Download n= n= n=1

Atomic, molecular and laser physics.
Solution set 4
Solution 4: Atomic spectroscopy and fine structure of hydrogen
1. An electron in the H atom is in the n = 3, l = 0, m = 0 state. It decays to the ground
states. What decay routes are open to it?
Solution
The only allowed transition path is shown below.
n= 3
3s
3p
n= 2
2s
2p
n= 1
1s
3d
2. The valence electron of Na is in the 3s state, when the atom is in the ground state. The
valence electron can be excited to higher levels. Indicate on the energy-level diagram
shown below, all allowed transitions among the levels. (It is sufficient to show only
the upward transitions).
5s
4p
3d
4s
3p
3s
Solution
All the allowed (upward) transitions are shown in the Figure below. Note that the
transitions ∆n = 0 between non-degenerate levels are allowed by selection rules.
Due date: November 23, 2011
1
Atomic, molecular and laser physics.
Solution set 4
5s
4p
3d
4s
3p
3s
3. The spectrum of an one-electron ion of an element showed that its ns-orbitals were
at 0, 2057927 cm−1 , 2439156 cm−1 , and 2572563 cm−1 for n = 1, 2, 3, 4, respectively.
Identify the species and predict the ionization energy of the ion.
Solution
The Rydberg formula for the wave number for any hydrogen like element is,
(
)
1
1
1
2
− 2
where, nf < ni ,
ν̄ = = R∞ Z
λ
n2f
ni
when a transition takes place from the ni to the nf level.
For n = 1, ν̄ = 0 ( this is a reference value and all wave numbers are referred to this.
See the Figure below.)
For n = 2, ν̄ = 2057927 cm−1 = 205792700 m−1
For n = 3, ν̄ = 2439156 cm−1 = 243915600 m−1
For n = 4, ν̄ = 2572563 cm−1 = 257256300 m−1 .
Using the given data,
Due date: November 23, 2011
2
Atomic, molecular and laser physics.
Solution set 4
n= 4
reference + 2572563 cm-1
n= 3
reference + 2439156 cm-1
n= 2
reference + 2057927 cm-1
n= 1
205792700 =
Z =
=
243915600 =
Z =
=
2572956300 =
Z =
=
(
)
1
R∞ Z 1 − 2
2
√
205792700 × 4
3 × 1.0974 × 107
5
(
)
1
2
R∞ Z 1 − 2
3
√
243915600 × 9
8 × 1.0974 × 107
5
)
(
1
2
R∞ Z 1 − 2
4
√
2572956300 × 16
15 × 1.0974 × 107
5.
2
Therefore, the element is boron (B) with Z = 5 . Please remember that in the above
calculations, we must use all the significant figures in
1
λ
in order to obtain meaningful
results.
Due date: November 23, 2011
3
Atomic, molecular and laser physics.
Solution set 4
The first ionization energy is,
1
= R∞ Z 2
λ
Eion
= R∞ Z 2
hc
Eion
(
(
1
1
− 2
2
nf
ni
1
1
− 2
2
nf
ni
(1)
= R∞ Z 2 hc
4
= 85 eV.
)
)
We take nf = 2 and ni = ∞ for the first ionization energy because the most loosely
bound electron of B is in the n = 2 (L) shell. Our value neglects screening effects
from low lying electrons.
4. When the spin-orbit interaction is taken into account, it is sometimes said that ml
and ms are no longer “good quantum numbers.” Explain why this terminology is
appropriate. What are the good quantum numbers for the one-electron atom when
the spin-orbit interaction is taken into account?
Solution
⃗ and spin angular momenIn spin-orbit interaction, the orbital angular momentum L
⃗ of an electron interact with each other through their magnetic moments and
tum S
⃗ . L.
⃗ Since Sz and Lz do not commute
the correction to their energies depends upon S
⃗ . L,
⃗ therefore, they are not conserved quantities. As a result, ml and ms are not
with S
⃗ . L,
⃗ J⃗2 ] = 0 and [S
⃗ . L,
⃗ Jz ] = 0, the good quantum
good quantum numbers. Since [S
numbers are now n, l, j, mj .
5. Calculate the spin-orbit interaction energy for a muonic atom, i.e., an electron is
replaced by a muon.
Solution
Since there is no spin-orbit coupling when l = 0, therefore, we can assume n = 2, l = 1
for a muon with rest mass m0 = 207 × me = 1.88 × 10−28 kg. The reduced mass of
Due date: November 23, 2011
4
Atomic, molecular and laser physics.
Solution set 4
the muon in the muonic atom is,
mo M
,
where M is the mass of nucleus (proton),
mo + M
1.88 × 10−28 1.67 × 10−27
=
1.88 × 10−28 + 1.67 × 10−27
= 1.68 × 10−28 kg.
m =
Since the muon is a spin half particle, j can have two values, equals to 1/2 and 3/2.
Consequently, the spin-orbit interaction energy for n = 2, l = 1, j = 1/2 is,
(
)
nEn2 j(j + 1) − l(l + 1) − 3/4
∆Es−o =
mc2
l(l + 1/2)(l + 1)
(
)
2E22 1/2(1/2 + 1) − 1(1 + 1) − 3/4
=
mc2
1(1 + 1/2)(1 + 1)
2 (
)
2E2
=
−1/2
mc2
(
)
2(3.39 × 1.6 × 10−19 J)2
−1/2
=
−28
8
2
(1.68 × 10 kg)(3 × 10 m/sec)
= 1.94 × 10−26 J.
The spin-orbit interaction energy for n = 2, l = 1, j = 3/2 is,
(
)
2E22 3/2(3/2 + 1) − 1(1 + 1) − 3/4
∆Es−o =
mc2
1(1 + 1/2)(1 + 1)
2 (
)
2E2
=
1/4
2
mc
( )
2(3.39 × 1.6 × 10−19 J)2
=
1/4
(1.68 × 10−28 kg)(3 × 108 m/sec)2
= 9.72 × 10−27 J.
Comparing it with the spin-orbit interaction energy of an electron in the n = 2, l = 1
state,
∆Es−o = 3.60 × 10−24 J
(for j = 1/2)
∆Es−o = 1.80 × 10−24 J
(for j = 3/2).
These results show that for the muonic atom, spin-orbit effects are three orders of
magnitude smaller than the electronic atom.
6. For a given n, what are the allowed values of j? What is the energy separation between
the maximum and minimum values of j for a given n?
Due date: November 23, 2011
5
Atomic, molecular and laser physics.
Solution set 4
Solution
The variation of j is,
|l − s| ≤ j ≤ l + s .
For any value of n, l = 0, 1, 2 . . . n − 1. Therefore, the variation of j will be,
|0 − s| ≤ j ≤ n − 1 + s
s ≤ j ≤ n − 1 + s.
For any n, the minimum value of j is s(= 1/2) while the maximum is n − 1 + s.
The total fine-structure correction to Bohr energies are,
En2 ( 3
2n )
∆Etot =
−
mc2 2 j + 1/2
En2 ( 3
2n )
Therefore, (Etot )min =
−
mc2 2 s + 1/2
)
En2 ( 3
2n
(Etot )max =
−
·
mc2 2 n + s − 1 + 1/2
Hence, the required energy separation is,
)
En2 ( 3
2n
En2 ( 3
2n )
(Etot )max − (Etot )min =
−
−
−
2
2
mc 2 n + s − 1/2
mc 2 s + 1/2
2(
)
1
2nEn
1
=
−
2
mc s + 1/2 n + s − 1/2
(
)
n−1
2nEn2
·
=
mc2 (n + s − 1/2)(s + 1/2)
7. Consider the states in which l = 4 and s = 1/2. For the state with the largest possible
⃗ and S,
⃗ (b) the angle
j and largest possible mj , calculate (a) the angle between L
between µl and µs , and (c) the angle between J⃗ and +z axis.
Solution
Largest
j = 4 + 1/2 = 9/2
Largest
mj = j = 9/2.
The magnitude of the vectors are,
√
√
99
J = j(j + 1)~ =
~
2
√
√
L = l(l + 1)~ = 20~
√
√
3
S = s(s + 1)~ =
~.
2
Due date: November 23, 2011
6
Atomic, molecular and laser physics.
Solution set 4
Z
S
m j h < Jz
θ
J
φ
L
µL
µS
Using the law of cosines, (refer to the accompanying Figure)
J 2 = L2 + S 2 − 2LS cos (π − θ)
J 2 = L2 + S 2 + 2LS cos θ √
√
99 2
3
3
~ = 20~2 + ~2 + 2 20~
~ cos θ
4
4
2
√
4 = 60 cos θ
4
θ = cos−1 ( √ )
60
= 58.9 ◦ .
⃗ and S
⃗ for maximum |J|
⃗ and |J⃗z |.
This is the angle between L
⃗ and µs is parallel to −S,
⃗ the angle between µ⃗l and µ⃗s
(b) Since µl = is parallel to −L
will also be 58.91 ◦ .
(c) The angle between J⃗ and the z-axis is,
cos ϕ =
Jz
=
J
9
~
2
√
99
~
2
( 9 )
ϕ = cos−1 √
99
◦
= 25.2 .
Due date: November 23, 2011
7
Atomic, molecular and laser physics.
Solution set 4
8. Evaluate the following commutators: (a) [L · S, L], (b) [L · S, S], (c) [L · S, J], (d)
[L · S, L2 ], (e) [L · S, S2 ], (f) [L · S, J2 ], where L, S and J are representing orbital, spin
and total angular momentum respectively. Hint: L and S satisfy the fundamental
commutation relations for angular momentum, but they commute with each other.
Solution
(a)
[L · S, Lx ] = [Lx Sx + Ly Sy + Lz Sz , Lx ]
= Sx [Lx , Lx ] + Sy [Ly , Lx ] + Sz [Lz , Lx ]
= Sx (0) + Sy (−i~Lz ) + Sz (i~Ly )
= i~ (Ly Sz − Lz Sy )
= i~ (L × S)x .
Similar is the case with other two components of L, i.e.,
[L · S, Ly ] = i~ (L × S)y
Therefore, in general, we get,
[L · S, L] = i~ (L × S).
(b)
[L · S, Sx ] = [Lx Sx + Ly Sy + Lz Sz , Sx ]
= Lx [Sx , Sx ] + Ly [Sy , Sx ] + Lz [Sz , Sx ]
= Lx (0) + Ly (−i~Sz ) + Lz (i~Sy )
= i~ (Lz Sy − Ly Sz )
= i~ (S × L)x .
Similar is the case with other two components of S, i.e.,
[L · S, Sy ] = i~ (S × L)y
Therefore, in general, we get,
[L · S, S] = i~ (S × L).
Due date: November 23, 2011
8
Atomic, molecular and laser physics.
Solution set 4
(c) L2 and S2 commute with the isotropic Hamiltonian L · S. This can be easily
verified.
(d) See answer to (c).
(e) Similarly, we have,
[L · S, S2 ] = 0
(f)
[L · S, J2 ] = [L · S, L2 + S2 + 2L.S]
= [L · S, L2 ] + [L · S, S2 ] + 2[L · S, L · S]
= 0+0+0
= 0.
9. The most prominent feature of the hydrogen spectrum in the visible region is the red
Balmer line, coming from the transition n = 3 to n = 2. First of all, determine the
wavelength and frequency of this line, according to the Bohr theory. Fine structure
splits this line into several closely spaced lines; the question is: How many, and what
is their spacing? Hint: First determine how many sublevels the n = 2 splits into,
and find Efs1 for each of these, in eV. Then do the same for n = 3. Draw an energy
level diagram showing all possible transitions from n = 3 to n = 2. The energy
released (in the form of a photon) is E3 − E2 +∆E, the first part being common to
all of them, and the ∆E (due to fine structure) varying from one transition to the
next. Fine ∆E (in eV) for each transition. Finally, convert to photon frequency,
and determine the spacing between adjacent spectral lines (in Hz)−not the frequency
interval between each line and the unperturbed line (which is, of course, unobservable),
but the frequency interval between each line and the next one.
Solution
The wavelength of red Balmer line, coming from n = 3 to n = 2 transitions, is given
Due date: November 23, 2011
9
Atomic, molecular and laser physics.
Solution set 4
by,
(
)
1
1
1
= R∞ 2 − 2
λ
nf
ni
(
)
1
1
= (1.097 × 10 m ) 2 − 2
3
2
7
−1
= (1.097 × 10 m )(−5/36)
7
−1
λ = 660 nm.
The frequency is thus,
c = fλ
c
f =
λ
3 × 108 m/s
=
660 nm
= 4.5 × 1014 Hz.
For n = 2, l could have values 0 and 1, so j = 1/2 and 3/2. The n = 2 level splits into
the following sub levels, (which are also shown in the accompanying diagram)
|n, l, j⟩ = |2, 0, 1/2⟩
|n, l, j⟩ = |2, 1, 1/2⟩
|n, l, j⟩ = |2, 1, 3/2⟩.
According to fine-structure correction, the energy of each level is,
(
)
En2
4n
En,j =
3−
.
2mc2
j + 1/2
Since the energy depends on n and j, |2, 0, 1/2⟩ and 2, 1, 1/2⟩ will have the same
energy.
Similarly, for n = 3, l = 0, 1, 2 and the possible values of j could be 1/2, 3/2 or 5/2.
For n = 2, (E2 and E3 are the unperturbed Bohr energies,)
(
)
E22
E|2,0,1/2⟩ = E|2,1,1/2⟩ = E2 +
3 − 8 = E2 − 5.6 × 10−5 eV
2mc2
(
)
E22
8
E|2,1,3/2⟩ = E2 +
3−
= E2 − 1.13 × 10−5 eV.
2mc2
2
Due date: November 23, 2011
10
Atomic, molecular and laser physics.
3d5/2 = |3,2,5/2>
Solution set 4
∆Ε
5
3d3/2 = |3,2,3/2>
3p3/2 = |3,1,3/2>
3p1/2 = |3,1,1/2>
3s1/2 = |3,0,1/2>
∆Ε
3
∆Ε
4
∆Ε
2
1
1
∆Ε
∆Ε
3
∆Ε
2s1/2 = |2,0,1/2>
2p3/2 = |2,1,3/2>
2p1/2 = |2,1,1/2>
For n = 3,
E|3,0,1/2⟩ = E|3,1,1/2⟩
E|3,1,3/2⟩ = E|3,2,3/2⟩
E|3,2,5/2⟩
(
)
E32
= E3 +
3 − 12 = E3 − 2 × 10−5 eV
2
2mc
(
)
E32
8
= E3 +
3−
= E3 − 0.67 × 10−5 eV
2mc2
2
(
)
E32
12
= E3 +
3−
= E3 − 0.2 × 10−5 eV.
2mc2
2
Only those transitions are possible for which ∆j = 0, ±1 and ∆l = ±1. Therefore,
the possible transitions and corresponding energy differences are,
∆E1 = E|3,0,1/2⟩ − E|2,1,1/2⟩ = E|3,1,1/2⟩ − E|2,0,1/2⟩ = 3.6 × 10−5 eV + ∆
∆E2 = E|3,0,1/2⟩ − E|2,1,3/2⟩ = −8.7 × 10−6 eV + ∆
∆E3 = E|3,1,3/2⟩ − E|2,0,1/2⟩ = E|3,2,3/2⟩ − E|2,1,1/2⟩ = 4.9 × 10−5 eV + ∆
∆E4 = E|3,2,3/2⟩ − E|2,1,3/2⟩ = 4.6 × 10−6 eV + ∆
∆E5 = E|3,2,5/2⟩ − E|2,1,3/2⟩ = 9.3 × 10−6 eV + ∆,
where ∆ = E3 − E2 is the separation between unperturbed levels. Defining ∆f =
∆/(2π~), the frequencies of the emitted photons become,
Due date: November 23, 2011
11
Atomic, molecular and laser physics.
f1 =
f2 =
f3 =
f4 =
f5 =
Solution set 4
∆E1
2π~
∆E2
2π~
∆E3
2π~
∆E4
2π~
∆E5
2π~
= 8.7 × 109 Hz + ∆f
= 2.1 × 109 Hz + ∆f
= 1.19 × 1010 Hz + ∆f
= 1.11 × 109 Hz + ∆f
= 2.2 × 109 Hz + ∆f .
Finally, the frequency spacing is,
|f5 − f4 | = 2.2 × 109 Hz
|f4 − f3 | = 1.1 × 109 Hz
|f3 − f2 | = 9.8 × 109 Hz
|f2 − f1 | = 6.6 × 109 Hz.
Due date: November 23, 2011
12