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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 1985 Problem Set #2 ANSWERS
1. The number of possible spin eigenfunctions for a single particle of spin I is 2I + 1.
(a) How many linearly independent spin eigenfunctions are possible for two equivalent particles of spin I ?
ANSWER (2I + 1)(2I + 1) spin eigenfunctions (b) For a particle with I = 1, denote the three spin eigenfunctions by α, β, and γ, corresponding to the eigen
values Mz = +, 0, −. How many linearly independent symmetric and how many linearly independent
antisymmetric spin states are there for two equivalent particles with I = 1?
ANSWER: 6 symmetric eigenfunctions, 3 antisymmetric eigenfunctions
symmetric:
αα� , ββ� , γγ� ,
1
√ (αβ� + βα�) ,
2
1
√ (αγ� + γα�) ,
2
1
√ (βγ� + γβ�)
2 antisymmetric:
1
√
2
1
√
2
1
√
2 (αβ� − βα�) ,
(αγ� − γα�) ,
(βγ� − γβ�) 5.76 Problem Set #2 ANSWERS 1985 Page 2 2. Atomic eigenfunctions contain a factor exp(iM φ). When the atom is a magnetic ﬁeld B, the quantum number
M represents the projection of the J –vector on B (− J ≤ M ≤ + J ). The usual selection rules for L, S , and
J still hold for moderate B, and in addition a selection rule governing the values of Δ M becomes important.
The dipolemoment operators for transitions involving M are ce�� and c� e�⊥ cos φ. The coeﬃcients c and c�
are nonzero constants (for the purposes of this problem); e is the charge on the electron; and �� and �⊥ are the
components of the electric ﬁeld of the radiation parallel and perpendicular to B. Derive the selection rules for
Δ M for radiation polarized parallel and perpendicular to B.
ANSWER:
Electric ﬁeld of radiation � to B ﬁeld
dipole moment operator = ce�� (c is a constant)
� 2π
� 2π
�
�
∴
e−iM φ ce�� eiMφ dφ = ce��
ei( M− M )φ dφ
0 0 = ce�� δ M, M�
∴ ΔM = 0 Electric ﬁeld ⊥ to B ﬁeld applied: ∴ dipole moment operator = c� e� cos φ
�� 2π
�
� 2π
� 2π
1�
−iM � φ �
iM φ
i( M − M � +1)φ
i( M − M � −1)φ
e
c e�⊥ cos φe dφ = c e�⊥
e
dφ +
e
dφ
2
0
0
0
�
�
1
= c� e�⊥ δ M� , M+1 + δ M� , M−1
2
∴ transition allowed for M � = M ± 1 (or Δ M = ±1) 5.76 Problem Set #2 ANSWERS 1985 Page 3 3. Calculate the Zeeman pattern to be expected for the sodium D–lines at 10,000 Gauss (G). You may neglect
nuclear hyperﬁne interactions. Indicate the polarization of each Zeeman line, that is, whether the electric
vector of the emitted radiation is parallel to the applied magnetic ﬁeld (π–component) or perpendicular to it
(σ–component).
(a) Show qualitatively the Stark eﬀect to be expected for the sodium D–lines. The splittings are proportional
to what power of the electric ﬁeld strength?
ANSWER: B = 10kG
�
�
(1)
SS
EZeeman = 2e� B M J 1 + J ( J +1)+2 J(( J+1)−L(L+1)
mc
+1)
Na D lines → 2 S 1/2 − 2 P1/2
2 (1)
EZeeman (2 S 1/2 ) S 1/2 − 2 P3/2 �
�
�1 1
� 0± 1
�
→  JS L M J � = �
22
2 1
(1) → EZeeman (2 S 1/2 , M J = +1/2) = (4.76 × 10−5 cm−1 /Gauss)(10kG)(1/2)(2) = 0.467 cm−1
2
1
(1)
M J = − → EZeeman (2 S 1/2 , M J = −1/2) = −0.467 cm−1
2
MJ = 2 P1/2 �
�
�1 1
� 1± 1
→�
�2 2
2
(1)
EZeeman (2 P1/2 , M J = +1/2) = (0.467 cm−1 )(1/3) = +0.156 cm−1
(1) EZeeman (2 P1/2 , M J = −1/2) = −0.156 cm−1 2 P3/2 �
�
�1 1
� 1 ± 3 ± 1 �
→�
22
22
(1) EZeeman (2 P3/2 , M J = +3/2) = (0.467 cm−1 )(4/3)(3/2) = +0.934 cm−1
(1) EZeeman (2 P3/2 , M J = −3/2) = −0.934 cm−1
(1) EZeeman (2 P3/2 , M J = +1/2) = (0.467 cm−1 )(4/3)(1/2) = +0.311 cm−1
(1) EZeeman (2 P3/2 , M J = −1/2) = −0.311 cm−1 Allowed Transitions: Δ M J = 0, ±1
�
Δ M J = 0 for Radiation E ﬁeld � B applied
�
Δ M J = ±1 for Radiation E ﬁeld ⊥ B applied 5.76 Problem Set #2 ANSWERS 1985 Page 4 ANSWER: 3(a), continued 2P 3/2 2P 1/2 2S �
��
��
� 1/2 3/2
1/2
−1/2
−3/2 1/2
−1/2 �
�
�� 1/2
−1/2 Unperturbed: 2 S 1/2 − 2 P3/2 ⇒ λ = 5889.963Å ⇒ ν = 16978.035 cm−1
2S
2
−1
1/2 − P1/2 ⇒ λ = 5895.930Å ⇒ ν = 16960.853 cm
With 10 kG Field
.311 2 2 ������������������������������
→ 16960.853 ± (0.467 − 0.156) = 16960.542, 16961.164 cm−1 2 ������������������������������
: Perpendicular Polarization → 16960.853 ± (0.467 + 0.156) = 16960.230, 16961.476 cm−1 2 ������������������������������
→ 16978.035 ± (0.467 + 0.311) = 16977.257, 16978.813 cm−1 2 ������������������������������
: Perpendicular Polarization → 16978.035 ± (0.467 − 0.311) = 16977.879, 16978.191 cm−1 S 1/2 − P1/2 : Parallel Polarization .623 2 S 1/2 − P1/2 .778 2 S 1/2 − P3/2 : Parallel Polarization .156 2 S 1/2 − P3/2 1.401 ������������������������������
also 16978.035 ± (0.467 + 0.934) = 16976.634, 16979.436 cm−1
2S − 2 P1/2 2S Spectrum unperturbed → − 2 P3/2 with B–ﬁeld →
polarization → ⊥ � � ⊥
E ⊥
� �⊥ ⊥� ⊥ 5.76 Problem Set #2 ANSWERS 1985 Page 5 Answer 3a, continued
2S 1/2 − 2 P1/2 2S Stark eﬀect ∝ � 2 unperturbed → 1/2 − 2 P3/2 with � –ﬁeld →
polarization → � or ⊥ ��
⊥⊥
E � (b) What do you think might happen to a beam of groundstate sodium atoms passing through a strong
inhomogeneous magnetic ﬁeld? a strong inhomogeneous electric ﬁeld?
Answer:
Ground State Na Atoms ⇒ 2 S 1/2
Strong Inhomogeneous Magnetic Field ⇒ Splits Atoms into 2 Beams: M J = ± 1
2
Strong Inhomogeneous Electric Field ⇒ 1 Beam that is bent by ﬁeld.
4. An atom is in a (3d)2 3 P0 state.
(a) List all L−S−J terms to which an electric dipole allowed transition might occur.
Answer:
(3d)2 3 P0 state → ?
Selection rules: even↔odd, ΔS = 0, ΔL = 0, ±1, Δ J = 0, ±1 (0 � 0) ∴ State must be of opposite parity ⇒ must have ◦ superscript (3dnp or 3dn f ), must be triplet state ⇒ 3 P◦ , 3 D◦
1
1
(b) List all twoelectron conﬁgurations into which electric dipole allowed transitions can occur from (3d)2
3P .
0
Answer:
(3d)2 → (3d)(np)
n>3
2 → (3d )(n f )
(3d)
n≥4 5.76 Problem Set #2 ANSWERS 1985 Page 6 5. The “transition moment,” or the probability of transition, between two rotational levels in a linear molecule
may be assumed to depend only on the permanent electric dipole moment of the molecule and thus to be the
same for all allowed purerotational transitions. In the purerotational emission spectrum of H35 Cl gas, lines
at 106.0 cm−1 and 233.2 cm−1 are observed to have equal intensities. What is the temperature of the gas? The
rotational constant B for H35 Cl is known to be 10.6 cm−1 , and the ratio hc/k has the value 1.44 cm·K.
Answer:
For emission:
ν = B( J � ( J � + 1) − J � ( J � − 1)] = 2 BJ �
∴ from ν = 106 cm−1 and 233.2 cm−1 we have J � = 5 and J � = 11
I5 = (2 J � + 1)e−1.44(30B)/T
I11 = (2 J � + 1)e−1.44(132B)/T
11e−1.44(30B)/T = 23e−1.44(132B)/T
∴ T = 2110 K
6. What would happen to the BirgeSponer extrapolation scheme for a molecular potential that correlates with
ionic states of the separated atoms?
Answer:
The BirgeSponer extrapolation would underestimate D0 because the electrostatic interaction between the
two atoms would cause the actual curve to approach zero much more slowly than the linear BirgeSponer
extrapolation. ...
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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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