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Solutions for various questions related to atomic and molecular physics 2 (pc 4243) as taught in the academic year ay06/07, semester 2. Topics covered include hyperfine splitting, ionization energy, selection rules, linewidths, doppler shift, and entropy. Students can use these solutions as a reference for understanding the concepts and solving related problems.
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The use of Prof Christian’s official solutions as reference is acknowledged.
Li has 3 electrons, and its complete electronic configuration is 1s^22 s, with spec- troscopic notation 2 S 1 / 2.
The levels in the P and D column appear in pairs because of the spin-orbit interation.
The hyperfine interaction leads to a splitting of the ground state.
For 7 Li with I = 3/ 2 , J = S = 1/2, F = 1, 2, so that EHF S = A/2(F (F +1)− J(J + 1) − I(I + 1)) gives EHF S = 3/ 2 A for F = 2 and EHF S = − 3 / 4 A for F = 1.
For 6 Li with I = 1, F = 3/ 2 , 1 /2 with corresponding values of EHF S = A/ 2 A and −A respectively. The physical origin of the hyperfine splitting is due to the contact interaction of s-electrons at the location of the nucleus.
The ground state splitting between the 2 levels are given by
∆EHF S =
where F +^ denotes the larger F value. Now A ∝ gI so that ∆EHF S ∝ gI F +, and we have
g^6 I =
The ionisation energy is proportional to the reduced mass, so that
E 1 = E 2 μ 1 μ 2
with
μ 1 = meMi Mi + me = me
me Mi + me
≈ me
me Mi
λ =
62350cm−^1 − 44043cm−^1 = 18307cm−^1
(^3) P 2 means S = 1, L = 1, J = 2. 3 S 1 means S = 1, L = 0, J = 1. Yes, the
transition follows the common selecton rules for dipole transition as ∆L = 1.
The homogenous linewidth is due to exponential decay :
A(t) = eiω^0 t^ · e−t/^2 τ
whose fourier transform
a(ω) =
ei(ω^0 −ω−^1 /^2 τ^ )t^ dt
=
i(ω − ω 0 ) − 1 / 2 τ |a(ω)|^2 =
(ω − ω 0 )^2 + (1/ 2 τ )^2
Thus the FWHM δω = 2 (^21) τ = (^) τ^1 , giving
δfF W M H =
2 π
τ = 19.89MHz
The Doppler shift is given by ∆f = v¸, so thatf
v = ∆f c f
= ∆f λ
. Assume a velocity distribution
p(vx) ∝ e−^
(^12) mv (^2) x/kT = e−^
∆ 2 σf 22
with
σ =
kT m
λ
. To find the Half Width at Half Maximum, we set
1 2 = e−^
x^2 2 σ^2
⇒ ln
x^2 2 σ^2 ⇒ x = σ
2 ln 2
so that
∆fF W HM = 2
2 ln 2σ
= 2
2 ln 2
kT m
λ = 495 .99MHz
From P V = N kT we define the number density
ρ =
kT
. For hard core collisions, the collisional cross sectional area σ = D^2 π.
From kinetic theory, the mean velocity v =
2 kT m. The mean free path s = (^) σρ^1 = (^) P DkT (^2) π , so that the mean time between collisions
t = s v = kT P πD^2
m 2 kT
=
P πD^2
kT m 2 = 8. 6 × 10 −^5 s−^1
Then ∆f =
t = 11.6kHz.
The interaction Hamiltonian
H(t) = B〈F = 4|μˆ|F = 3〉
where ˆμ = ˆμs = gs sˆz μB /ℏ. Here,
|F = 3〉 =
(|mi = 1/ 2 , mj = − 1 / 2 〉 + |mi = − 1 / 2 , mj = 1/ 2 〉)
so that
〈 4 |μˆz | 3 〉 =
gs(〈− 1 | − 〈+|)ˆμz (|−〉 + |+〉)
=
gs(〈−|sz |−〉 − 〈+|sz |+〉)
=
gsμB (−hbar/ 2 − ℏ/2)
= gs μB ℏ
= μB
Now
ℏΩR = B 0 μB ⇒ ΩR =
B 0 μB ℏ so that
B 0 =
μB
2 πfR μB
ρe =
No, because √^12 (〈−|±〉 + |) is an eigenstate of ˆsx = 12 (ˆs+ + ˆs−) so 〈−|sˆx|+〉 = 0 implying no transfer.
