 1... n lnhng nghim c lp, tnguyn l chng cht trng thi, ta c:

= C 1. 1+ C 2. 2+ ... C nn cng l nghim ca phng trnh v thu ccc gi trnng lng tng ng l E1, E2, ...., Enca ht vi m ng vi cctrng thi .

Ch : Ch nhn nhng hm ph hp vi iu kin chun ho hmsng (phng trnh (2.11)).

3. H1 E, 1 HT NHN (nguyn tH v cc ion ging H)3.1. Mt skhi nim cn bit3.1.1. Hta cu


44/275



a)Hta ccC cc trc x, y, z tng ng vi cc bin sx, y, z.

HNH 2.1.Hta cu v ta cc

b) Hta cuC ba bin s:Gc : c to bi chiu dng ca trc Oz vi vc tvtr r

Gc : c to bi hnh chiu ca vc tr

trong mt phng xOy vichiu dng ca trc Oxdi rr hay r (cn gi l moun ca vc t )

Trsca cc bin s:

====

0

20

0

rr

(2.16)

c) Mi lin hgia ta cc v ta cu

===

cos

sinsin

cossin

rz

ry

rx

(2.17)

3.1.2. Trng lc i xng xuyn tmTrng lc c gi l trng lc i xng xuyn tm khi lc tc dng ln

ht chuyn ng trong trng i qua mt im cnh c chn lm tmca trng (im cnh ny c chn lm gc ta ).

Trong trng lc i xng xuyn tm, ln ca lc tc dng ch phthuc vo khong cch tvtr ca ht v tm ca trng m khng phthuc

vo phng.Do thnng )(rU r ca ht chl hm ca khong cch rr , tc l:

U = U(r) (2.18)

x

y

z

O

r


45/275



3.1.3. H1e, 1 ht nhnNguyn tH v nhng ion m lp vchc 1e ging nhH, nh: He+,

Li2+, He3+, nhng hny chkhc nhau vsin tch ht nhn (Z).V d1: in tch (+) ca ht nhn nguyn tH l +e, cn cc ht nhn

khc l +Ze.Nhn xt: Bi ton vchuyn ng ca electron trong nguyn thiro lbi toncbn nht ca chc lng tvcu to nguyn t. Nhng kt quthu c ca bi ton ny sl cscho vic xy dng l thuyt chung vcuto nguyn t.

3.2. Bi ton h1e, 1 ht nhnHgm : + 1 ht nhn tch in dng vi sn vdin tch dng l Z

Ze0(e0l in tch nguyn t). Vi nguyn tH ht nhn chc 1 proton, ccht nhn khc c Z proton.

+ Mt electron, c in tch -e0, c k hiu l e chuyn ngquanh ht nhn. Electron c khi lng l:

kgmmm Hpe31.10,19

1840

1

1840

1 ===

HNH 2.2.M hnh h1e, 1 ht nhnr

: vc tvtr ca electron

C thm thny nhshnh 2.2. Trong hnh ny, n gincoi ht nhn ng yn v nm ti gc ta (gc O). Cn electron chuyn ngxung quanh ht nhn vi v tr ca n so vi gc ta c xc nh

bng vc tvtr rr

.Thnng U ca hc tnh theo biu thc:

r

erUU

20)( == (2.19a) (vi nguyn thiro, Z = 1)

r

ZeU

20= (2.19b) (vi ion ging hiro, Z = 2, 3, 4,...)

Trong rr r l moun hay di ca vc t v tr, r l di ca

khong cch (ca vtr) tvtr ca electron n ht nhn.Nh vy thc cht ca h ny l: Xt chuyn ng ca electron trong

trng lc ht nhn c sn vin tch dng Ze0.

Phng trnh Sringca hc dng:

Ze0

e

r


46/275



EUzyxm

h=

+

+

+

22

2

2

2

2

2

2

8 (2.20)

V nguyn tc tnh cht i xng cu nn vic gii phng trnh (2.20)trong hta cc gp nhiu kh khn nn chuyn (2.20) sang hta cu

vi cc bin smi r, , sthun tin hn.Khi chuyn sang ta cu (2.20), c dng:

0)(811

)(2

2

22

2=+

+

rUEhm

rrr

rr (2.21)

Trong : (lam a) l phn gc ca ton tLaplace (2)

+

=+

=2

2

2

22

22

sin

1sin

sin

1

11

rr

rrr

(2.22)

T(2.21) v (2.22) phng trnh Sring (2.21) c vit gn thnh:

( ) 0)r(UEh

m82

22 =

+ (2.23)

3.3 Nguyn tc gii phng trnh Sringy l mt trong st cc phng trnh ca chc lng tc c li

gii chnh xc. gii c phng trnhg Sring(2.23), cn tin hnh theocc bc:

Bc 1: chp nhn

),().(),,( YrRr = (2.24)Trong : ),( Y l phng trnh phthuc gc

)(rR L phng trnh phthuc bn knhBc 2: Thay (2.24) vo (2.23) ri bin i ton hc ta s c mt

phng trnh mi l tch ca hai phng trnh. Mt phng trnh ph thucgc-phng trnh hm gc v mt phng trnh phthuc vo bn knh-phngtrnh hm bn knh.

Bc 3: Gii tng phng trnh+ Gii phng trnh hm gc, ta c v s nghim l cc hm gc

, )( Y + Gii phng trnh hm bn knh, ta c v snghim l cc hm bn

knh )(rR Bc 4: Ly tch cc hm ),().( YrR mt cch ph hp vmi quan h

gia cc slng t(trnh by phn sau), ta sc v snghim l cc hmsng ),,( r .

Ch : Chly cc nghim ph hp vi u kin chun ho ca hm sng

=tn nguygiankhngTon

12

d (2.11)T cc nghim ca phng trnh Stringsc gi trhu hn.

3.4. Kt qugii bi ton 1e, 1 ht nhn


47/275



Gii phng trnh Sringcho h1e, 1ht nhn thu c

a) Hm bn knh

Phng trnh hm bn knh (chphthuc khong cch r) c dng:[ ] =+ )(2 22 rUEmrRR

rr

h (2.25)

Vi: )1( += ll Vic gii phng trnh ny cho ta nghim l hm bn knh )(, rRn l hay

)(rRnl , c dng:

[ ]

+

= +

+

on

naZr

n na

Zre

na

Zr

a

Z

nn

nrR L

22

)!1(

)!1(4)(

12

10

2

3

034

0

ll

l

l (2.26)

n l slng tchnh; cn l l slng tph(slng tobitan)Z l sn vin tch dng ca ht nhn

a0l bn knh Bo thnht, 00820

2

0 53,,5290.10,5290 AoAcmmea ===

h

e l cslgarit tnhin

+

+ 0

12

1

2

na

ZrL

n

l

l a thc Lagher

r l bin schkhong cch tht nhn n vtr ca electron ang xt

Hm bn knh Rnl(r) l cc hm ton hc c chun ho, cc hm nyl cc hm thc phthuc vo bn knh r v hai slng tchnh n v slngtph l. ng vi mi mt bhai slng tn, l ph hp nhau vmi quanhta sc mt hm Rnl(r) phthuc vo bn knh r. C v sbhai sn,l nnsc v shm bn knh Rnl(r).

Mt shm Rnl(r) cchun ho ca hmt electron, mt ht nhn

+ n = 1; l = 0: 0.2)(2

3

0

110a

Zr

s e

a

ZRrR

==

+ n = 2; l = 0: 0.2.22

1)(

0

2

3

0220

a

Zr

s ea

Zr

a

ZRrR

==

+ n = 2; l = 1: 0..62

1)(

2

5

0221

a

Zr

p era

ZRrR

==

+ n = 3; l = 0: 0320

2

0

2

3

030 21827

381

2)( a

Zr

ea

rZ

a

Zr

a

ZrR

+

=

+ n = 3; l = 1: 0320

2

0

23

0331 6

681

4)( a

Zr

p ea

rZ

a

Zr

a

ZRrR

==


48/275



+ n = 3; l = 2: 0320

222

3

0332

3081

4)( a

Zr

d ea

rZ

a

ZRrR

==

* Slng tchnh n+ S lng tchnh (n) l cc snguyn dng, khc 0; n nhn cc gi

tr: n = 1, 2, 3, 4...., .+ nha ca slng tchnh n:

- Xc nh khong cch trung bnh ca electron n ht nhn.

++=

202 )1(1

2

11.

n

ll

z

anr

- Xc nh cc mc nng lng trong nguyn t:

2

0

22

40

22

)4(

1...2

=

hn

ezmEn cc e cng mt lp c E nhnhau.

- Xc nh tng smt nt ca Obitan = (n-1) khng ksmt nt xav cc.

Mt nt: L khu vc khng gian (thng l mt phng) m hm sngtrit tiu, )(q = 0 0)( 2 =q . Tc l ngay trong khng gian nguyn tcngc nhng khu vc khng tm thy electron.

- c trng cho lp Obitan (lp electron): Tt ccc obitan nguyn t(AO) c cng slng tchnh (n) th thuc cng 1 lp.

Trsca n = 1 2 3 4 5 6 7 .....

Lp electron K L M N O P Q .....

* Slng tphl (slng tobitan)+ Slng tphl l cc snguyn, l nhn cc gi tr:

l = 0, 1, 2, ..., (n-1).+ Quan hgia slng tphl v slng tchnh nng vi 1 gi trca slng tchnh n, th slng tph l nhn n

gi trt: l= 0 (n-1), tc l 0 l (n-1)V d1: 01 == ln c 1 gi trca l = 0

1;02 =

=

= lln c 2 gi trca l= 0, 12;; 103 ==== llln c 3 gi trca l = 0, 1, 2+ ngha ca slng tphl:

- Xc nh phn lp electron trong nguyn t:

Trsca l 0 1 2 3 4 ...

K hiu ca phn lp electron s p d f g ...

Dthy: Vi 1 gi trca slng tchnh n th c n phn lp electron.V d2: 01 == ln c 1 phn lp (k hiu l phn lp 1s)1;02 === lln c 2 phn lp (k hiu l phn lp 2s, 2p)


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2;; 103 ==== llln c 3 phn lp (k hiu l phn lp 3s,3p.3d)Cc hm bn knh tng ng ln lt c k hiu l: R30(r); R31(r); R32(r)Ch : Chcc AO c cng slng tchnh nv slng tphl mi

thuc cng mt phn lp.

- Xc nh hnh dng hay tnh cht i xng ca cc AOAO-s : hnh cu, i xng cu (c tm i xng)AO-p : hnh s8, i xng trc (c trc i xng)

- Xc nh momen ng lng Obitan (M) ca electron:2

).1(h

llM +=

V d3: Khi electron obitan c l = 0, thM= 0; obitan c l = 1 th,

2h

2M=

Ch : 1, Tmi quan hca n v l phng trnh (2.26), ta thy

cho cn bc hai c ngha th biu thc trong cn bc hai phi dng, nn101 nn ll

2, Vic gii hm bn knh dn n vic xut hin hai slng tn,l mt cch hon ton tnhin.

b) Hm gc(hm cu)Phng trnh hm gc (chphthuc v gc , ) c dng:

0=+= YYY

Y (2.27)

Vic gii phng trnh (2.27) thu c nghim l hm gc ),(, ll mY hay),(

llmY .

Hm gc ),(, ll mY : - Phthuc vo hai bin sgc , - Phthuc vo hai tham sl slng tph l v

slng tt lm * Slng tt+

lm l cc snguyn nhn cc gi tr:

lm = -l , (-l+1),..., -1, 0, 1,...., ( l -1), l .+ Mi quan hgia l v lm :

l = 0 l

m = 0l = 1

lm = -1, 0, 1

l = 2 lm = -2, -1, 0, 1, 2Vy: ng vi mt gi trca l , th lm snhn (2l + 1) gi tr.

Nh th th ng vi mt gi trca s lng tchnh n sc bao nhiugi trca lm ?

n = 1 l = 0 lm = 0 c 12= 1 gi trca lm

n =2 l = 0 lm = 0

l = 1 lm = -1, 0, 1 c 2

2

= 4 gi trca lm


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Vy: ng vi 1 gi tr ca s lng t chnh n s c n2 gi tr ca slng tt lm

+ ngha ca slng tt lm - Xc nh sObitan nguyn t(AO) c trong 1 phn lp: Vi 1 gi tr

ca l c (2 l +1) gi trca lm . Tc lphn lp l c (2 l +1) AO.- Xc nh hnh chiu ca m men ng lng Obitan trn trc z

2

hmMz l=

Biu thc ca mt shm gc (hm cu) chun ha:

+ l= 0; lm = 0 4

1),(00 =Y (1)

+ l=1; lm = 0

cos4

3

),(10 =Y (2)

lm = 1

ieY .sin

8

3),(11 =

lm = -1

ieY = .sin8

3),(11

+ l= 2; lm = 0 )13cos(165

),( 220 = Y

lm = 1

ieY .cossin

815),(21 =

lm = -1

ieY = .cossin8

15),(12

lm = 2

ieY 2222 .sin32

15),( =

lm = -2

ieY 2222 .sin32

15),( =

Cc hm (1), (2) c k hiu: 00Ys = ; 10Yzp = Cc hm cu cha so i (so i2= - 1 1=i ) l hm phc. Trong

trng hp sphc, khi thp tuyn tnh cc hm ny mt cch thch hp thc cc hm thc. Cc hm khng cha so i lun l hm thc nn khngcn thp tuyn tnh, l cc hm cu c lm = 0.

Chuyn cc hm cu phc thnh hm thc v k hiu tng ng(n gin khng ghi bin s)

+ l= 0; lm = 0 s =4

1Y00 =


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+ l= 1; lm = 0 cos

4

3= 10 =Ypz

+ l= 1; lm = 1 cossin43

)(2

11111 =+= YYpx

+ l= 1; lm = -1 sinsin43

)(2

11111 == YY

ipy

+ l= 2; lm = 0 )13cos(165 2

202 == Ydz + l= 2; lm = 1 coscossin4

15)(

2

11221 =+= YYdxz

+ l= 2; lm = -1 sincossin415

)(2

11221 == YYidyz

+ l= 2; lm = 2 )sin(cossin1615

)(2

1 222222222 =+= YYd yx

+ l=2; lm = -2 sincossin415

)(2

1 22222 == YY

idxy

c) Hm sng: ),,( rmn ll Hm sng m ttrng thi ca electron trong h1e, 1ht nhn l tch ca

hm bn knh )(rRnl v hm cu ),( llmY :),().(),,(

ll ll lmnmnYrRr =

Vic ly tch ca hm bn knh v hm cu phi ph hp vmi quan hgia 3 slng tn, l, lm .

V d4:

00

2

3

0

2

3

000101100

1

4

1

.2.

a

Zr

a

Zr

s ea

z

ea

z

YR

=

==

=

Lm tng tta c

02

0

2

3

000202200 2

24

1. a

Zr

s ea

Zr

a

ZYR

===

Vi hiro Z=1, xt trong h n v nguyn t th a0 = 1 khi

re= 1100 Vi 200 ,... lm tng tnhtrn.


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d) Cc mc nng lng tm cGii phng trnh hm bn knh, cn thu c biu thc tnh nng lng

20

22

4 20

2

20

22

40

2

)4(

12

)4(

1

2

==

hn

Zme

n

emZEn

h(erg) (2.28)

Vi:2h=h

Trong : n l slng tchnhm l khi lng ca 1 electronh l hng sPlng

Trong biu thc tnh nng lng tt ccc i lng u tnh theo n vca hcgs. n vca nng lng l ec hay erg: 1ec = 1erg = 10-7J.

Du (-) biu thnng lng ca e cn chu tc dng ca lc ht ht nhn

nguyn tng vi in tch Z . Encng thp (cng m), tc l slng tchnhn cng b th hcng bn. Hbn nht khi Enthp nht (Encc tiu), ng vi n= 1.

Nu xt trong h n v nguyn t (vn hay au), vi quy c cc ilng sau u bng n v:

+ in tch cbn e0= 1+ Khi lng ca 1electron bng 1; me= 1+ Hng sPlng bng rt gn+ Bn knh Bo thnht a0= 0,529A

0= 1

+ Hng sin mi trong chn khng 0 v tch 14 0 = Nu tnh theo hn vnguyn t, th:

2

2

2

1

n

ZEn = (vn) (2.29)

Nu nng lng tnh ra eV, th:

2

2

,613n

ZEn = (eV) (2.30)

V d5: Hy tnh E1, E2, E3,... theo n veV cho nguyn tH. Vi n bngbao nhiu th Enc gi trm nht.Li giiVi nguyn thir Z=1, th:

)(1

6,132

2

eVn

En = Vy: E1= -13,6 (eV); E2= - 3,4 (eV); E3= - 1,15 (eV); E4= - 0,85 (eV);

E5= - 0,544 (eV); ...; E = 0.

Vi n = 1 th E1= -13,6 (eV) l m nht.


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Nhn xt:- Nng lng Enchnhn nhng gi trgin on, tc l nng lng b

lng tho.- E1c gi trm nht (hay mc nng lng ng vi n = 1 l thp nht).

y l trng thi c bn ca electron trong nguyn t hiro, trng thi nyelectron lin kt bn nht vi ht nhn nguyn t. V vy c thni:"Hbn lhc nng lng cc tiu".

- Vi n = 2, 3, 4, ... l cc trng thi kch thch ca electron trong nguynthiro.

- Khi n = , En= 0, electron chuyn sang trng thi khng lin kt vi htnhn, khi ny electron khng bht nhn ht na, tc l electron tch ra khinguyn tv chuyn ng tdo, cn nguyn tth bin thnh ion dng.

3.5. p dng kt quca bi ton h1e, 1 ht nhn gii thch quang phvch ca nguyn thiro

3.5.1. Cc trng thi nng lng ca e trong nguyn tHNng lng Enphthuc vo slng tchnh n.n = 1 lp K (trng thi cbn) E1= - 13,6eVn = 2 lp L (trng thi kch thch) E2 = - 3,4eVn = 3 lp M (trng thi kch thch) E3= - 1,5eVn = 4 lp N (trng thi kch thch) E4= - 0,8eV. .............................................. ...................

n = En= 0Khi ny electron tch ra khi trng lc ht ca ht nhn, nguyn t

schuyn thnh ion dng (+).

3.5.2 M tquang phca nguyn thiro

Quang phca nguyn thiro - mt quang phpht xn gin nht,c thto ra c bng cch: Phng in qua mt ng mao qun cha kh hirop sut rt thp, cvi mmHg, nhmt my quang phlng knh m thu c

quang phvch.mt vng vi bc sng xc nh, ta c mt dy tng ng gm cc

vch phdi nhau. Cc vch phdi nhau l du hiu c trng ca quang phnguyn t- quang phvch

Trong vng nh sng nhn thy (vng khkin), thc nghim cho kt qudy Banme c hnh nh nhsau:


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HNH 2.3. Cc vch ca dy Banme trong vng nhn thy

ca quang phnguyn thiro3.5.3. Gii thch quang phvch ca nguyn thiro

T biu thc tnh nng lng (2.28) tm c khi gii phng trnhSring. Vi nguyn thiro Z = 1, ta c:

22

402

2 hnemZEn = (2.31a)

V2h

=h nn22

40

22

hn

meEn

= (2.31b)

Nh vy, khi electron trong nguyn t hiro trng thi ng vi hmsng ( c l p-xi) c gi trn xc nh xc mt nng lng xc nh.

Gisc hai trng thi sau:- Trng thi c n nh(thp), k hiu l tn , nng lng tng ng l tE - Trng thi c n ln (cao), k hiu l cn , nng lng tng ng l cE Theo (2.31b), ta c:

2

40

22

hn

meE

cc

= ;

2

40

22

hn

meE

tt

=

cE

Tbiu thc (2.28), ta thy n cng cao th nng lng Entng ng cngcao. Vy Ec cao hn Et. Do khi electron mc nng lng Ec chuyn v

trng thi c mc nng lng Etthp hn sgii phng ra mt nng lng E.tE E=

Nng lng ny c pht ra di dng sng in t c tn s , theoPlng

hE =

M:

c

= , vi c l vn tc nh sng trong chn khng. Thay cc biu

thc ny vo biu thc tcEE E= , ta c: == 222

42

1121tc

otc nnh meEEhc


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Vy

=223

42 1121

ct

o

nnch

me

(3.18a)

K hiu: 13

40

2

1096782 == cm

ch

meRH

RHc gi l hng sRitbe (v tt c cc i lng v phi u l hngs),do :

==22

111

ctH nn

R

(3.18b)

Trong c gi l ssng.

Ch :Vi ion ging H th:

=22

11..

1

ctH nn

ZR

(3.19)

V d6: Hy tnh trs, c km theo n vca hng sRitbe t:a) Cc sliu ca hng sc trong bng 1.6 (trang 12)

b) Thc nghim cho bit vch c 6565 A0Li giia) Tbng 1.6 (trang 12), ta c:h = 6,625.10-27erg.s; c = 2,99792458.1010cm.s-1; e0= 4,803.10

-10ues cgs;m 9,1093897.10-28g

Thay cc hng sny vo biu thc tnh RH, c:

110323

410282

,351097721099792458.2,.)10.,6256(

10 )803.,410 .(1093897.9,.),1416(32

= cmRH

b) Vch trong quang ph vch ca hiro ng vi s chuyn dielectron t nc= 3 vnt= 2.

Theo biu thc (3.18b):

=

22 3

1

2

11HR

Vy: 18

109622105636..5

36 == cmRH

Nhvy, RHtnh theo thc nghim nhhn tnh theo l thuyt mt t.Da vo mi lin hgia schuyn di electron vi bc xnng lng

km theo,ta c scc dy vch quang phpht xca nguyn thiro hnh2.4 (trang 56).

+ Dy Laiman gm cc vch ng vi schuyn di electron tmc nnglng c n 2 vmc n = 1. Dy ny trong vng tngoi (tm).

+ Dy Banme gm cc vch ng vi schuyn di electron tmc nnglng c n 3 vmc n = 2. Dy ny trong vng khkin.

+ Dy Pasen gm cc vch ng vi schuyn di electron tmc nng

lng c n 4 vmc n = 3. Dy ny nm trong vng hng ngoi.


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Trong mi dy li gm c mt svch tch ri nhau, v cn c mt svch khng tch di nhau, to ra vng phlin tc (hnh 3.1). Min lin tc nynm vng c bc sng ngn hn trong gii hn ca dy. Chng ta chxt

phn gm cc vch tch di nhau.

V d7: Hy tnh bc sng v ssng ca vch phu v vch phcui ca dy Laiman, ca dy Banme.Li giiXt dy LaimanVch u: nt= 1; nc= 2. Theo biu thc (3.18b), tnh c:

122

5,. 82258,7502

1

1

1 =

= cmRR HH

HNH 2.4.Mt sdy quang phvch ca hiro

081

121510.1215,582258

11Acm

cm===

Vch cui cng: nt= 1; nc= . Vn sdng (3.18b)

122

22

22 109678111

111 ==

= == cmRRR HHH


57/275



081

7,18 9117.,91110678

11Acm

cm===

Xt dy Banme: Lm tng t, th c (vi nt= 2)

Vch u: 1,0615233 cm ; 06560A

Vch cui: 1,527419 cm ; 03640A 3.6. Cc khi nim cbn c rt ra tvic gii bi ton h1e, 1 ht nhn3.6.1. Cc Obitan nguyn t, ngha, cch biu din.

a) Obitan nguyn t (AO):L nhng hm sng m t trng thi chuynng ca e trong nguyn t.

b) ngha: 2 cho bit mt xc sut c mt e ti mt ta no

trong khng gian nguyn t.c) Cch biu din:C thdng nhiu cch

* C 2 cch phbin:1. Dng thphn gc ca cc AO biu dinchnh cc AO .2. Dng thhm mt xc sut theo gc: C

dng tng tphn gc ca cc AO nhng khc chchtt ccc min ca thu mang du dng hoc bng khng (0) v cng dng biu din ccAO.

* Hnh dng cc AO v du ca nAO-s: Dng hnh cu

HNH 2.5:Hnh dng cc AO-1s, AO-2s, AO-3s

HNH 2.6.Hnh dng hm cu 2p, v hm mt xc sut ,biudin cc AO-p


58/275



AO-p: Dng hnh s8 ni (cc AO-px, AO-py, AO-pzhng theo trc x,y, z tng ng).

cos4

310 == zpY

Nu coi

41 lm n v, ta c: cos310 == zpY , khi ny hm mt xc

sut tng ng l:

( )2210 cos3 =Y Xt sto thnh hnh nh khi kho st hm Y10:Khi = 0 900, hnh nh ca hm ny l hnh cu ng knh 3 . Hnh

cu ny nm phn dng ca trc z.Khi = 9001800, hnh nh hm Y10 tng t trn, nhng nm phn

m ca trc z.Hai hnh tip xc vi nhau ti gc to, nhvy trong mt phng

xOy (mt phng vung gc vi trc z) khng c im biu din ca hm pz; tani hm pztrit tiu (bng 0) trong mt phng xOy. Dng khi nim mt nt nu trn,vy mt phng xOy l mt phng nt hay mt nt ca hm Y10hayhm pz.

HNH 2.7.Hnh dng cc AO-2px, AO-2py, AO-2pz, biu dintheo hm mt xc sut

Cc AO-2px, AO-2py, AO-2pz c hnh dng ging

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