Aufgabe 1

Aufgabe 1
Consider boron‐like
boron like carbon ion Ca7+ (nuclear charge Z=12,
Z 12
number of electrons N=5) in which one of electrons is in
Rydberg state with n=50.
• Find ionization energy for the outer (Rydberg)
electron (in state n
n=50).
50).
• Find wavelength of the photons emitted in the
transition between n=51 and n=50 levels of this ion.
Aufgabe 1: Lösung
Rydberg electron can be considered to be in Coulomb field of
effective charge Z effff = 12 − 4 = 8 .
Therefore,, ionization p
potential can be obtained as:
I 50 = − E50 = R
Z eff2
n2
82
= R 2 ≈ 0.35 eV
50
In order to find energy of n=51 ‐ n=50 transition:
1 ⎞
⎛ 1
hω = E51 − E50 = 64 R⎜ − 2 + 2 ≈ 0.0135 eV
⎝ 51 50 ⎠
The wavelength, corresponding to this energy is:
λ=
2πc
ω
≈ 10 − 4 m = 0.1 mm
Aufgabe 2: Normaler Zeeman‐Effekt
Wie viele unterschiedliche Linien
inien sind beim Übergang
von einem d‐Niveau in ein p‐Niveau?
Wie viele bei einem beliebigen Übergang?
Hinweis: Auswahlregel
g ∆m=+1,0,‐1
Aufgabe 2: Lösung
B=0
B≠0
d level (l=2)
ml
2
1
0
‐1
‐2
Energies of Zeeman levels (d‐state)
Emd l = E d + µ B ml B, ml = −2,−1,0,+1,+2
Energies of Zeeman levels (p‐state)
p level (l=1)
ml
m’
1
0
‐1
Empl′ = E p + µ B ml′B, ml′ = −1,0,+1
here E
p ,d
Z2
= −R 2
n
Transition energies:
∆E = Emd l − Empl′ = E d − E p + µ B B(ml − ml′ ) = const + µ B B∆ml
just a constant independent on projections ml
Selection rule: !!!
∆ml = ml − ml′ = 0, ± 1
Only 3 transitions will be observed in experiment!
Aufgabe 3
Given hydrogen atom A in a 4d state and atom B in a
3p state. Find the j values for each atom and then
combine them to find the J values for the entire
system.
Aufgabe 3: Lösung
Atom A: 4d state
Atom B: 3p state
l A = 2, s A = 1 / 2
lB = 1, sB = 1 / 2
lA − sA ≤ jA ≤ lA + sA
lB − sB ≤ jB ≤ lB + sB
j A = 3 / 2, 5 / 2
jB = 1 / 2, 3 / 2
j A − jb ≤ J ≤ j A + jB
J = 0,1, 2, 3, 4

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