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Unformatted text preview: MIT OpenCourseWare
http://ocw.mit.edu 5.80 SmallMolecule Spectroscopy and Dynamics
Fall 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms. MASSACHUSETTS INSTITUTE OF TECHNOLOGY Chemistry 5.76 Spring 1994 Problem Set #1 ANSWERS
1. (a) Make the necessary conversions in order to ﬁll in the table:
Wavelength (Å)
420 Wavenumber (cm−1 )
100 Energy (J) Energy (kJ/mole)
490 Frequency (Hz)
8.21 × 1013 Answer:
10−2 m 108 cm−1
1
1
·
=
= λ λ[Å] · 10−10 mÅ−1 1 cm
λ[Å] hc 6.626 · 10−34 J s · 2.998 · 108 m s−1
=
E=
λ
λ[Å] · 10−10 mÅ−1
1.988 · 10−15 J
=
λ[Å]
1.988 · 10−15 J/photon
E=
· 6.022 · 1023 photon/mole · 10−3 k J/ J
λ[Å]
1 . 1 9 7 · 1 0 6 k J /m o l e
=
λ[Å]
c
ν is not angular frequency, but the reciprocal of the period of oscillation
ν= ;
λ
2.998 · 108 m s−1
2.998 · 1018 Hz
=
=
λ[Å] · 10−10 mÅ−1
λ[Å]
ν=
˜ (b) Name the spectral region associated with each of the last four columns of the table.
Answer:
XUV
Far IR
UV
Mid IR
Name for region of the spectrum
6
wavelength/Å
420
1 .0 0 ·1 0
2440
3 .6 5 · 1 0 4
−1
5
wavenumber/cm
2 .3 8 · 1 0
100
40900
2740
−18
−21
−19
energy/J
4 .7 3 · 1 0
1 .9 9 · 1 0
8 .1 4 · 1 0
5.44 · 10−20
energy/kJ mole−1
2850
1 .2 0
490
3 2 .8
frequency/Hg
7.14 · 10−15 3.00 · 1012 1.23 · 1015 8.21 · 1013 Problem Set #1 ANSWERS Spring, 1994 Page 2 2. A 100W tungsten ﬁlament lamp operates at 2000 K. Assuming that the ﬁlament emits like a black
body, what is the total power emitted between 6000 Å and 6001 Å? How many photons per second
are emitted in this wavelength interval?
Answer: This problem makes use of the StefanBoltzman Law, I = σT 4 . You can either derive it (problem 8a) or look it up. 100 W tungsten ﬁlament lamp operating at 2000 K. What is the total power emitted between 6000 Å and 6001 Å? We can calculate this quantity by integrating the Planck function from 6000 Å to 6001 Å, and comparing that the value of the Planck function integrated over all λ’s. v2 Power = v1
∞
0 ρ(v, T )d v
ρ(v, T )d v × 100W ; v1 = c
6001Å , v2 = c
6000Å . ρ(v, T ) is approximately constant between 6000 Å and 6001 Å.
d v = − λc2 d λ, d v ≃ λc2 Δλ
v2
ρ(v, T )d v ≃ ρΔv
v
1 hν
8πh 3 − kT ν e , hν ≫ k T
c3
∞
3
ρ(v, T ) = 8cπ3h 0 hν d v
ν
0 e kT −1
hν
let x = kT , d ν = kh d ν
T ρ≃ ∞ ∞ ρ(v, T )d ν = 0 8πh kT
c3 h
4 = ρΔν
∞ 0 ρ(v)d v hν = ν3 e− kT Δν
π4
15 kT 4
h = 8πh π
c3 15 4 ∞
0 kT
h 4 x3
dx
ex − 1
StefanBoltzman Law (4.996 × 1014 s−1 )3 (1.611 × 105 )−1 (8.326 × 10+10 s−1 )
(6.494)(4.168 × 1013 s−1 )4 = 3.29 × 10−6
3.29 × 10−6 × 100W = 329 µW 3.29 × 10−6 J s−1
3.313 × 10−19 J photon−1
= 9.93 × 1012 photons s−1 = Problem Set #1 ANSWERS Spring, 1994 Page 3 3. (a) What is the magnitude of the electric ﬁeld for the beam of a 1−mW heliumneon laser with a
diameter of 1 mm?
Answer: Use Bernath (1.4b) E = 27.4I 1/2 = 988 V/m
(b) How many photons per second are emitted at 6328 Å?
Answer:
1 mW = 1 × 10−3 J s−1 @ 6328 Å 1.988 × 10−15 J photon−1
−3
−1
= 1 × 10 J s
6328
= 3.18 × 1015 photons s−1
(c) If the laser linewidth is 1 kHz, what temperature would a blackbody have to be to emit the
same number of photons from a equal area over the same frequency interval as the laser?
Answer: The HeNe beam has d = 1mm, A =
A = AHeNe = 0.785mm2. c
A I (v, T )d v ≃ Aρ(6328Å, T ) 4 Δv = 1mW hv 1 π
ρ(6328Å, T ) = 8λ3h e kT λ − 1
c
Aρ(λ, T ) 4 Δv = P
hv 1
A 8πh c
Δv = e kλ T − 1
P λ3 4 π
4 mm2 = 0.785mm2. For calculation of T bb , −1 A 8π hc
0.785 × 10−6 m2 (8π)6.626 × 10−34 J s 3.000 × 108 m s−1
× 103 s−1 =
×
×
−3 J s−1
−7 m)3
1 × 10
(6.328 × 10
4
P λ3 4
= 396
ex ≃ 1 + x, x ≪ 1
hc 1
≃ 3.86 × 10−6
hλ T
6.626 × 10−34 J s × 3.00 × 108 m s−1
1
×
1.381 × 10−23 J K −1 × 6.328 × 10−7 m 3.86 × 106
= 5 .8 9 × 1 0 9 K T= Problem Set #1 ANSWERS Spring, 1994 Page 4 4. The lifetime of the 32 P3/2 → 32 S1/2 transition of the Na atom at 5890 Å is measured to be 16 ns.
Na 32 P3/2 ← 32 S1/2 λ = 5 8 9 0 Å, τ = 1 6 n s (a) What are the Einstein A and B coeﬃcients for the transition?
Answer: The radiative lifetime of the state is related to the Einstein spontaneous emission
coeﬃcients
τ−1 =
Ai j
i
j The 32 P3/2 state can radiate only to 32 S1/2 (the ground state), so A= 1 1
=
= 6.3 × 107 s−1
τ 16n s From Eq. (1.22) λ3 A
(5.890 × 10−7 m)3 =
8πh 8π6.626 × 10−34 J s(16 × 10−9 s)
= 7.67 × 1020 J −1 m3 s−2 B= (b) What is the transition dipole moment in debye?
Answer: from Eq. (1.52) 16π3 ν3 2
A10 =
µ = 3.136 × 10−7 ν3 µ20
˜1
3 10
3ǫ0 hc
1
5890 Å
1
=
−4 3.136 × 10−7
16 × 10
108 Å cm−1
µ10  = 6.38 D
µ20
1 ν = [cm−1 ] ˜
−3 = 4 0 . 7 D2 (c) What is the peak absorption cross section for the transition in Å2 , assuming that the linewidth
is determined by lifetime broadening? Answer: Lorentzian line proﬁle is due to only lifetime broadening g(ν − ν0 ) 4
1
;γ =
γ
τs p
g(0) = 4τs p
g(0) = Problem Set #1 ANSWERS Spring, 1994 Page 5 Answer: 4c, continued
From Eq. (1.57)
Aλ2 g(ν − ν10 ) λ2 g(ν − ν10 )
1
σ = =
;A =
8π
8πτs p
τs p
2
2
λ 4τ s p λ
σmax =
=
; ν = ν0
8πτs p
2π
(5.890 × 10−5 cm)2
= 5.52 × 10−10 cm2
=
2π
5. (a) For Na atoms in a ﬂame at 2000 K and 760 Torr pressure calculate the peak absorption cross
section (at line center) for the 32 P3/2 − 32 S1/2 transition at 5890 Å. Use 30 MHz/Torr as the
pressurebroadening coeﬃcient and the data in Problem 4.
Answer: Na atoms, T = 2000K, p = 760 torr. Calculate peak absorption crosssection for 32 P3/2 − 32 S1/2 5890Å. Δν1/2 = (30 MHz/torr) p
3 0 M Hz
=
(760 torr)(29979 MHz/cm−1 )−1
torr
= 0.76 cm−1
From Eq. (1.75), for a Lorentzian lineshape
Δν1/2 = γ
2π ; γ = 2πΔν1/2 From Eq. (1.57),
4
Aλ2 g(0) Aλ2
=
, g(0) =
8π
2πγ
γ
2
Aλ
1
(5.890 × 10−5 cm)2
=
=
(2π)2 Δν1/2 16 × 10−9 s (2π)2 (22900 × 106 s−1 )
= 2.41 × 10−13 cm2 σmax = Problem Set #1 ANSWERS Spring, 1994 Page 6 (b) If the path length in the ﬂame is 10 cm, what concentration of Na atoms will produce an
absorption (I /I0 ) of 1/e at line center?
Answer: N1 ≈ 0, N0 ≈ N
ln N
N0 169.78 = 4 e− 1340 = 1.0 × 10−5
2 I
= −σN ℓ = −1
I0
N = [2.41 × 10−13 cm2 × 10cm]−1 = 4.15 × 1011 cm−3 (c) Is the transition primarily Doppler or pressure broadened?
Answer:
−7 Δν0 = 7.1 × 10 ν0
˜
˜ T
M 1/2 ν0 = 16978 cm−1
˜
T = 2000 K
M = 23 amu
Δν0 = 0.11 cm−1
˜
This compares to the pressure broadened linewidth of 0.76 cm−1 , as determined in part (a). The transition is primarily pressure broadened. NOTE: lifetime broadening contributes < 0.001 cm−1 to homogeneous broadening. The fact Δν L
˜
that ΔνD ≃ 7 means that the crosssection calculated in part (a) is slightly overestimated. ˜
(d) Convert the peak absorption cross section in cm2 to the peak molar absorption coeﬃcient ǫ .
No Answer given Problem Set #1 ANSWERS Spring, 1994 Page 7 6. For Ar atoms at room temperature (20◦ C) and 1–Torr pressure, estimate a collision frequency for
an atom from the van der Waals radius of 1.5 Å. What is the corresponding pressurebroadening
coeﬃcient in MHz/Torr?
Answer: Collision rate was determined using equations from P. W. Atkins, Physical Chemistry, 2nd edition. From Bernath, Eq. (1.79) 1
; T 2 is the average time between collisions
πT 2 T 2 = z−1 ; z = collision rate √
N
z = 2σ c
V
σ = πr2 = π[1.5 × 10−8 cm]2 = 7.07 × 10−16 cm2 Δν1/2 = T 1/2
cm s−1 = 3.80 × 104 cm s−1
M 1torr × 710 atm torr−1 × 10−3 ℓcm−3
p
p/torr
N
6
=
=
×
6.022 × 1023 molecules · mole−1
V RT
T /K
1 K
× 0.08206 ℓ atm mole−1 K −1 p molecules cm−3
= 9.656 × 1018
T c = 14551 Δν1/2 z
1 √
1 T 1/2 9.656 × 1018
2
= = p
=
2(πr )14551
π T2 π
π M
T 1/2
√
9.656 × 1018
−8
2 293
14551
p
= 2(1.5 × 10 cm)
40 293 = 0.41 MHz/torr × p/torr 7. Solve the following set of linear equations using matrix methods
4 x − 3y + z = 11
2 x + y − 4z = −1
x + 2y − 2z = 1.
Answer: 11 4 −3 1 2 1 −4 x = −1 1 1 2 −2 Problem Set #1 ANSWERS Spring, 1994 Page 8 Answer: #7, continued A x = b x = A−1 b The diﬃcult part of this problem is calculating A−1 . Bernath, Eq. (3.28) (A−1 )i j =
  = 4 A 2 −4
2 1 1 −4 − (−3) + (1) 2 −2 1 −2 1 2 = 4[−2 − (−8)] + 3[−4 − (−4)] + [4 − 1]
= 27
M11 = (−1)2
M12 = (−1)3 1 −4
= 6 2 −2 2 −4
= 0 1 −2 M13 = (−1)4 2 1 = 3 1 2 M21 = (−1)3 −3 1
= −4 2 −2 M22 = (−1)4 4 1 = −9 1 −2 M23 = (−1)5 4 −3
= −11 1 2 M31 = (−1)4 −3 1
= 11 1 −4 M32 = (−1)5 4 1 = 18 2 −4 4 −3
M33 = (−1)6
= 10 1 1 6 −4 11 1 −1
0 −9 18 A b= 27 3 −11 10 6 −4 11 11 3 1 −1 0 −9 18 −1 = 1 A b= 27 2
3 −11 10 1 3 x = 1 2 M ji A Problem Set #1 ANSWERS
8. Spring, 1994 Page 9 (a) Find the eigenvalues and normalized eigenvectors of the matrix
A= 2
4−i
.
4 + i −14 Answer: It will be much easier to ﬁnd the eigenvalues and eigenvectors if we rewrite A as
follows:
A= ǫ◦ V ⋆
V −ǫ ◦ +
+ B 1
ǫ = (2 − 14)
2
◦
ǫ =8 ǫ0
0ǫ
ǫI = −6 V ⋆ V = V 2 = (4 + i)(4 − i) = 17
Eigenvalues of A are found by solving d et[B − E I] = 0 and adding ǫ
d et[B − E I] = (ǫ ◦ − E )(−ǫ ◦ − E ) − V 2 = 0
E 2 = (ǫ ◦ )2 + V 2
E = ±9 , ǫ = −6 The eigenvalues of A are +3 and −15. Determine the eigenvectors of A by substituting the
appropriate values of ǫ into A − ǫ I
ǫ1 = +3 A − ǫ1 I = −1 4 − i
4 + i −17 Normalize it to unit length 1 4−i
ǫ1 = +3, u1 = √
18 1
ǫ2 = −15 A − ǫ2 I = 17 4 − i
4+i
1 Same as for ǫ1 , but 1
−1
ǫ2 = −15, u2 = √
4+i
18 Problem Set #1 ANSWERS Spring, 1994 Page 10 (b) Construct the matrix X that diagonalizes A and verify that it works.
Answer: X, which diagonalizes A, consists of the column vectors u1 and u2 .
X−1 AX = Λ
1
−1 4 − i
X= √
1
18 4 + i
−1
⋆T
X = X = (X )
Let a = 4 + i, a⋆ = 4 − i, a⋆ a = 17
1
18
1
=
18
1
=
18 X−1 AX = −1 a⋆ 2 a⋆ −1 a⋆
a 1 a −14 a 1
−1 a⋆ 15 3a⋆
a 1 −15a 3 −15(1 + 17)
0
−15 0
=
0
3(17 + 1)
0 3 X diagonalizes A. 9. Given the matrices A and B as −1
3 A= 2 3 √ 2
3 2
3 0
1
√
3 √ 2 3 1
√ 3 2
−3 5 3 √ B = 16 1 − √
32 1
√
6
3
2
1
√
23 1
− 3 √2 1 √ . 2 3 11 6 Show that A and B commute. Find their eigenvalues and eigenvectors, and obtain a unitary trans
formation matrix U that diagonalizes both A and B.
Answer:
A=
−.33333 .81650
.4 7 1 4 0
.8 1 6 5 0 .0 0 0 0 0
.5 7 7 3 5
.47140 .57735 −.66667
B=
1.66667 .40825 −.23570
.4 0 8 2 5 1 .5 0 0 0 0
.2 8 8 6 8
−.23570 .28868 1.83333 Problem Set #1 ANSWERS Spring, 1994 Answer: #9, continued
COMMUTATOR (AB − BA) :
.0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0
.0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0
.0 0 0 0 0 .0 0 0 0 0 .0 0 0 0 0
MATRIX BEFORE DIAGONALIZATION:
1
2
3
1 =
1.66667
.40825 −.23570
2 =
.40825
1 .5 0 0 0 0
.2 8 8 6 8
3 =
−.23570
.2 8 8 6 8
1 .8 3 3 3 3
EIGENVALUES AND EIGENVECTORS:
#1
#2
#3
Value:
1.0000
2 .0 0 0 0
2 .0 0 0 0
Vector:
1 =
.57735 −.70711
.4 0 8 2 5
2 =
.22116 −.34588 −.91184
3 =
.78597
.61674 −.04331
MATRIX BEFORE DIAGONALIZATION:
1
2
3
1 =
−.33333
.8 1 6 5 0
.4 7 1 4 0
2 =
.81650
.0 0 0 0 0
.5 7 7 3 5
3 =
.47140
.57735 −.66667
EIGENVALUES AND EIGENVECTORS:
#1
#2
#3
Value: −1.0000 −1.0000 −1.0000
Vector:
1 =
−.22217 −.34509
.9 1 1 9 0
2 =
.78569 −.61718 −.04214
3 =
.57735
.70711 −.40825 Page 11 Problem Set #1 ANSWERS Spring, 1994 Page 12 10. Obtain eigenvalues and eigenvectors of the matrix 1
2α
0 2 3α H = α 2 + α 0
3α 3 + 2α
to second order in the small parameter α.
Answer: 1
2α
0 3α H = 2α 2 + α 0
3α 3 + 2α H = H◦ + H′ 1 0 0 H◦ = 0 2 0 0 0 3 Energy corrected to second order: 0 2 0 , H′ = α
2 1 3 032 E i = E i◦ + iH′ i +
1 H ′ 1
1 H ′ 2
1 H ′ 3
2 H ′ 2
2 H ′ 3
3 H ′ 3  i H ′  j 2
E i◦ − E ◦
j
i� j =0
= 2α
=0
=α
= 3α
= 2α (2α)2
= 1 − 4α2
1−2 (2α)2 (3α)2
E 2 = 2 + α + + = 2 + α − 5α2 2 − 1 2 − 3 (3α)2
E 3 = 3 + 2α +
= 3 + 2α + 9α2
3−2 Hi j
j ◦
 = i ◦ +
i
E i◦ − E ◦
j
i� j
E1 = 1 + 0 + 1 = 1 ◦ − 2α 2 ◦
2 = 2 ◦ + 2α 1 ◦ − 3α 3
3 = 3 ◦ + 3α 2 ◦ ◦ Problem Set #1 ANSWERS Spring, 1994 Page 13 Answer: # 10, continued
Checking orthogonality:
1 2 = 0 13 = −6α2
2 3 = 0 ; α ≪ 1, so approaches zero 11. A particle of mass m is conﬁned to an inﬁnite potential box with potential
∞,
x < 0, x > L,
x
k 1 − L , 0 ≤ x ≤ L. V ( x) = Calculate the ground and fourth excitedstate energies of the particle in this box using ﬁrstorder
perturbation theory. Obtain the ground and fourth excitedstate wavefunctions to ﬁrst order, and
sketch their appearance. How do they diﬀer from the corresponding unperturbed wavefunctions?
Answer: Particle of mass m conﬁned a perturbed square well potential.
ψ◦ ( x)
n 2
=
L 1/2 sin ∞,
x
k 1− L
x
H′ = k 1 −
L V ( x) = nπ
x
L
x < 0, x > L
, 0≤x≤L k 2 L 2 nπ
x sin2 xd x
L L nπ
0
2k (nπ)2 1
=k
=k−
(nπ)2 4
2
L
2k
nπ
mπ
= k ψ◦ ψ◦ − 2
x sin x sin
x dx
nm
L0
L
L ψ◦ H′ ψ◦ = k −
n
n ψ◦ H′ψ◦
n
m sin A sin B = 1 [cos(A − B) − cos(A + B)]
2
L
L
−k
(n − m)π
cos(n + m)
x cos
x
xd x −
xd x
2
L
L
L
0
0
(n−m)π
(n+m)π
L2
L2
−k
x cos xd x −
x cos xd x
=2
L (n − m)2 π2 0
(n + m)2 π2 0 ψ◦ H′ ψ◦ =
n
m n ± m = even, ψ◦ H′ψ◦ = 0
n
m Problem Set #1 ANSWERS Spring, 1994 Page 14 Answer: #11, continued
n ± m = odd, ◦ ′ ◦ +2k
1
1 2k
4nm
−
= 2
ψn H ψm = 2
π (n − m)2 (n + m)2
π (n + m)2 (n − m)2 2k 8 ψ◦ H′ ψ◦ = 2
1
2
π9 2k 16 ◦
′ ◦ ψ1 H ψ4 = 2
π 225 2k 48 ◦
′ ◦ ψ4 H ψ3 = 2
π 49 2k 80 ◦
′ ◦ ψ4 H ψ3 = 2
π 81 2k 112 ◦
′ ◦ ψ4 H ψ7 = 2
π 1089 Δn ≥ 3 perturbations contribute extraordinarily little relative to Δn = 1 ◦
E 1 = E 1 + ψ◦  ′ ◦
1 H ψ1 1
h2
= α + k
; α = 2 8mL2 ◦ ′ ◦ ◦
E 4 = E 4 + ψ4   4
Hψ
1
= 16α + k
2 To ﬁrst order
 1
ψ
 4
ψ 2k 8 ◦ =
− 2
ψ
π α 27 2 2k
48
80 ◦ ◦ ◦ = ψ4 + 2
ψ − ψ
π α 343 3
729 5 1
ψ◦ We can rewrite the ψ’s in terms of a single parameter, β 1
 1 = ψ◦ − 0.296β ψ◦
ψ
2 4
 4 = ψ◦ + 0.140β ψ◦ − 0.110β ψ◦
ψ
3
5
2k
β  ≪ 1
;
β= 2
πα
Establishing the qualitative eﬀect on ψ1 is simple 1
ψ◦ =
0 , L 1
ψ◦ =
0 L
Mixing in −0.296β ψ◦ makes ψ1 asymmetric, with a slightly increased probability of ﬁnding the particle
2
on “L–side” of the well. Problem Set #1 ANSWERS Spring, 1994 Answer: #11, continued
n ± m = odd, 1
 1 = ψ◦ − 0.296β ψ◦
ψ
2
β = 0, 0.05, 0.1, 0. 2 4
 4 = ψ◦ + 0.140β ψ◦ − 0.110β ψ◦
ψ
3
5
β = 0, 0.2, 0.5, 1 Page 15 ...
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This note was uploaded on 11/28/2011 for the course CHEM 5.74 taught by Professor Robertfield during the Spring '04 term at MIT.
 Spring '04
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