Exercise 4
November 16, 2015
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Problems
1. Calculate the Zeeman energy shif ∆EMJ = gµB Bz MJ for Na atom in state
F4 with MJ = 1 in a magnetic flux density of Bz = 2.964. Here µB is the Bohr
magneton with value 9.274 × 10−24 TJ and g the Lande factor with value 1.250.
Enter your answer in kcal / mol.
The correct answer is 0.00494540929213.
Solution: ∆EMJ = gµB Bz MJ = 1.250 ∗ 9.274 × 10−24 (J/T ) ∗ 2.964(T ) ∗ 1 ∗
6.0221367 × 1023 (1/mol)/4184.0(J/kcal) = 0.00494540929213 kcal/mol
2. Fur ein Atom werden die Ubergange 2 P1/2 →2 S1/2 und die Ubergange
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P3/2 →2 S1/2 bei λ1 = 4552.552 Å und λ2 = 2087.229 Å gefunden. Berechnen Sie die absolute Energiedifferenz zwischen den Zustanden 2 P1/2 und 2 P3/2 .
Geben Sie Ihre Antwort in eV.
The correct answer is 3.2212E+00
Solution: E(2 P1/2 →2 S1/2 ) = hν1 = E(2 P1/2 )−E(2 S1/2 ) = hc/λ1 , E(2 P3/2 )−
2
E( S1/2 ) = hc/λ2 , so E(2 P3/2 ) − E(2 P1/2 ) = hc(1/λ2 − 1/λ1 ) = 6.62607004 ×
10−34 (m2 ·kg·s−1 )∗299792458(m/s)∗(4552.552−2087.229)/(4552.552∗2087.229)(Å−1 )∗
1.0 × 1010 (Å/m) ∗ 6.242 × 1018 (eV /J) = 3.2212E + 00eV
# Note: 1 J = 1 kg·m2 /s2
3. Calculate the maximum possible value of the LS coupling total angular
momentum J of the Term 2 P .
The correct answer is 1.5000E+00.
Solution: For the Term 2 P , we have L=1 and S=1/2, so the possible values
of J are |1 − 1/2| and |1 + 1/2|, while the latter is the maximum value.
4. Specify which of the following terms for N Atom with electron configuration 1s2 2s2 2p3 is possible.
1.3 D1 , 1 D2
2.2 P3/2 , 2 P1/2
3.3 D1 , 3 P0
The correct answer is 2.
Solution: for configuration p3 , the allowed values of L ans S are 2,1,0 and
3/2,1/2, respectively. This will give us only half-integer number of J when
coupling L and S.
5. Berechnen Sie die Anzahl von Microzustanden (microstates) des C Atomes.
The correct answer is 15.
Solution: for C, its electronic configuration is p2 , there are 6 possible states
↑ ↓ ↑ ↓ ↑ ↓
(px , px , py , py , pz , pz ) for 2 electrons to be filled, so one could get the results by
C62 = 15.
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6. Indicate which of the following statements is no postulate of quantum
mechanics.
1.Jeder Messung einer Groesse definiert durch Operator  korrespondiert mit
den Eigenwerten des Operators, d.h. Âψ = aψ.
2.Die Zeitentwicklung der Wellenfunktion gehorcht ∂t ψ = c∇2 ψ with c ∈ R.
3.Die Wellenfunktion The wavefunction evolves in time according to Schrodinger
equation Ĥψ = ih̄∂t ψ.
The correct answer is 2.
Solution: The answer should be obvious.
7. Welche der folgenden Aussagen ist richtig
1.[Ĥ, L̂2 ] 6= 0
2.[Ĥ, Ŝz ] 6= 0
3.[Ĥ, Ŝx ] 6= 0
The correct answer is 3.
Solution: Just remember, when neglecting spin-orbital coupling, the complete set of operators that commute with each other include Ĥ, L̂2 , Ŝ 2 , L̂z , Ŝz ,
8. Wer hat die folgenden Prinzipien oder Gesetze eingefuehrt?
1.Ausschlussprinzip: Stern und Gerlach
2.Ausschlussprinzip: Pauli
3.Hydrogen spektrum ν ∝ 19 − n12 , n ≥ 4: Einstein
The correct answer is 2.
Solution: The answer should be obvious.
9. Geben Sie gemaess den Hundschen Regeln die korrekte energetische Reihenfolge der Termsymbole fuer ein Atom mit Elektronenkonfiguration 1s2 2s2 2p4
an.
1.3 P1 >2 P1/2 >1 D2 >3 P0 >2 S1/2
2.1 D2 >3 P1 >2 P1/2 >2 P3/2 >2 S1/2
3.3 P2 <3 P1 <3 P0 <1 D2 <1 S0
The correct answer is 3.
Solution: According to Hund’s rule, for a given electron configuration, the
term with maximum multiplicity has the lowest energy. With this rule alone,
we could tell only 3 is right.
#10. non-graded problem: Zeigen Sie, wie man alle moeglichen Terme and
their energetic order for an atom with the electron configuration 1s2 2s2 2p4 .
Solution:
For two electronic configurations (nl)x and (nl)y , if x + y = 2(2l + 1), we call
these two configures complementary configurations, which share the same term
symbols, though the corresponding order of energies are different. That is to
say, there is a one-to-one mapping between every microstate of the configuraP
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tion
We could illustrate
this by noticing that ML [(nl)x ] = x ml =
P p and p .P
P
y
2(2l+1) ml −
y ml = −
y ml = −ML [(nl) ] (the same is also true MS ).
Since for a given S (L), there are always both negative and positive MS (ML )
with the same absolute value, so this reversion of sign doesn’t matter at all,
resulting in the same term symbols as p2 for its complementary electronic configuration p4 . The term symbols have all been derived at the PChem class, so
it’s not presented here. We just list the term symbols here, they are 1 S,1 D, 3 P
and their corresponding energetic order is 3 P < 1 D < 1 S according to Hund’s
rule. With spin-orbital coupling, each term will be further split into levels, and
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we could label these levels by further introducing another quantum number J
with possible values ranging from |L − S|, |L − S| + 1, ...|L + S|. For term 3 P ,
the possible levels are 3 P2 ,3 P1 ,3 P0 , the corresponding energy order is 3 P2 < 3 P1
< 3 P0 (according to Hund’s rule 3, see the #Note section); For term 1 D, the
only possible level is 1 D2 ; For term 1 S, the only possible level is 1 S0 .
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Notes
1. Hund’s rule
It refers to three rules:
(a) For a given electron configuration, the term with maximum multiplicity
has the lowest energy. The multiplicity is equal to 2S + 1, where S is the
total spin angular momentum for all electrons. Therefore, the term with lowest
energy is also the term with maximum S.
(b) For a given multiplicity, the term with the largest value of the total
orbital angular momentum quantum number L, has the lowest energy.
(c) For a given term, in an atom with outermost subshell half-filled or less,
the level with the lowest value of the total angular momentum quantum number
J, (for the operator J = L + S) lies lowest in energy. If the outermost shell is
more than half-filled, the level with the highest value of J, is lowest in energy.
These rules specify in a simple way how usual energy interactions dictate the
ground state term. The rules assume that the repulsion between the outer electrons is much greater than the spin–orbit interaction, which is in turn stronger
than any other remaining interactions. This is referred to as the LS coupling
regime.
Full shells and subshells do not contribute to the quantum numbers for total S, the total spin angular momentum and for L, the total orbital angular
momentum. It can be shown that for full orbitals and suborbitals both the
residual electrostatic term (repulsion between electrons) and the spin–orbit interaction can only shift all the energy levels together. Thus when determining
the ordering of energy levels in general only the outer valence electrons must be
considered.
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