Problemas Quimica Fisica
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problemas quimica...
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λ λ E c h ¯
·10
34
−
·10
·
19
−
v = 106
1
−
(∆ p/p = p/p = 10
8)
−
Φ λ = 540
λ
34
−
·
1
−
1
1
−
−
·10
14
−
·10
1
−
19
−
1
−
n = 1 mH
mD
→ n = 2
I (H) (H) +
2+
3+
a0 /Z λ
·1016
1
−
·106
·10
17
−
1
−
·106
1
−
1
−
1
−
1
−
d2 /dt2 d2 /dt2 z 2 d2 /dz 2 dn /dxn
a cos(bt cos(bt + c) exp(ibt exp(ibt)) z exp( z 2 /2) c exp( ax) ax)
− −
−b2 −b2
−
( a)n
−
ax
−
e
sin3x sin3x 3cos x
sin x +cos3x +cos3x
pˆx
−h¯ 2a2
h ¯2
h ¯2
2
( h ¯ 2 /m) /m)
−
ˆ , ˆ [H x]
−
ˆ , p [H ˆx ]
ˆ= P
ˆ , p ih ¯ [∂V ( ∂V (x)/∂x] /∂x ] [H ˆx ]
× ∂/∂x
ˆ V
i
∂ ∂Q
i
× i = −1
Ψ(x, Ψ(x, t) = Ψ1
x ˆ P ] ˆ] [Q, P
ˆ 2 = Q ˆ O
−
− iP ˆ √ i = −1
+ 3 cos cos x
ˆ H
ψ (Q) ˆ 1 = Q ˆ + iP ˆ O
e
h ¯2 [ˆx, x, d/dx] [ˆx, p x, p ˆx ] pˆx = ih ¯ (∂/∂x) ∂/∂x)
ih ¯
ax
−
sin x + cos x
ˆ 1 , O ˆ2] [O
i
√ 1 [Ψ1(x, t) + 2Ψ2(x, t)] 5
Ψ2
E 1
E 2
ψ1 = N 1 (ϕ1 + ϕ2 ) ψ2 = N 2 (ϕ1 ϕ2 )
−
ϕ1
ˆ H
ϕ2 N 1 ψ1 ψ1 ψ1
N 2
ψ2 ˆ H
ψ2
ψ1|ψ2 =
ˆ ϕ2 = E H = E 2 ϕ2
∗
ψ1 ψ2 dτ = 0
ψ2
a
= C sin nπx ψ = C a
(2/a (2/a))1/2
ˆ ϕ1 = E 1 ϕ1 H
C
a/4 a/4, 3a/4 a/4
a/2 a/2
h2 /4a2
m
L
−
ψn (x) = ∆A = A2
A ∆x∆ px
2/L sin (nπx/L (nπx/L)) 1 / 2 A2 ψ = c = c 1 φ1 + c2 φ2 + c3 φ3
n
φn
E n
c1 = 1/2
c2 =
ψ
−1/2
c3 ˆ H
E 1 E 2
E 3
ψ = N = N sin( πx a )
a
m
< px > a
≤x≤
a (2/a (2/a))1/2
·105
≤ x ≤
a
2
√
L
sin
≤ x ≤ L)
πx , L
ϕ L π
− x)
ψ
n
a
1
ϕ(x) =
a
≤ x ≤
−
L (0
ψ = N x(a
a
2n
V = 0
π a
b
b>a
π
L ψ (0) = ψ = ψ (L)
dC
λ1 = C =
λ2 =
λ3 =
−
ψ (x) = (β/π) β/π )1/4 exp β = mω/ mω /h ¯
ω = 2πν
−
βx 2 /2 ,
2 T = 2 V = E
hmω/2 ¯hmω/2
1/(2α (2α)
x2
T
1
−
2 2
re < x2 >1/2
√ µk/h¯ α = µk/¯
ψ0 = (α/π) α/π )1/4 exp( αx2 /2)
−
√ mk/h¯ β = = mk/¯
ν ν˜
µ = m A mB /(mA + mB ) x = r = r
ψ0 = (β/π) β/π )1/4 exp ω = 2πν
·
− re
− βx 2 /2 ] ψ1 = 4β 3 /π
1
1
−
−
1/4
x exp
−
βx 2 /2 ]
± 2
1
−
≡
re
T x2
−
α ψ0 (x) = π k
1/4
exp
αx2 2
,
(µ k )1/2 α = , µ h ¯
V
x = r = r
− re
ψ (θ, φ) = 120 < 120 < θ < 180
Y ( Y (θ, φ) =
3 4π
1/2
sin θ cos φ 0 < θ < 60
60 60 < < θ < 120
1/2
15 4π
sin θ cos θsin φ
E = = 6¯h2 /(2I (2I ) L2 = l( l (l + 1)¯h2 = 6¯ h2
l
dyz
px = C 1 senθ sen θ cos φ
py = C 2 senθ sen θ senφ sen φ
√ 2¯h
C 1 = C = C 2 = (3/ (3/4π )1/2
ψ = Ne
r/(2 r/(2a a0 )
−
rmp
ˆz L
ψ
ψ210
z
√
1 = 4 2π
1 a0
3/2
r exp a0
− r 2a0
cos θ
z
−1/8 E h
E h = e = e 2 /(4π (4π0 a0 )
−1/4 E h
√ −
1 ψ = 4 2π
1 a0
3 2
r a0
2
e
4a 0
r/2 r/2a0
−
V
E
−1/4E h
r r = 3 a0/ 2
T
1/8E h
ψ =
−1/8E h √ 1π a0 3/2e −
r/a0
−
,
rmp a0 rmp = a 0
√ 2¯h
−E h = −e2/(4π (4π 0 a0 ) = −h ¯ 2 /(me a0 )
2 pz ψ2 pz =
1 32 32πa πa 30
rmp rmp = 4a0
r = 5a0
r exp( r/2 r/2a0 )cos θ, a0
−
r
1/r
1/(4a (4a0 )
ψ =
ψ (r,θ,φ) r,θ,φ) =
L2
1/2
√ 1π a0 3/2e −
r/a0
−
1/2
1/8 E h
,
5/2
Z a0
T
1/(5a (5a0 )
1 32 32π π
1/ r
r exp
−
Zr cos θ. 2a0
Lz E h rmp
1 r
2 pz
¯h2
3+
Z
a0
E h
E h
1/a0 Z
ψ200 =
√ √ − √ √ 1 4π
1
2 2
|ψ(r)|2
ψ320 (r,θ,φ) r,θ,φ) = R(r ) =
Z a0
3/2
(2
ρ) exp( exp( ρ/2); ρ/2);
1 1 Z 4π 2 2 a0 P ( P (r ) = r 2 R2 (r)
ρ =
−
Zr a0
3/2
Y ( Y (θ, φ)
√ × − × 4 27 270
Z a0
3/2
Zr a0
2
exp
Zr 3a0
5 16 16π π
1/2
3cos2 θ
− 1
R(r ) Y ( Y (θ, φ)
E n V
Lz
T
1/r = 1/ r
|L2|
Z ψ1 = N ρ e ψ2 = N ρ e ρ = Zr/a0
ρ/2 ρ/2
−
ρ/2 ρ/2
−
N = ψ1
cos θ sin θ cos φ
√ 1 4 2π
Z a0
3/2
ψ2 z
r 1/r
rmp
ψ1
V
T
θ
ψ2px
φ
ψ2py ◦
◦
◦
1 ψ (r,θ,φ) r,θ,φ) = ψ200 (r,θ,φ) r,θ,φ) + 3
√
2 ψ210 (r,θ,φ) r,θ,φ) 3
{200} {210} ψ
ψ200
ˆ2 L ˆz L
ψ ψ
ψ210
L2 Lz
ˆ2 L
ϕ = exp( λx2 )
λ
−
λ = (k m)1/2 /(2¯h) E = = ¯h/2 h/2
× (k/m) k/m)1/2 = ¯hω/2 hω/2 = h ν/2 ν/ 2 ϕ = exp( λr2 /a20 )
λ
λ = 8/(9π (9π ) E = =
−
−h2/(3π (3π 3 ma0 ] =
18
−
−e2/(2π (2π 0 r )
V ( V (r) =
ψ (r,θ,φ) r,θ,φ) = N exp a0
− ξr a0
N
ξ ξ ξ
N =
ξ 3 πa 30
1/2
.
e2 ¯h2 = . 4π0 me a 0
m
L
ϕ(x) = c 1 φ1 (x) + c2 φ2 (x) donde φ1 =
30 x(L L5
− x)
y φ2 =
630 2 x (L L9
− x)2
ϕ 2
S 11 11 = S 22 22 = 1;
H 22 22 =
6¯h ; mL2
S 12 12
√ ¯h 21 2
H 12 12 = H 21 21 =
H 11 11
mL2 c1
c2 3
−
S 12 (27/28)1/2 H 11 h2 /(mL2 ) 12 = (27/ 11 = 5¯ c2
c1
h2 /(mL2 )
ε1 =
φ(α) α
− || − φ(α) =
2α 2 α πa 20
3/4
ˆ φ(α) = E (α) = φ(α) H
exp
2
α r 2 a0
¯h2 3α 2 me a0 2
2 2πα
1/2
α
E 100 100 =
2
− 2 m¯he a2 0
1,5a0 ˆ 2 ψ = 6 ¯h2 ψ S ˆ2 ψ = 2 ¯h2 ψ J ˆ2 ψ = 6 ¯h2 ψ L ˆ z ψ = 2 ¯hψ ˆz ψ = L hψ S
+
a
Z (P) (P)
p1 s1
−hψ ¯hψ
ˆz ψ = ¯hψ J hψ
np2 , n > 2 [np1 nd1 ]
[np1 (n + 1)s 1)s1 ]
np2
np1 nd1
ψ (1, (1, 2)
L2 ψ (1, (1, 2) = 2 h ¯ 2 ψ (1, (1, 2) ˆ z ψ (1, (1, 2) = ¯h ψ (1, (1, 2) L
J J 2 ψ(1, (1, 2) = 6 h ¯ 2 ψ(1, (1, 2) ˆz ψ (1, J (1, 2) = 2 h ¯ ψ (1, (1, 2)
S S 2 ψ (1, (1, 2) = 2 h ¯ 2 ψ (1, (1, 2) ˆz ψ (1, S (1, 2) = ¯h ψ (1, (1, 2) L S J M L M J J M S S
2
2
ψ1 (1, (1, 2) ψ2 (1, (1, 2) ψ3 (1, (1, 2) ψ4 (1, (1, 2)
= = = =
∗
∗
[1σg (1)1σ (1)1σu (2) + 1σ 1σu (1)1σ (1)1σg (2)][α (2)][α(1)β (1)β (2) (2) α(2)β (2)β (1)] (1)] [1σg (1)1σ (1)1σu (2) 1σu (1)1σ (1)1σg (2)][α (2)][α(1)α (1)α(2)] [1σg (1)1σ (1)1σu (2) 1σu (1)1σ (1)1σg (2)][β (2)][β (1)β (1)β (2)] (2)] [1σg (1)1σ (1)1σu (2) 1σu (1)1σ (1)1σg (2)][α (2)][α(1)β (1)β (2) (2) + α(2)β (2)β (1)] (1)] ∗ ∗ ∗
− − −
−
∗ ∗ ∗
S
M S S
+ 2
De
+ 2
2
+ 2
2
+ 2
2
−
2
2
2
2
2
+
−
+ + 2
+ 2
2
+ 2 2
+ 2
+
C2+ O2+ CN 2
−
NO+
O2+ 2 +
−
2
ψs
∝ 1/(2√ π) ; ◦
120
ψ px
∝ (1/ (1/2)
3/π sin /π sin θ cos φ ; ψ py
∝ (1/ (1/2)
3/π sin sin φ /π sin θ θ sin φ
2
yz
σ (yz) yz )
x
1sO , 2sO , 2 pxO , 2 pyO , 2 pzO , 1sHa , 1sHb
ψ1 ψ2 ψ3 ψ4 ψ5
= 0,99421 99421ssO + 0, 0 ,0152 0152ssO + 0, 0 ,00342 pzO 0,004 004ϕ ϕ1 ; E 1 = 20 20,,252 = 0,0271 0271ssO + 0, 0 ,8202 8202ssO + 0, 0 ,1322 pzO + 0, 0 ,152 152ϕ ϕ1 ; E 2 = 1, 1 ,258 = 0,6242 pyO + 0, 0 ,424 424ϕ ϕ2 ; E 3 = 0, 0 ,594 = 0,0261s 0261sO 0,5022 5022ssO + 0, 0 ,7872 pzO + 0, 0 ,264 264ϕ ϕ1 ; E 4 = 0, 0 ,460 = 2 pxO ; E 5 = 0, 0 ,393
−
− −
ϕ1 = 1sHa + 1sHb
−
−
ϕ2 = 1sHa
− 1sH
b
− −
− −
2
2
2
+ 3
1s
1s 1s
+ 3
x3
3
− 3x + 2 = (x ( x − 1)(x 1)(x2 + x − 2) π
2 =
E des des =
2
π
2
+
2 =
−
2
−
β
2
E des des =
−
/
\
2
β E des = β des = β
− −
2
π π
pz
π
pz
β =
−75
π π
pz
π
pz
β
π π β π 3
π
∗
3
pπ π π
pz
π
pz π φ 1 φ2
pz
φ3 x = 4
−2
x = +1
4
π x π
2 pz
π
2 pz E g
⊕ A2u ⊕ B2u π
pz π
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