Atomic & Molecular Physics Solutions to Problem Sheet 2224.4 Issued Thursday 6 March 2008, due Thursday 13 March 2008 1. For a diatomic molecule to exhibit a rotation spectrum it must possess an electric dipole moment on which the applied electric field can exert a torque. Homonuclear molecules [1] typically have an equal probability of finding the electrons more closely associated with each nucleus (no separation of charge), i.e. no dipole moment, unlike heteronulcear molecules. [1] Purely rotational transitions typically occur in the microwave region of the spectrum. [1] 0 0 The frequencies of purely rotational transitions from J → J are E(J) − E(J ), which for J 0 = J − 1 is E = 2BJ, where B is the rotational constant. A graph of frequency against J (the higher rotational quantum number) should therefore be a straight line, as in figure 1 below.
Figure 1: graph of transition frequency against higher rotational quantum number, J For the right graph
[4]
The gradient of the graph is 2B = 12566 MHz, giving the rotational constant B = 6283 MHz [1] The rotational constant is related to the bond length through the moment of inertia: h ¯2 B= 2Ie (in energy units, where Ie = µRe2 with µ the reduced mass which is:
µRbF =
1 1 + 85 19
−1
× 1.66 × 10−27 kg = 2.59 × 10−26 kg 1
In frequency units (Hertz) we have that: 1 h ¯ B =B= 2π 2µRe2 h giving for the bond length: s
Re =
1 h ¯ 2π 2µB
Inserting values: s
Re =
1 1.054 × 10−34 = 2.27 × 10−10 m 2π 2 × 2.59 × 10−26 × 6.283 × 109 [2]
2. The wavefunctions and electron distributions are shown in figure 2
Figure 2: Wavefunctions and electron distributions for H+ 2 To find the electronic energies we must evaluate: ˆ el |ψ± i = hψ± |E± (R)|ψ± i hψ± |H = E± hψ± |ψ± i and so the energy of the ψ± states is: ∗ ˆ ψ± Hel ψ± dτ ∗ ψ± ψ± dτ A± = N± ˆ el expectation value of H = . normalization constant
R
E± (R) =
R
2
[2]
[1] Then the normalization term is: 1Z [Φ∗1s (rA ) ± Φ∗1s (rB )] [Φ1s (rA ) ± Φ1s (rB )] dτ 2 Z 1 = 1 + 1 ± 2 Φ∗1s (rA )Φ1s (rB )dτ 2 = 1 ± I(R) ,
N± =
since:
Z
and
Z
I(R) =
Φ∗1s (rA,B )Φ1s (rA,B )dτ = 1
Φ∗1s (rB )Φ1s (rA )dτ
=
Z
Φ1s (rB )Φ∗1s (rA )dτ
I(R) is the overlap between Φ1s (rA ) and Φ1s (rB ). It is non-zero since they are not orthogonal as they are centered on different origins. [2] Now for the expectation values, A± . We use the electronic Hamiltonian: ˆ el = − 1 ∇2r − 1 − 1 + 1 H 2 r A rB R First we recall from the treatment of atomic hydrogen that: 1 1 Φ1s (rA ) = E1s Φ1s (rA ) − ∇2r − 2 rA
and 1 1 − ∇2r − Φ1s (rB ) = E1s Φ1s (rB ) , 2 rB
where E1s is ground state energy of H atom. We want: 1 √ [Φ∗1s (rA ) ± Φ∗1s (rB )] 2 1 ˆ el √ [Φ1s (rA ) ± Φ1s (rB )] dτ . × H 2
A± =
Z
that we will re-write as: ˆ el i = 1 [HAA + HBB ] ± HAB A± = hH 2
3
where HAA =
Z
ˆ el Φ1s (rA )dτ Φ∗1s (rA )H
= E1s (Energy of H atom) Z 1 + Φ1s (rA )Φ1s (rA )dτ (internuclear repulsion) R Z 1 − Φ∗1s (rA ) Φ1s (rA )dτ (Coulomb Integral) rB 1 = E1s + − J(R) R where J(R) is due to the interaction with the other atom, B. There is a similar term HBB from the other atom, B: 1 HBB = E1s + − J(R) R [2] Also: HAB = HBA =
Z
ˆ el Φ1s (rB )dτ Φ∗1s (rA )H
and
HAB
Z 1 1 ∗ I(R) − Φ1s (rA ) Φ1s (rB )dτ . = ± E1s + r R A | {z }
‘K’ exchange integral
So
HAB = ± E1s +
1 I(R) ∓ K(R) . R
[2] Putting these results together, the electronic energy is:
E± = A± /N±
=
E1s +
1 R
− J ± E1s +
1 R
I ∓K
1±I 1 −J ∓ K = E1s + + . 1±I | {z R} H atom + Coulomb repulsion [1]
Note: The numbers in square brackets are intended to show the approximate credit for each part of the working, although as this question is bookwork all students should be getting 10 marks. 4
Figure 3: The Morse potential 3. The Morse potential, VMorse is sketched in figure 3 below: For the right sketch, with De and Re marked correctly:
[3]
If we expand the Morse potential:
VMorse = De e−2α(R−Re ) − 2e−α(R−Re )
for small displacements from equilbrium, i.e. α(R − Re ) 1, we find: VMorse ≈ De
1 1 1 − 2α(R − Re ) + (−2α(R − Re ))2 − 2De + 2α(R − Re ) − 2(α(R − Re ))2 2 2
or:
1 VMorse ≈ De −1 + α2 (R − Re )2 = De −1 + k(R − Re )2 2 i.e. the Morse potential is approximately harmonic for small displacements with spring constant, k = 2α2 De . [2]
The molecular state terms 1 Σg and 3 Σu tell us: • Σ → the projection of the electronic angular momentum onto the molecular axis is 0, i.e. Lz = 0. • the superscripts 1 and 3 give the spin multiplicity, i.e. there is a singlet and triplet term with net spin S = 0, 1 • the subscript gives the symmetry with respect to exchange of nuclear labels: u (ungerade) odd with respect to exchange, g (gerade) even with respect to exchange. [1] From the given parameters and the above expression for the spring constant, we have: De = 240 cm−1 = 4.77 × 10−21 J and so: k = 2 × 240 cm−1 = 4.77 × 10−21 × (0.720 × 1010 )2 = 0.494 Nm−1 . 5
q
The vibrational frequency is related to the spring constant as ω = µk , with µ the reduced mass. For 133 Cs2 the reduced mass is µ = 12 mCs = 12 ×133×1.67×10−27 kg = 1.11×10−25 kg. This gives a vibrational frequency: s
ω=
0.494 = 2.11 × 1012 rad s−1 1.11 × 10−25
or: ν˜ =
1 1 ω = 11.2 cm−1 100 2π c
From the given equilibrium bond length and that Ie =
µRe2 ,
[2] we get that:
Ie = 1.11 × 10−25 × (6.496 × 10−10 )2 = 4.68 × 10−44 kg Since we know that the rotational constant B =
h2 ¯ 2Ie
we calculate:
(1.054 × 10−34 )2 B= = 4.69 × 1024 J −44 2 × 4.68 × 10 or in wavenumbers: 4.69 × 1024 ˜= 1 B = 1 = 5.97 × 10−3 cm−1 B 100 hc 100 6.626 × 10−34 × 2.998 × 108 [2]
6