Solutions Atomic Physics for FYSC11 16-03-21.
1.
Z = 5 i.e. B+4
2a. Yes Lˆz   . Eigenvalue .
2
2b. No Lˆ2  
5
2c.
 Lˆz  
2d.
26
 Lˆ2  
5
2
(Y1,1  6Y2,1 )    
2
3a. 5s is more tightly bound since an s-electron penetrates the core more that a d-electron, larger
quantum defect.
3b. Tnd  Eion  End 
End 
RM ζ 2
RM ζ 2

δ

n

. ζ  1. RM  R
(n  δ )2
Tnd
1
1

in cm 1.
λ(3s  3 p ) λ(3 p  nd )
From this we approximate δ(8d) = 0.015 giving T (8d )  1721 cm1  E (8d )  39728 cm1 
λ(3 p  8d )  4393.5 Å. Experimental value is 4393.4 Å.
4a. The Doppler width is given by:
7
f D  7.16  10 (g/mol) K
4b.  
2
0
2nt cos  
1/2
1/2
3  108 m/s
2000 K


 0.296 GHz
6
2.776  10 m 137 g/mol
4 t
f with n = 1 and θ = 0.
c
dδ 4πt
c
c 1 R
3  108 m/s 0.1

 Δf 
Δδ 

 1.68 GHz.
df
c
4πt
2πt R
2π  3  103 m 0.9
The experiment would be limited by the resolution of the Fabry-Perot interferometer
Thus
5ab. Parallel to the field we only observe the (circularly polarized) σ± components arising when
ΔMJ = ±1
5c. Emin  3 /4  B B  9.27  1024 J  57.6 μeV = 13.93 GHz. Yes we can resolve the Zeeman
structure with our Fabry-Perot.
6a.
6b.
6c. ΔE(5/2-3/2) = 1860 + 892 = 2752 MHz. Using the Landé interval rule this gives
A = 2752/2.5 = 1101 MHz.
ΔE(3/2-1/2) = 2600 - 892 = 1708 MHz. Using the Landé interval rule this gives
A = 1708/1.5 = 1139 MHz.
Best estimate is the average A = 1120 MHz.