Physique Chimie - MP - Fiches méthodes (Proetudes.blogspot.com)

MP
Fiches-méthodes
Physique
Chimie
Exercices corrigés
Sébastien Abry
Christophe Bernicot
Antoine Billaud
Pauline Boulleaux-Binot
Stéphanie Calmettes
Florence Depaquit-Debieuvre
Tommy Kopp
Gabrielle Lardé
MP
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Une collection dirigée
par Sylvain Rondy
Physique
Chimie
Sébastien Abry
Stéphanie Calmettes
Professeur au lycée la Martinière Duchère
à Lyon
Professeur au lycée Saint-Denis à Annonay
Christophe Bernicot
Professeur au lycée Dupuy de Lôme à Lorient
Antoine Billaud
Florence Depaquit-Debieuvre
Professeur au lycée Roosevelt à Reims
Tommy Kopp
Diplomé de l’école des Mines de Paris
Professeur au lycée Jacques Prévert
à Boulogne-Billlancourt
Pauline Boulleaux-Binot
Gabrielle Lardé
Professeur au lycée Jean-Baptiste Corot
à Savigny-sur-Orge
Professeur au collège Marcel Mariotte
à Saint Siméon de Bressieux
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3RXU −a < x < a RQ D Ep = 0 GRQF G2 φ
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2
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0 = 2B VLQ ka
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2 k 2
π )LQDOHPHQW SRXU N SDLU RX LPSDLU RQ D EN = 2mN = N 2 8ma
2
πx
6L ka = (2n + 1) π2 B = 0 HW φ2n+1 = A FRV 2n+1
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a π
πx
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a
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2
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2
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2 2 2
51. Déterminer les états stationnaires d’une particule pour un puits de potentiel infini
301
52
Déterminer les états stationnaires d’une
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2
2
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0)
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φ=0
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2
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+ k2 φ = 0
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0)
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0
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2
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0
2
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2
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DXFXQH UDLVRQ G¶rWUH UpIOpFKLH DSUqV x = 0 /D SDUWLH Be−ik2 x Q¶D GRQF SDV GH VHQV
SK\VLTXH HW DLQVL B = 0 2Q D HQVXLWH FRQWLQXLWp GH φ(x) HQ x = 0 GRQF A0 + Ar = At
304
Mécanique quantique
'H SOXV Gφ
Gx HVW FRQWLQXH HQ x = 0 2Q D � ∂φ
ik1 x − ik A e−ik1 x
1 r
∂x (x < 0) = ik1 A0 e
∂φ
ik
2x
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0)
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ik
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e
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2
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At
1
1
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1
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1
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6XU TXHOOH SURIRQGHXU FDUDFWpULVWLTXH FHWWH RQGH VXEVLVWHWHOOH " (VWLPHU FHWWH SURIRQ
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PDUFKH G¶pQHUJLH V0 = 6 H9 2Q GRQQH � 10−34 -V HW e � 2 × 10−19 &
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V0 VLWXpH HQ x = 0 2Q VXSSRVH E < V0 /¶pTXDWLRQ GH 6FKU|GLQJHU YpULILpH SDU OD SDUWLFXOH
V¶pFULW
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V
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x
VL x < 0
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G φ
− 2m
− V0 φ = Eφ
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2
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k12 = 2mE
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2
306
Mécanique quantique
√
2mE
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SDUW
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k12 = 2mE
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2
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3RXU x ≥ 0 RQ D FRPPH SUpFpGHPPHQW G2 φ
0)
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φ=0
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2
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= δi HQ SRVDQW δ = √ 2m(V0 −E)
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x
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2m(V0 −E)
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DYHF k12 = 2mE
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0
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2
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IRUPH 52. Déterminer les états stationnaires d’une particule pour une marche de potentiel
φ2 (x) = At eik2 x + Be−ik2 x
307
φ1 (x) = A0 eik1 x + Ar e−ik1 x
√
DYHF k12 = 2mE
,FL E > 0 GRQF k1 = 2mE
2
3RXU x ≥ 0 RQ D FRPPH SUpFpGHPPHQW G2 φ
0
+ 2mV
φ = − 2mE
φ
Gx2
2
2
2
2m(E+V
)
G φ
0
+
φ=0
Gx2
2
2
G φ
2
+ k2 φ = 0
Gx2
0)
DYHF k22 = 2m(E+V
,FL E + V0 > 0 GRQF k2 =
2
IRUPH √
2m(E+V0 )
/HV VROXWLRQV VRQW GH OD
φ2 (x) = At eik2 x + Be−ik2 x
$SUqV x = 0 OD SDUWLFXOH Q¶HVW SDV UpIOpFKLH GRQF B = 0 2Q D DORUV φ2 (x) = At eik2 x
(Q XWLOLVDQW OD FRQWLQXLWp GH φ(x) HW ∂φ
∂x HQ x = 0 RQ WURXYH FRPPH GDQV O¶H[HPSOH
WUDLWp �
A0 + Ar = At
A0 − Ar = kk21 At
8Q FDOFXO DQDORJXH j FHOXL GH O¶H[HPSOH WUDLWp GRQQH �
At
1
t= A
= k22k
+k1
0
k1 −k2
r
r=A
A0 = k2 +k1
/D GHQVLWp GH SUREDELOLWp GH FRXUDQW V¶pFULW →
−
−
→
j = |φ(x)|2 mk
/HV DPSOLWXGHV GHV GHQVLWpV GH SUREDELOLWp GH FRXUDQW SRXU O¶RQGH LQFLGHQWH J0 SRXU
O¶RQGH UpIOpFKLH Jr HW SRXU O¶RQGH WUDQVPLVH Jt V¶pFULYHQW 1
J0 = A20 k
m
1
Jr = A2r k
m
2
Jt = A2t k
m
308
/HV FRHIILFLHQWV GH WUDQVPLVVLRQ HW GH UpIOH[LRQ VRQW DORUV �
T = JJ0t = t2 kk21
R = JJr0 = r2
⎧
⎨ T = 4k12 k2
(k1 +k2 )2 k1
2
⎩ R = (k1 −k2 )2
(k1 +k2 )
Mécanique quantique
�
1 k2
T = (k4k
2
1 +k2 )
1
Jr = A2r k
m
2
Jt = A2t k
m
/HV FRHIILFLHQWV GH WUDQVPLVVLRQ HW GH UpIOH[LRQ VRQW DORUV �
T = JJ0t = t2 kk21
R = JJr0 = r2
⎧
⎨ T = 4k12 k2
(k1 +k2 )2 k1
2
⎩ R = (k1 −k2 )2
(k1 +k2 )
�
1 k2
T = (k4k
2
1 +k2 )
2
1 −k2 )
R = (k
(k1 +k2 )2
⎧
√
E(E+V0 )
⎪
⎨ T = 4×2m
√
√
2m( E+ E+V )2
0
√
√
⎪
⎩ R = 2m(√E−√E+V0 )2
2m( E+ E+V0 )2
⎧
√
⎪
4 E(E+V0 )
⎪
⎪
2
T = ⎪
⎪
V
⎪
E 1+ 1+ E0
⎨
2
V
1− 1+ E0
⎪
⎪
⎪
⎪
2
R= ⎪
⎪
V
⎩
1+ 1+ E0
Solutions
⎧
V
⎪
4
1+ E0
⎪
⎪
2 = 75 T
=
⎪
⎪
V
⎪
1+ 1+ E0
⎨
2
V
⎪
1− 1+ E0
⎪
⎪
⎪
2 = 25 R
=
⎪
⎪
V0
⎩
(1+
1+ E
52. Déterminer les états stationnaires d’une particule pour une marche de potentiel
309
53
Déterminer les états
stationnaires
d’une
particule
%©UFSNJOFS
MFT ©UBUT
TUBUJPOOBJSFT
EVOF
QPVS VOF
CBSSJ¨SF
pourQBSUJDVMF
une barrière
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t HW GH UpIOH[LRQ r GpILQLV SDU
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=
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IRQFWLRQ G¶RQGH GDQV OD EDUULqUH GH SRWHQWLHO HVW XQH RQGH pYDQHVFHQWH TXL V¶DWWpQXH VXU
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TX¶XQH RQGH WUDQVPLVH H[LVWH DSUqV OD EDUULqUH F¶HVW O¶HIIHW WXQQHO 8QH SDUWLFXOH FODVVLTXH
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OH IDLUH j FRQGLWLRQ TXH OD KDXWHXU GH OD EDUULqUH HWRX VRQ pSDLVVHXU VRLHQW VXIILVDPPHQW
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310
Mécanique quantique
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GLVFRQWLQXLWp GX SRWHQWLHO
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WUDQVPLVVLRQ HW GH UpIOH[LRQ
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6DYRLU H[SOLTXHU O¶LQIOXHQFH GH OD KDXWHXU HW GH O¶pSDLVVHXU G¶XQH EDUULqUH VXU O¶HIIHW WXQQHO
$POTFJMT
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&YFNQMF USBJU©
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SK\VLTXHV G¶pQHUJLH E HW GH PDVVH m GDQV OH VHQV G¶XQ D[H (Ox) GpFURLVVDQW ,OV VRQW VRXPLV
j XQH EDUULqUH G¶pQHUJLH SRWHQWLHOOH Ep (x) GH KDXWHXU E0 > E VWDWLRQQDLUH GH ODUJHXU L
Ep
E0
1
2
3
L
E
x
2Q FKHUFKH OHV pWDWV VWDWLRQQDLUHV G¶pQHUJLH VRXV OD IRUPH φ DYHF
φ1 (x) = A1 eikx + B1 e−ikx SRXU x < 0
φ2 (x) = A2 eαx + B2 e−αx SRXU 0 ≤ x ≤ L
φ3 = A3 eikx + B3 e−ikx SRXU x > L
4XHOOHV VRQW OHV UpJLRQV FODVVLTXHPHQW DFFHVVLEOHV "
([SULPHU k HW α
3DUPL OHV FRQVWDQWHV Ai HW Bi ODTXHOOH HVW QXOOH "
1
/H UDSSRUW GHV DPSOLWXGHV B
B3 YDXW B1
2e−ikL
1
B3 = eαL +e−αL +iM (eαL −e−αL ) DYHF M = 2
α
k
k − α
53. Déterminer les états stationnaires d’une particule pour une barrière de potentiel
311
2
1
/H IDFWHXU GH WUDQVPLVVLRQ T = B
B3 HQ LQWHQVLWp j WUDYHUV XQH EDUULqUH pSDLVVH αL � 1
VH PHW VRXV OD IRUPH T = T0 eηαL ([SULPHU T0 HQ IRQFWLRQ GH M HW GpWHUPLQHU OD YDOHXU
GH η
0 −E)
2Q SHXW PRQWUHU TXH T0 = 16E(E
6DFKDQW TXH eηαL = 10−4 T = 0, 04 HW
E02
E = 5 H9 GpWHUPLQHU O¶H[SUHVVLRQ GH E0 HW FDOFXOHU VD YDOHXU
4XHO HVW DORUV OH GpELW qf G¶REMHWV SK\VLTXHV WUDYHUVDQW OD EDUULqUH "
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40-65*0/
/HV SDUWLFXOHV VRQW pPLVHV GDQV OD ]RQH DYHF XQH pQHUJLH E < E0 &ODVVLTXHPHQW
HOOHV GHYUDLHQW rWUH UpIOpFKLHV VXU OD EDUULqUH HW Q¶DFFpGHU TX¶j OD ]RQH 3RXU x > L RQ D G φ
= Eφ
− 2m
Gx2
2
2
G2 φ
+ 2mE
φ=0
Gx2
2
G2 φ
+ k2 φ = 0
Gx2
√
DYHF k = 2mE
3RXU 0 < x ≤ L RQ D G φ
+ E0 φ = Eφ
− 2m
Gx2
2
2
G2 φ
0)
+ 2m(E−E
φ=0
Gx2
2
G2 φ
2
+ k2 φ = 0
Gx2
0)
DYHF k22 = 2m(E−E
,FL E
2
√
2m(E0 −E)
α=
< E0 GRQF k2 = i
2m(E0 −E)
= iα 2Q D GRQF LFL
$SUqV x = 0 OD SDUWLFXOH QH UHQFRQWUH DXFXQH GLVFRQWLQXLWp GH SRWHQWLHO HOOH Q¶D GRQF
DXFXQH UDLVRQ G¶rWUH UpIOpFKLH DSUqV x = 0 /D SDUWLH A1 eikx Q¶D GRQF SDV GH VHQV
SK\VLTXH HW DLQVL A1 = 0
2Q D T
=
=
=
=
=
312
√
Mécanique quantique
|2e−ikL |
2
|eαL +e−αL +iM (eαL −e−αL )|2
4
|2FK(αL)+2iM VK(αL)|2
4
4(FK2 (αL)+M 2 VK2 (αL))
1
1+VK2 (αL)+M 2 VK2 (αL)
1
1+(1+M 2 )VK2 (αL)
2αL
&RPPH VK2 (αL) � e 4 VL αL � 1
T =
1
2αL
1+(1+M 2 ) e 4
�
1
2αL
(1+M 2 ) e 4
4
−2αL
= 1+M
2e
4
2Q LGHQWLILH T0 = 1+M
2 HW η = −2
2 E0
2
E ( −1)
E0
&RPPH T0 = 16 EE2
= 16 EE0
E − 1 RQ D 0
E
E0
E
E0
2 E
E0
2
2 E0
E −1
E0
E −1
= T160
= 16eTηαL = 14
− 2 EE0 × 12 + 14 = 0
2
E
1
−
=0
E0
2
E0 = 2E = 10 H9
/H GpELW G¶REMHWV WUDYHUVDQW OD EDUULqUH HVW qf = T qs = 40 V−1
&YFSDJDFT
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DSSDUDLVVDQW GDQV O¶pTXDWLRQ GH 6FKU|GLQJHU
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(Q H[SORLWDQW OD FRQWLQXLWp GH φ HW Gφ
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WUDQVPLVVLRQ t HQ DPSOLWXGH
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FRHIILFLHQWV GH WUDQVPLVVLRQ T HW GH UpIOH[LRQ R SRXU OD GHQVLWp GH SUREDELOLWp GH FRXUDQW
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FKLH ]RQH HW O¶RQGH WUDQVPLVH ]RQH 53. Déterminer les états stationnaires d’une particule pour une barrière de potentiel
313
(WDEOLU O¶H[SUHVVLRQ GHV FRHIILFLHQWV T HW R
4XHOOHV VRQW OHV YDOHXUV H[WUpPDOHV GH T " &RPPHQWHU
1PVS WPVT BJEFS
E©NBSSFS
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&9&3$*$& )DLUH OH OLHQ DYHF OHV FRHIILFLHQWV GH UpIOH[LRQ HW GH WUDQVPLVVLRQ HQ DP
SOLWXGH
4PMVUJPOT EFT FYFSDJDFT
&9&3$*$& 'DQV OD ]RQH RQ D DYHF k3 =
√
G2 φ
+ 2mE
φ=0
Gx2
2
2
G φ
2
+ k3 φ = 0
Gx2
2mE
/D IRQFWLRQ G¶RQGH V¶pFULW GRQF φ3 (x) = A3 eik3 x + B3 e−ik3 x
'DQV OD ]RQH RQ D DYHF k2 =
√
G2 φ
0)
+ 2m(E−E
φ=0
Gx2
2
G2 φ
2
+ k2 φ = 0
Gx2
2m(E−E0 )
/D IRQFWLRQ G¶RQGH V¶pFULW GRQF φ2 (x) = A2 eik2 x + B2 e−ik2 x
'DQV OD ]RQH RQ D DYHF k1 =
√
G2 φ
+ 2mE
φ=0
Gx2
2
G2 φ
2
+ k1 φ = 0
Gx2
2mE
/D IRQFWLRQ G¶RQGH V¶pFULW GRQF φ1 (x) = A1 eik1 x + B1 e−ik1 x
2Q UHPDUTXH TXH k1 = k3 'H SOXV OD SDUWLFXOH Q¶HVW SDV UpIOpFKLH HQ x < 0 GRQF RQ
D A1 = 0
314
Mécanique quantique
3RXU x = 0 HW x = L φ HVW FRQWLQXH GRQF φ1 (0) = φ2 (0)
GRQF
B1 = A2 + B2
φ2 (L) = φ3 (L)
A2 eik2 L + B2 e−ik2 L = A3 eik3 L + B3 e−ik3 L
3RXU x = 0 HW x = L Gφ
Gx HVW FRQWLQXH GRQF Gφ1
Gφ2
Gx (0) = Gx (0)
Gφ2
Gφ3
Gx (L) = Gx (L)
HW
−ik1 B1 = ik2 A2 − ik2 B2
GRQF
ik2 A2 eik2 L − ik2 B2 e−ik2 L = ik3 A3 eik3 L − ik3 B3 e−ik3 L
2Q D DORUV DYHF k1 = k3 ⎧
B1 = A2 + B2
⎪
⎪
⎨ A eik2 L + B e−ik2 L = A eik3 L + B e−ik3 L
2
k
2
3
3
− 3 B1 = A2 − B2
⎪
⎪
⎩ k2 k2 ik2 L
− B2 e−ik2 L ) = A3 eik3 L − B3 e−ik3 L
k3 (A2 e
�
�
⎧
k3
⎪
=
1
−
2A
2
⎪
k2 � B 1
⎪
�
⎪
⎨
2B2 = 1 + kk32 B1
⎪
⎪
2A3 eik3 L = (1 + kk23 )A2 eik2 L + (1 − kk23 )B2 e−ik2 L
⎪
⎪
⎩
2B3 e−ik3 L = (1 − kk23 )A2 eik2 L + (1 + kk23 )B2 e−ik2 L
Solutions
⎧
−k3
B1
A2 = k22k
⎪
⎪
2
⎪
k
⎪ B2 = 2 +k3 B1
⎨
2k2 �
�
e−ik3 L k2 +k3 k2 −k3
ik2 L + k3 −k2 k2 +k3 B e−ik2 L
A
=
B
e
3
1
1
⎪
2 �
k3
2k2
k3
2k2
⎪
�
⎪
⎪
⎩ B = eik3 L k3 −k2 k2 −k3 B eik2 L + k3 +k2 k2 +k3 B e−ik2 L
3
1
1
2
k3
2k2
k3
2k2
⎧
k2 −k3
A2 = 2k2 B1
⎪
⎪
⎪
+k3
⎨
B2 = k22k
B1
2
ik L
−ik2 L
e−ik3 L
⎪
A3 = 2k2 k3 B1 (k22 − k32 ) e 2 −e
⎪
2
⎪
�
�
⎩
eik3 L
2 eik2 L + (k + k )2 e−ik2 L
B3 = 4k
B
−
k
)
−(k
1
3
2
2
3
k
2 3
⎧
k2 −k3
⎪
=
B
A
2
1
⎪
2k2
⎪
⎪
⎨ B2 = k2 +k3 B1
2k2
e−ik3 L
2
2
A
=
⎪
3
2k2 k3 (k2� − k3 )B1 i VLQ k2 L
⎪
�
⎪
ik
L
⎪
⎩ B3 = e 3 B1 −(k 2 + k 2 ) eik2 L −e−ik2 L + 2k3 k2 eik2 L +e−ik2 L
2
3
2k2 k3
2
2
⎧
k2 −k3
A2 = 2k2 B1
⎪
⎪
⎪
⎨ B2 = k2 +k3 B1
2k2
−ik L
⎪ A3 = e2k2 k33 (k22 − k32 )B1 i VLQ k2 L
⎪
⎪
�
�
⎩
eik3 L
B3 = 2k
B1 −i(k22 + k32 ) VLQ k2 L + 2k3 k2 FRV k2 L
2 k3
53. Déterminer les états stationnaires d’une particule pour une barrière de potentiel
315
⎧
−k3
B1
A2 = k22k
⎪
⎪
2
⎪
k
+k
⎪
2
3
⎨ B2 = 2k B1
2
e−ik3 L
2
2
A
=
3 � 2k
⎪
� 2 k3 (k2 − k3 )B1 i VLQ k2 L
⎪
⎪
⎪
1�
⎩ t = �� B
√ 2 2 2 22k2 k3 2 2 2
B3 � =
(k2 +k3 ) VLQ k2 L+4k3 k2 FRV k2 L
⎧
k2 −k3
A2 = 2k2 B1
⎪
⎪
⎪
+k3
⎪
⎨ B2 = k22k
B1
2
−ik
3L
e
2
2
A3 =
⎪
� 2k� 2 k3 (k2 − k3 )B1 i VLQ k2 L
⎪
⎪
⎪
1�
⎩ t = �� B
√ 2 2 2 2 2k2 k3 2 2
B3 � =
(k2 +k3 ) VLQ k2 L+4k3 k2 (1−VLQ2 k2 L)
⎧
−k3
A2 = k22k
B1
⎪
⎪
2
⎪
k
+k3
⎪
⎨ B2 = 22k
B1
2
e−ik3 L
A3 =
(k22 − k32 )B1 i VLQ k2 L
⎪
2k
�
�
2 k3
⎪
⎪
⎪
1�
⎩ t = �� B
√ 4 4 2 2 2k2 k23 2 2
2 2
B �=
3
(k +k +2k k −4k k ) VLQ k L+4k k
2
2
3
2 3
2 3
3 2
⎧
k2 −k3
=
B
A
⎪
2
1
2k2
⎪
⎪
⎪
k2 +k3
⎪
B
=
B
2
⎨
� 2 � 1 �� �
� 2k
� A3 � � A3 � � B1 � k22 −k32 √
2k2 k3
r = � B3 � = � B1 � � B3 � = 2k2 k3
VLQ k2 L
2 −k 2 )2 VLQ2 k L+4k 2 k 2
⎪
(k
⎪
2
2
3
3 2
�
�
⎪
⎪
⎪
1�
⎩ t = �� B
√ 2 2 22k2 k23
2 2
B3 � =
(k2 −k3 ) VLQ k2 L+4k3 k2
&9&3$*$& /D GHQVLWp GH SUREDELOLWp GH FRXUDQW V¶pFULW →
−
−
→
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r
Rs
ne P−3
3, 5 × 1014
2 × 1014
9 × 1013
5 × 1013
3 × 1013
(VWLPHU OD WHPSpUDWXUH GH OD FRXURQQH HW OD FRPSDUHU j FHOOH WURXYpH GDQV O¶H[HUFLFH
SUpFpGHQW
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ne
r
Rs
P−3
3, 5 × 1014
2 × 1014
9 × 1013
5 × 1013
3 × 1013
(VWLPHU OD WHPSpUDWXUH GH OD FRXURQQH HW OD FRPSDUHU j FHOOH WURXYpH GDQV O¶H[HUFLFH
SUpFpGHQW
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/D GHQVLWp YROXPLTXH GH SDUWLFXOHV GHYLHQW E
n1 (z) = n0 e
− k pT
B
= n0 e
− kmgTs z
B
z
= n0 e− H
BT
DYHF H = kmg
s
3RXU z = Rs OD GHQVLWp YDXW n = n0 × 10−3 2Q HQ GpGXLW GRQF Rs
10−3 = e− H
OQ 10−3 = − RHs
Rs
8
H = − −3ROQs 10 = 3 OQ
10 � 10 P
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A
ne = nN
V = kB T
326
Thermodynamique statistique
Gp
Gr = −ρg(r)
2
Gne
kB T Gr = −ne mgs Rr2s
mgs Rs2 1
Gne
Gr = −ne kB T r 2
ρ = ne m
T = mgkBs H � 1, 6 × 106 .
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A
ne = nN
V = kB T
ρ = ne m
Gp
Gr = −ρg(r)
Rs2
e
kB T Gn
Gr = −ne mgs r 2
mgs Rs2 1
Gne
Gr = −ne kB T r 2
mgs Rs2 Gr
Gne
n e = − kB T r 2
2
OQ nn0e = RHs 1r − R1s
2
Rs
1
− R1
H
r
s
ne = n0 e
/D GHQVLWp YROXPLTXH GH SDUWLFXOHV V¶pFULW DXVVL Rs Rs
ne = n0 e H ( r −1)
OQ ne = RHs Rrs − 1 + OQ n0
Solutions
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LQYHUVDQW 2Q SHXW DLQVL WUDFHU OD FRXUEH OQ ne = f ( Rrs ) j OD PDLQ RX ELHQ JUkFH j XQ
WDEOHXU
2Q REWLHQW XQH GURLWH GH SHQWH α � 10, 4 /D SHQWH V¶LGHQWLILH j RHs GRQW RQ DYDLW WURXYp
XQH HVWLPDWLRQ /¶RUGUH GH JUDQGHXU UHVWH OH PrPH OH PRGqOH HVW ELHQ FRKpUHQW
3DU LGHQWLILFDWLRQ RQ D s Rs
α = mg
kB T
6
s Rs
T = mg
kB α � 10 .
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56
Étudier un système à spectre
&UVEJFS VOd’énergies
TZTU¨NF discret
TQFDUSF
E©OFSHJF EJTDSFU
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V¶pFULW − E
− E
DYHF
Z = e kB T
p(E) = Z1 e kB T
E
Z HVW DSSHOpH IRQFWLRQ GH SDUWLWLRQ
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SUREDELOLWpV − E
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E
E
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−βE
�E� = Z1 Ee−βE = Z1 − ∂(e∂β ) = − Z1
E
E
∂ OQ Z
�E� = − Z1 ∂Z
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−βE
e
)
∂(
E
∂β
'H PrPH RQ PRQWUH TXH σ 2 = ∂ ∂βOQ2Z 2
OQ Z
�E� = − ∂ ∂β
328
Thermodynamique statistique
σ 2 = ∂ ∂βOQ2Z
2
3RXU XQ V\VWqPH FRPSRVp GH N SDUWLFXOHV LQGpSHQGDQWHV O¶pQHUJLH PR\HQQH WRWDOH GX
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2 VXU O¶pQHUJLH GX V\VWqPH HVW OD VRPPH GHV pFDUWV
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2
TXDGUDWLTXHV PR\HQV σi GHV SDUWLFXOHV 3RXU GHV SDUWLFXOHV LGHQWLTXHV FHV VRPPHV VH
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G¶K\GURJqQH μ = 1, 4 × 10−26 -7−1 VRXPLV j XQ FKDPS PDJQpWLTXH B0 = 1 7
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μ B
=e
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B
μ B
� 1 − 2 kpB T0
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k T
−
μp B0
k T
B
e B −e
1−e
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η0 = N
μp B0 =
N− +N+ = μp B0
−
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e kB T +e kB T
1+e
μp B 0
η0 = kB T � 3 × 10−6
μp B0
kB T
μp B0 �
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μ B
2 kp T0
B
2
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N VLQK (βμp B0 )
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=
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N
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Solutions
C
56. Étudier un système à spectre d’ énergies discret
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57
Déterminer la capacité thermique
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C = ∂�E�
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70. Optimiser un processus
465
Les ouvrages de cette collection ont pour objectif de faciliter l’acquisition
et la maîtrise des notions fondamentales du programme. Le but est de
faire en sorte que chacun sache « quoi faire », même lorsqu’il pense se
trouver face à un obstacle insurmontable.
Chaque fiche de ce livre est conçue de la façon suivante :
„ Quand on ne sait pas !
Les raisons expliquant pourquoi on ne sait pas, avec parfois des rappels
de cours et les premières pistes à explorer afin de s’en sortir.
„ Que faire ?
Les méthodes permettant de solutionner le type de problème étudié,
assorties des rappels de cours essentiels à leur mise en œuvre.
„ Conseils
Les conseils de rédaction et une ou deux astuces pratiques.
„ Exemple traité
Mise en pratique et en lumière de ce qui a été vu précédemment.
„ Pour vous aider à démarrer
Les idées permettant de démarrer sereinement les exercices proposés.
„ Solutions des exercices
Les solutions complètes et détaillées des exercices.
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Illustration de couverture : © diego1012 - Fotolia.com
„ Exercices
Énoncés choisis soigneusement afin de balayer largement le thème étudié,
certains étant extraits de sujets de concours.