ا)('%ن ا #$%
"! ا ا و (دة ا ء وا +ء
ا م ا ا ا را 2008/2007
ا "#1+ء )6ن(
ا -1-F+
ها ا ول ددات ا ت ا . 4' – Aا :3!+ *+ا ) ا ا ( ' & $%ن ! ،ا ر *+ - Kا .1 , (1أ /ا (1) (%و 89 :6آ 7ر 56آ 3ا 6ب ; ا )& > 0,5) .( Y2ن( (2او -ا ; د ا (B) , 7ا * ?$ا ) 0,5) . u AB (tن(
*=6& :6 u ABا ( )& > ( Y1و ا E
−t @$ (3أن ; A u AB (t ) = E 1 − e RC د ا 0,5) .(B) , 7ن( & (4ذا :$% )C 6 τ = RCا ( ؟ أ 59ا ? A HEد &; ا ;& G9د ا ;6Fد و 0,75) . ? Aن( (5ا J (2) (%ات ) L 6 u AB (tا A ،:&Kد &) τ ! /؟ ) 0,25ن(
; 3!+( <#= -Bدارة : RL 9ا N/ t=0 M$ر ! O-ا ر إ Gا ،2 ,ا J (3) (%ات ) *=6& :6 u AB (tا ( L 6ا .:&K M/ & (6م ا 66ت ؟ ا3ح
ذا و ا ) & T u AB (tا 0,5) .:&Kن( (7او -ا ; د ا , 7ا * ?$ا ) A HE u AB (tد ا ار ا NCول > :9د ا 66ت 0,5) .ن( (8ا! ا &Uأن ه Wا ارة :( Lا;
? $ا ، Laذا؟ ا!ح ) 9 Cا G9 &U 0,25) .H?(%& Aن( AB – Cا @? ا >. ا MAL ،&Uا ا :& 8/ا
( :إ ?- + ,ز ارة ا / X ;Cا 66ت. (9ا3ح دور ها ا ? ز & :ا Aا = !؟ ) 0,25ن( Z7/ );/ (10ا د ا *+ ;Cا ارةJ & ،ات ) L 6 u AB (tا / [ ;6 :&Kا 66ت 0,25) .ن( (11أ ); 59ا ور ا ص T0و ا دد ا ص 025) . N 0ن( (12ا ارة ا )( 6 =)& 66ا Xت ا ي $ل ا -ا (? ] 6إ [ - & Gذات دد 0,25) . N 0ن( أ -ا ، N 0 `CAه ا @+ه ا ا دد ا 0,5) L= 0.1H *=;/ .Laن( ب & -ا آ) ا( aو /ا * ( X$ ? ! 5), :ل G9ا La؟ - 9ا 0,25) .b6ن( ج 9 L 5)d/ -ا & ، 232mHا ا ا Xدرة :9ا ? ز 0,5) .ن( (' Eا 4#%
(ن4) تH ا I اه#KL 4 د4+ I λ = 768nm 1%H( از ?ل%( د ا نP أB (P AR زرN ز1H K "ا1>T اI FS'@; D >>a2 ، D>>a1 VP 4+R ا4( D=2,5m ;R( I WX .a 1>a2 VP a2 وa1 ا% ا
.Y"اH F ء؟X >%R ر ا% [ ( أ اE>'% Fن أ( ه0,75 .Y F+ E; ا اF+ )أ(و)ب(ا4+ ا4" 4( دP ] ؟P^ ة ا#ن ب( ( ه ا ]ه0,75
(1
. آ، L = 20mm 1B# B >" اI ;@ هa1 `#KL يb اYR " c%' وa2 ويR `#KL يb اYR اF (2 .2 >T اd@=
. ه#= A أT ، θ ف# ( ن أ0,5 . ( ب1 :الfR ا$% 4( آg= T D وλ ، L [ " a1 #= دP ، ة#hi θ ر% " (ب ن1 . a1 L k%@%ج( ا ن1 ( )ن3ا ء ا @و . Pb صi# ة اI إPo 'ل ة ا م%= (1 . n%% (د اo%ن أ( اآ0,5 :K . n%% ا4 $=@ اLK اMeV : بT ل$ " oRP ن ب( ا0,5 1u = 1,66 × 10 −27 Kg ، c = 3 × 10 8 m / s ، e = 1,6 × 10 −19 C . m( Pb) = 206,0385u ، m(α ) = 4,00394u . ln a = f (t ) ا اoRP 210 84 Po @ ةa ) ا @ ط ا#h% (2 206 82
، m( Po) = 210,04824u ،
210 84
. e ln a = a : K . a #= ( دP ( ن أ0,5 . dRP اT t1 / 2 3S@ ا# ف# ( ن ب0,5 . mo = 10 g : هt = 0 ]' @ ا1%% آ210 84 Po ة ا ن4( @ #% (3 t = 1h ]' > @ ا% ا ن ا% آoRP ا، F%+ R@ " ) ن ا @ ط اL ل%" ( ن أ0,5 . t = 1h ]' > @ ا% د ا @ى اk%@% ن ب(ا0,5
ا ( ء )7ن(: ? ف ها ا :إ Gدرا j A 9 7ا b / aوا / &Lك & ا
ء و= ر > j A 5ا b / aوا / &Fك *+ ا
ء. ا ت Hا /زه 9در -ا $ارة . 25°C 6 Eا d $ ر&Kه ا Kدو 9 ! -ةj A/ − pK A = 4,7 أ ن ا / aات j A/اb / a CH 3COOH / CH 3COO + pK A = 9,2 ا / &Fك/أ ن ا / &Fم NH 4 / NH 3 pK A = 0 ا
ء/أ ن اFآ / Cم H 3O + / H 2 O pK A = 1,4 أ ن ا ? روآ/ Cا
ء H 2 O / HO −
-1ا $اء ا rء: :1-1أ ; 59ا اء ا */ Fء 0,25) .ن( A:2-1د ! 8ا :& !U=/ا ;= ت ا 0,25) .6 Cن(
sP F = -2ا)! W( Yا ء. j A d/ :1-2ا b / aا Tإ Gا
ء $& G9 X$+ ،ل V1 = 10ml 8 A S1وآWK 0,25) . C1 = 2.10 −2 mol / lن( ;=* ! س pHا $ل . pH 1 = 3,2 : S1 :1-2اآ` &; د j A 9 7ا & b / aا
ء 0,25) .ن( (2-2ار 5=& Hا ? Kدو -أ ن ا / aات j A/ا b / aوا nا ع ا ? *+ :ا $ل 0,5) . S1ن( A (3-2د ا م ا ? ]* j A 9 7 x1 fا & b / aا
ء :( ) .ا 6 / ;Lول ا م( 0,5) .ن( ! (4-2رن x1 fوا م ا x1 m GX!Fإذا آ ن ا $ل آ 0,25) .ن( (5-2ا )C/ nا م ا ? ]* 0,25) .9 7ن( (6-2ه ا & ا / & +ا NCال2-2؟ - 9ا 0,25) .b6ن(
F = (3ا (rك ( Wا ء: −2
r6 d$/ذا ' 6ز ا / &Fك *+ا
ء & V2 = 10ml 8 A S 2 L $وآ ! *=; . C 2 = 10 mol / l WKس pH
ا $ل S 2ا . pH 2 = 10,6 :1-3اآ` &; د 9 7ا / &Fك & ا
ء 0,25) .ن( (2-3ار 5=& Hا ? Kدو -ا / &Fك/أ ن ا / &Fم .وا nا ع ا ? *+ :ا $ل 0,5) . S 2ن( A (3-3د ا م ا ? ]* j A 9 7 x 2 fا & b / aا
ء :( ) .ا 6 / ;Lول ا م( 0,5) .ن(
ه ا & ا / & +ا NCال2-3؟ - 9ا.b6
K= (4ر sP Auا)! Yوا (rك ; ا ء:
، ك/ &F ا:& 10 −4 mol وb / a اj A :& 2.10 −4 mol / l ا
ءG إ+ ,r6 V = 20ml 8 A S L $& d$/ : ذي ا ; د9 7 6 [ $ ل ا$ ج ا/ CH 3 COOH + NH 3 ( aq )
( aq )
CH 3 COO − + NH 4 ( aq )
+
( aq )
( ن0,25) .] ) ا$ * ا+ 9 Qr ,i 9 7 ` > رج اCA ا:1- 4 ( ن0,5) ؟nC & ذا، Qr ,eq وQr ,i ! رن:2-4 ( ن0,5) .( ول ا م6 / ;L ا:( ) .9 7 x3 f *] ? ا م اL 6 Qr ,eq :9 )9 :3-4
:-دوK ا:& ( )C 6 ] ? ا$ * ا+ ا
دةXA :& !U=/ ا:9 آ ؟9 ل ا$ ) ر9 ا:( ه:4-4 + S ل$ اpH >s
ذاC+. S ل$ * ا+ ? اع ا ( ] ا/Fا، NH 4 / NH 3 وCH 3COOH / CH 3COO − ( ن0,5) . ا ازن9 pH = 4,7 ا
: > اX' ة ا# ( ا5 = ا6 رd/ . C a = 10m.mol / l WK آb / a اj A ل$& :& Va = 20ml A سs* آ+ d/ ل$ اpH ، & - pH = ا6 Z/ وCb = 20m.mol / l WK د م آX اC ه روآ:& Vb A A $ .+ ,; آ إ6 89 X$ ا ( ن0,25) . 9 اd $ ز ا ; ة ا/L ;C & ا ? ز ا )* ا:1-5 ( ن0,25) . ا ; ة9 7 اآ` &; د:2-5 ( ن0,25) .N+ ( اG9 لX$ 8+ , ا ! إVbE H $ ` اCA ا:3-5 . pH = 4,8 G9 X$/ د مX اC ل ه روآ$& :& Vb = 5ml + , إ9 & - pH ? ز- ;
ل6 :4-5 ( ن0,5) . pKe = 14 :*=;/ . ة9 !-j A ا ; ة9 7 *] ? ) ا م اC/ * & ه. = ا ةY b" وWL @ء اv إFH أ4( #$% 'ن ا%([ اWB أن !ا (اb(^% ة واb=r ا4( H# : اء SBIRO ABDELKRIM
E-MAIL
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msn :
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Correction E =
K>@ ا
ب$ ا
Lر الfR ا (1
0.5
Y2
Y1 0.5
0.5
0.75
E = u AB + u R : ات#=% ; اB ن إL dq q وu AB = ، u R = R.i : أن i= dt C du AB 1 E + u AB = :(B) B% إذن ا د ا dt RC RC −t t du AB E ( − RC ) = e وu AB (t ) = E 1 − e RC @ dt RC du AB (B) ; ا د وu AB (t ) و ض dt
(2
(3
(4
@ τ = RCو (5
(6 (7
] C = [I ][Tو ] [U ] [I ] [U
=R
إذن ا > ار " d τز(@.
و ] [I ][T ] [U ] = [T ] [U ] [I
=τ ) u AB (V
)t(ms ( τ = 0.001ms : $ ]م ا "b"bت dدوري. @> yو Wا "b"bت " oRوHد ا >و( . R Lن إ ;Bا #=%ات u AB = u R + u L :
0.25
τ 0.5 0.5 2
ا د ا : B%
(8 (9 (10
(11 أ- ب- ج-
d u AB R du AB u AB + + =0 L dt LC dt 2
du AB ا 4( L^Kا د ا o$ H #== I FS' + ، B%أن +ن ((F dt . L ≠ 0 V'" R=0و" z; %ن ا >و( Rه ا fRو u 4د ا "b"bت. ا K"#( 4" #=%ا S' H #v 3!+ل #= Iدد ( 4ا; Eا @? ا >. ا[#%Lاح ا 1H : #$%ز ا ) Sا ] #ا رس( دور ا 1$ز s= ; 4+ا LKا دة "ل Hل .
T0 = 2π LC N 0 = 502.8 Hzوه [ ا;#= Eدد ا @?
1 T0
أي
= N0
La
" أن ); N 0 = f(L ,Cن ا #آ ا 1KB 4+ %ه (( s#'= Fا أو ا .3!+ L=232mHو d z; N 0 =330Hzا; Eا @? ا > .Mi
0.25 0.25 0.25
0.5 0.5 0.25 0.5
ا Bع ا ! ; ا ء:ا Hت (1أ( [ ( E>'%أ ا[ %ر ا X >%Rء r،ن " ا%Hز` ا # %@ YRا Xء ; WHا[=$هت ;@' I FSا >0,75 W ا ه ة ; ا .1 %>T ........................................................................................................................................................................ ب(ه `bا ]ه#ة = : IRه#ة ا 'د. آ آن #KLا #hi YRا آ +ن #ض ا > ا #آ أآ.# وا ) F+ب( ا; Eا YRا bي R `#KLوي . 0,75 a1 و" : %ا )F+أ( ا; Eا YRا bي R `#KLوي . a 2 ..................................................................................................................................................................... (2أ( : θﺘﻤﺜل ﺍﻝﻔﺭﻕ ﺍﻝﺯﺍﻭﻱ ﺒﻴﻥ ﻭﺴﻁ ﺍﻝﺒﻘﻌﺔ ﺍﻝﻤﺭﻜﺯﻴﺔ ﻭﺃﻭل ﺒﻘﻌﺔ ﻤﻅﻠﻤﺔ .ﺍﻨﻅﺭ ﺍﻝﺸﻜل.
λ
= θوو 1= Pا #ادن rad
و=#ه ه: a1 ..................................................................................................................................................................... 2λ D λ L/2 L = a1 و(@: d = R@ " tgθوا ا #hSة = ب( (^u 4ل ا F+ا : E"R = tgθ ≈ θ 1 L 2 D a1 D ^ cPأن #KLا # W( R+ o@% YRض ا > ) آ آن #ض ا > ا #آ آ#ا آ آن #KLا #hi YRا(. L1 > L2و" %ا )F+أ( ا ا;r Eآ# #ض > ا; Eا YRذو أ #KL #hiأي . a 2وا ) F+ب( ا;. a1 E ....................................................................................................................................................................... 2λD 2. × 768.10 −9 = a1 = ج( ت.ع× 2,5 = 192.10 − 6 m . 1 −3 L 20.10 ....................................................................................................................................................................... ا Bع ا ! ; Vا ء:ا ء ا @و:
( (1د ا : n%%
A Po→ 206 82 Pb + Z x
⇐ @ :ا 'ظ د ا '@210 = 206 + A : ⇐ ا 7$/ظ 9د ا (84 = 82 + Z : 4 إذن ا ! ا ); ه* د! αوه* /اة ا ? م . 2 He :
0,5
210 84
A=4 Z =2
4 Po→ 206 &; د ا 82 Pb + 2 He : u7
210 84
---------------------------------------------------------------------------------------------------- $ )C 6 (2ل: 4 Po→ 206 82 Pb + 2 He
210 84
@ :
E = ∆m.c = [(m( Pb) + m( He) − m( Po)].c = [(206,0385 + 4,0039 − 210,0482)]u × c ≈ −5,8 × 10 u × c 2 −3
2
2
2
= −5,8 × 10 −3 × 1,66 × 10 −27 Kg × [3.10 8 m / s ] ≈ 8,67 × 10 −13 J 2
(3ﺍﳌﻨﺤﲎ ﺍﻟﺬﻱ ﳝﺜﻞ ﺗﻐﲑﺍﺕ ln aﺑﺪﻻﻟﺔ tﻋﺒﺎﺭﺓ ﻋﻦ ﻣﺴﺘﻘﻴﻢ ﻻ ﳝﺮ ﻣﻦ ﺃﺻﻞ ﺍﳌﻌﻠﻢ ﻣﻌﺎﻣﻠﻪ ﺍﳌﻮﺟﻪ ﺳﺎﻟﺐ ).ﺩﺍﻟﺔ ﺗﻨﺎﻗﺼﻴﺔ( ﻣﻌﺎﺩﻟﺘﻪ ﻛﻤﺎ ﻳﻠﻲ ln a = k .t + 6,9 :
ﺍﻷﺭﺗﻮﺏ ﻋﻨﺪ ﺍﻻﺻﻞ .6,9
(ln a) B − (ln a) A 6,9 − 5,9 = = −5 × 10 −3 Jours −1 tB − t A (0 − 200) J ln a = −5.10 −3.t + 6,9
"zدuل دا ا vر %ا #ي ;#? Iه `bا R%و =: S
=k )t +6,9
−3
e ln a = e ( −5.10
e a +b = e a × e b a = 992 × e − 5 .10
−3
.t
:ﻭﻣﻨﻪ
λ = 5.10 −3 Jours −1 : ﳒﺪ، ﺇﺫﻥ ﺑﺎﳌﻘﺎﺭﻧﺔ ﻣﻊ ﺍﻟﻌﻼﻗﺔ ﺍﻟﺴﺎﺑﻘﺔ
−3
t +6,9 )
a = e 6,9 × e −5.10
a = ao.e−λ.t
−3
.t
:ﺃﻱ
:إذن
: ( ﻧﻌﻠﻢ ﺃﻥ ﺍﻟﻨﺸﺎﻁ4
ln 2 = s ≈ 138 Jours : م/ ) اة اX ا9 λ 5.10 Jours −1 ---------------------------------------------------------------------
t1 / 2 =
λ=
ln 2
a = e ( −5.10
: ﻭﻧﻌﻠﻢ ﺃﻥ
ln 2 : ﻣﻊm = mo e − λ .t t1 / 2
=
−3
N = N O e − λ .t : ﻫﻮt ]' @ >% ( " أن د ى ا @ ا5 mo NO = .N A : ﻫﻮmo 1%% د ا @ى ; @ آ M m N= .N A : ﻫﻮm 1%%و د ا @ى ; @ آ M m m :ﺑﻌﺪ ﺍﻻﺧﺘﺰﺍﻝ .N A = o .N A .e −λ .t :I وr اL^ ; اs% " M M : ور#( " >% ا @ ا%آ −
( 140 رب
210 84
ln 2 ×t t1 / 2
−
ln 2 ×1 138
m = mo e = 10 g × e ≈ 9,95 g Po م/ ) X ا9 ن & دF 50mg أي0,05 g ى:&K ا:& 9 ;6 u7 H 8/أي أ
:ء+ ع اB( k = H 3O . HO − (1-1 (1 ............ .............................................................................................................................................................. + : H 2 O / HO − :H دوR@ " (2-1 H 2 O + H 2 O → H 3 O + HO −
[
(l )
(l )
( aq )
+
][
]
( aq )
k = k A = ke = 10 −14
............ .............................................................................................................................................................. CH 3COOH + H 2 O CH 3 COO − + H 3O + :1-2 (2 ............ .............................................................................................................................................................. . CH 3COOH s' ا41 ، PH < PK A ⇐ pK A = 4,7 و pH = 3,2 : @ (2-2 ............................................................................................................................................................................. : > م% ول اH (3-2 − + ا د CH 3 COOH + H 2 O CH 3 COO + H 3O ة#;" 0 0 ا ' ا C1V1 ة#;" 'ل% @ء اTأ C1V1 − x1 x1 x1 − PH − PH − PH ة#;" 1@ ا ' ا (C1 − 10 ).V1 10 .V1 10 .V1
x1 f = 10 − pH = 6,3.10 −6 mol ............................................................................................................................................................................. . آ#v 'ل% اx1 < x1 max : و" أنx1. max = C1 .V1 = 2.10 −4 mol. : 4-2 ............................................................................................................................................................................. x1 f τ1 = = 0,03 = 3% : 5-2 x1 max ............................................................................................................................................................................. .Y !) اsP ه41 إذن ا @ع ا، Y !) اsP 4( 3% A>; n = :6-2 ............................................................................................................................................................................. (3
+
NH 3 + H 2 O NH 4 + HO − (1-3 ............................................................................................................................................................................. . NH 3 ا > ة41= ، PH > PK A ⇐ pK A = 9,2 و pH = 10,6 :@ :2-3 ............................................................................................................................................................................. :3-3 + − ا د NH 3 + H 2 O NH 4 + HO
ة#;" ة#;" ة#;"
C 2V2 C 2V2 − x 2 Ke (C 2 − − pH ).V2 10
0
0
ا ' ا 'ل% @ء اTأ 1@ ا ' ا
x2 Ke.V2 10 − pH ke.V2 − pH ke τ 2 = 10 = − pH = 0,04 = 4% C 2 .V2 10 .C 2 ............................................................................................................................................................................. + CH 3 COOH + NH 3 (4 NH 4 + CH 3 COO − ( aq )
x2 Ke.V2 10 − pH x
1
( aq )
( aq )
Q r ,i
( aq )
[CH 3COOH ]i .[NH 4 + ]i = [CH 3COO − ].[NH 3 ]i
=0
:1-4
Qr ,i = 0 : إذن. ( م وأ ت ا)! ات ; ا 'ل ; ا ' اr ا ت اH= [
............................................................................................................................................................................. :2-4 K A1 .(1) أي# اI'@ ; ا$ ر اK%= Qr , éq = K = = 3,2.10 4 > Qr ,i KA2 ............................................................................................................................................................................. :3-4 + − ا د CH 3 COOH + NH 3 NH 4 + CH 3COO 0 0 ا ' ا 2.10 −4 10 −4 −4 −4 'ل% @ء اTأ x3 x3 2.10 − x3 10 − x3 x
2.10
−4
− x3 f
10
−4
− x3 f
1
x3 f Qr ,éq =
x3 f
x3 f 2
(2.10 − x3 f )(10 4
1@ ا ' ا
−4
− x3 f )
= 3,2.10 4
.o #v ( وx ≤ 10 −4 ) > م% ن ا+ أنo$ E; ا اF' ا4+ . 4P 1 ! اH ا ر4( (دI FS' . x3 f = 10 −4 mol : I FS' ............................................................................................................................................................................. . آF % اτ 3 = 1 ⇐ NH 3 ا ' هF % ن اr x3 f = 10 −4 mol = x max :4-4 @" انHا% Y !) اsP تH "@ أ ت ا)! ات و. آ#>= I%uن ا[( ك اr ( م ; ا 'لr أ ت ا41= PH = pK A1 = 4,7 Y b ; ا 'ل+ ا ............................................................................................................................................................................. : > اX' ة ا# ( ا5 . ا رس#] ا:1-5 ............................................................................................................................................................................. :ة# اF = :2-5 CH 3COOH + HO − → CH 3COO − + H 2 O ............................................................................................................................................................................. :@ f;+% @ ا:3-5
VBE =
C A .V A = 10ml CB
C AV A = C B .V BE
⇐
............................................................................................................................................................................. :I FS' . > م% ول ا$" %[ ا4+ :4-5 ni (CH 3COOH ) = C A .V A = 10.10 −3.20.10 −3 = 2.10 −4 mol / l n( HO − ) = C B .VB = 20.10 −3.5.10 −3 = 10 −4 mol / l CH 3 COOH 2.10 −4 2.10 −4 − x 2.10
−4
−x
+ HO
−
10 −4 10 −4 − x f
10
−4
x
− xf
CH 3COO − + H 2 O 0 x
ا د 0 x
ا ' ا 'ل% @ء اTأ
xf
1@ ا ' ا
. HO − ا ' هF % ⇐ ا
ni ( HO − ) < ni (CH 3COOH )
1
xf
. x max = ni ( HO − ) = 10 −4 mol : % "و 10 [HO ] = [H10 O ] = 10 = [HO ].(V + V ) + x −14
−
+
−14
− 4 ,8
= 6,31.10 −10 mol / l ⇐
pH = 4,8 : F % ا1 @
3
xf
−
A
B
f
= 10 −4 mol / l : d@(و
[HO ] = 10V
−4
−
A
− xf + VB
. ة آ# اF = % " وτ =
: ول$ ^ل اu 4(و xf x max
10 −4 =1 10 − 4 :م% و; ا
=
'S% " ا >مI 1 =وI Y bع وآB ا اb ها ; إ اد ه4b ا ا و اT ; ة ا ءb= أ#+ . $ ا . = ا ةY b" وWL @ء اv إFH أ4( #$% 'ن ا%([ اWB أن !ا (اb(^% ة واb=r ا4( H# : اء SBIRO ABDELKRIM
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