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PC3233 Atomic and Molecular Physics I
November 2005/
Note All diagrams are NOT necessarily drawn to scale. If there are any errors, please feel free to notify the physics society and provide your corrections.
1i)
For Hyperfine Splitting, i=1/2, j=1/2, f=0, i=1/2, j=3/2, f=1,
1ii) Answer is provided in lecture notes
1iii) Ground state of deuterium: 1s j=1/2, i=1 f=1/2, 3/
2P3/2^2
2S1/
1S1/
2P1/
1S1/
2P1/
2S1/
2P3/
1s
2s2p
Hyperfine Shifted
Lamb Shifted
f
Bohr Model
Dirac Model
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(^ 1)^ (^ 1)^ (^ 1)
HF 2
a V f f i i j j
Subst. f=3/2, f=1/2: , 2 218.256 , 436.
V HF a a
MHz MHz
Total HF Splitting = 654.768MHz
1iv)
a g^ i^ B^ BJ j j
Subst. given values: B (^) J = 28.9T
Na: 1s 2 2s 2 2p 6 3s 1
f=1/
f=3/
2a
a
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To calculate splittings, first determine the lande g-factor of each state
For 3P3/2: ( 1) ( 1) ( 1) 1 2 ( 1) 4 3
j j l l s s g j j
Zeeman Shifts:
m B j
B B
V g Bm
B B
For 3P1/2:
g = +2/3
m (^) 3 B
V B
For 3S1/2:
g = +2 Vm BB
3P1/
3S 1/
Total Splitting: 8/3 (^) B B
3P1/
3S 1/
Total Splitting: 10/3 B B
Note: All splittings are in units of B B
Crossed out boxes are forbidden transitions Highlighted boxes denote the transitions which represent the total splitting
3a)
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The wavefunction for a system of electrons must be anti-symmetric with respect to the exchange ofany 2 sets of space-spin coordinates.
Reasons: Refer to notes (4.2.11)
3b) Refer to Nov06/07 Question 1
4ai) For Rotational Spectrum,
Line spacing = 2B = 21.2cm - B = 10.6cm -
= 2 2
(^8) e
h c I
Substitute I (^) e Re^2
2 2
(^8) e
h c R
Substituting in the given values,
Re m
4aii)
106.0cm -1^ corresponds to J=5 spectral line 233.2cm -1^ corresponds to J=11 spectral line
exp
vJ vJ vJ vJ
Intensity N E N g kT
It is also given that,
2 1 ( 1)
vJ vJ
g J E J J B
For the most intense line, vJ 0
dN dJ
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2 2
[( '' ')( '' ') 1]
e e
e e e e
v J J B v J J B
v v J J J J B J J J J J J J B J B
For P branch, v 1, J 1
Similarly, we arrive at
2 2
[ ( '' ') 1]
e e
e e e e
v J J B v J J B
v v J J J J B J J J J J B J B
For P branch, J 1 J’’-J’=- If J’=0 J’’=-1 which is not possible
If J’=0, (^) e 2(0) B e
And so the e spectral line is absent since J’ 0.
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