Solutions to exercisesDiscussion questions E13.1b 1 The principal quantum number,n, determines the energy of a hydrogenic atomic orbital through eqn 13.13.. 2 The azimuthal quantum numbe
Trang 1Solutions to exercises
Discussion questions
E13.1(b) (1) The principal quantum number,n, determines the energy of a hydrogenic atomic orbital through
eqn 13.13
(2) The azimuthal quantum number,l, determines the magnitude of the angular momentum of a
hydrogenic atomic orbital through the relation{l(l + 1)}1/2¯h.
(3) The magnetic quantum number,m l, determines the z-component of the angular momentum of a hydrogenic orbital through the relationm l h.¯
(4) The spin quantum number,s, determines the magnitude of the spin angular momentum through
the relation{s(s + 1)}1/2¯h For a hydrogenic atomic orbitals, s can only be1/2 (5) The spin quantum number, m s, determines the z-component of the spin angular momentum through the relationm s¯h For hydrogenic atomic orbitals, m scan only be±1/2.
E13.2(b) (a) A boundary surface for a hydrogenic orbital is drawn so as to contain most (say 90%) of the
probability density of an electron in that orbital Its shape varies from orbital to orbital because the electron density distribution is different for different orbitals
(b) The radial distribution function gives the probability that the electron will be found anywhere
within a shell of radiusr around the nucleus It gives a better picture of where the electron is
likely to be found with respect to the nucleus than the probability density which is the square of the wavefunction
E13.3(b) The first ionization energies increase markedly from Li to Be, decrease slightly from Be to B, again
increase markedly from B to N, again decrease slightly from N to O, and finally increase markedly from N to Ne The general trend is an overall increase ofI1with atomic number across the period That is to be expected since the principal quantum number (electron shell) of the outer electron remains the same, while its attraction to the nucleus increases The slight decrease from Be to B is
a reflection of the outer electron being in a higher energy subshell (largerl value) in B than in Be.
The slight decrease from N to O is due to the half-filled subshell effect; half-filled sub-shells have increased stability O has one electron outside of the half-filledp subshell and that electron must pair
with another resulting in strong electron–electron repulsions between them
E13.4(b) An electron has a magnetic moment and magnetic field due to its orbital angular momentum It also
has a magnetic moment and magnetic field due to its spin angular momentum There is an interaction energy between magnetic moments and magnetic fields That between the spin magnetic moment and the magnetic field generated by the orbital motion is called spin–orbit coupling The energy of interaction is proportional to the scalar product of the two vectors representing the spin and orbital angular momenta and hence depends upon the orientation of the two vectors See Fig 13.29 The total angular momentum of an electron in an atom is the vector sum of the orbital and spin angular momenta as illustrated in Fig 13.30 and expressed in eqn 13.46 The spin–orbit coupling results in
a splitting of the energy levels associated with atomic terms as shown in Figs 13.31 and 13.32 This splitting shows up in atomic spectra as a fine structure as illustrated in Fig 13.32
Trang 2Numerical exercises
E13.5(b) The energy of the photon that struck the Xe atom goes into liberating the bound electron and giving
it any kinetic energy it now possesses
Ephoton= I + Ekinetic I = ionization energy
The energy of a photon is related to its frequency and wavelength
Ephoton= hν = hc
λ
and the kinetic energy of an electron is related to its mass and speed
Ekinetic= 1
2mes2
Sohc
λ = I +12mes2⇒ I = hc
λ −12mes2
I = (6.626 × 10−34J s) × (2.998 × 108m s−1)
58.4 × 10−9m
−1
2(9.11 × 10−31kg) × (1.79 × 106
m s−1)2
= 1.94 × 10−18J = 12.1 eV
E13.6(b) The radial wavefunction is [Table 13.1]
R3,0 = A6− 2ρ +1
9ρ2
e−ρ/6whereρ ≡ 2Zr
a0 , andA is a collection of constants Differentiating
with respect toρ yields
dR3,0
dρ = 0 = A(6 − 2ρ +19ρ2) ×−1
6
e−ρ/6+−2 +2
9ρAe −ρ/6
= Ae −ρ/6−ρ542 +5
9ρ − 3
This is a quadratic equation
0= aρ2+ bρ + c where a = −1
54, b = 5
9, and c = −3.
The solution is
ρ = −b ± (b2− 4ac)1/2
√ 7
sor =
15
2 ±3(71/2 )
2
a0
Z .
Numerically, this works out toρ = 7.65 and 2.35, so r = 11.5a0/Z and 3.53a0/Z Substituting
Z = 1 and a0= 5.292 × 10−11m,r = 607 pm and 187 pm.
The other maximum in the wavefunction is at r = 0 It is a physical maximum, but not a calculus
maximum: the first derivative of the wavefunction does not vanish there, so it cannot be found by differentiation
Trang 3E13.7(b) The radial wavefunction is [Table 13.1]
R3,1 = A4−1
3ρρe −ρ/6 where ρ = 2a Zr
0 The radial nodes occur where the radial wavefunction vanishes This occurs at
and when
4−1
3ρ= 0, or ρ
3 = 4, or ρ = 12
thenr = ρa0
2Z =
ρa0
2 =12a0
2 = 6a0 = 3.18 × 10−10m
E13.8(b) Normalization requires
|ψ|2
dτ = 1 =
∞ 0
π 0
2π 0 [N(2 − r/a0)e −r/2a0]2dφ sin θ dθ r2
dr
1= N2
∞ 0
e−r/a0(2 − r/a0)2r2dr
π 0 sinθ dθ
2π 0
dφ
Integrating over angles yields
1 = 4πN2
∞ 0
e−r/a0(2 − r/a0)2r2dr
= 4πN2
∞ 0
e−r/a0(4 − 4r/a0+ r2/a2
0)r2dr = 4πN2(8a3
0)
In the last step, we used
∞ 0
e−r/k r2dr = 2k3,
∞ 0
e−r/k r3dr = 6k4, and
∞ 0
e−r/k r4dr = 24k5
4
2πa3 0
E13.9(b) The average kinetic energy is
K =
ψ∗ˆEKψ dτ
whereψ = N(2 − ρ)e −ρ/2withN =1
4
Z3
2πa3 0
1/2 andρ ≡ Zr a
0 here
ˆEK= −¯h2
2m∇2 dτ = r2sinθ dr dθ dφ =
a3
0ρ2sinθ dρ dθ dφ
Z3
In spherical polar coordinates, three of the derivatives in∇2are derivatives with respect to angles, so those parts of∇2ψ vanish Thus
∇2ψ = ∂2ψ
∂r2 +2
r
∂ψ
∂2ψ
∂ρ2
∂ρ
∂r2
2 + 2Z
ρa0
∂ψ
∂ρ
∂ρ
∂r =
Z
a0
2
×
∂2ψ
∂ρ2 + 2
ρ
∂ψ
∂ρ
Trang 4
∂r = N(2 − ρ) ×
−1 2
e−ρ/2 − N e −ρ/2 = N1
2ρ − 2e−ρ/2
∂2ψ
∂ρ2 = N1
2ρ − 2×−1
2
e−ρ/2+1
2N e −ρ/2 = N3
2−1
4ρe−ρ/2
∇2ψ =
Z
a0
2
Ne −ρ/2 (−4/ρ + 5/2 − ρ/4)
and
K =
∞ 0
π 0
2π 0
N(2 − ρ)e −ρ/2
Z
a0
2
×
−¯h2
2m
× Ne −ρ/2 (−4/ρ + 5/2 − ρ/4) a
3
0dφ sin θ dθ ρ2dρ
Z3 The integrals over angles give a factor of 4π, so
K = 4πN2a0
Z
×
−h¯2
2m
∞
0 (2 − ρ) ×−4 +5
2ρ −1
4ρ2
ρe −ρdρ
The integral in this last expression works out to−2, using
∞ 0
e−ρ ρ ndρ = n! for n = 1, 2, and 3 So
K = 4π
Z3
32πa3 0
×a0
Z
×
¯
h2
m
= ¯h2Z2
8ma2 0 The average potential energy is
ψ∗V ψ dτ where V = − Ze2
4πε0r = −
Z2e2
4πε0a0ρ
and
∞ 0
π 0
2π 0
N(2 − ρ)e −ρ/2
− Z2e2
4πε0a0ρ
N(2 − ρ)e −ρ/2 a30ρ2sinθ dρ dθ dφ
Z3 The integrals over angles give a factor of 4π, so
2
− Z2e2
4πε0a0
×
a3 0
Z3
∞
0 (2 − ρ)2ρe −ρdρ
The integral in this last expression works out to 2, using
∞ 0
e−ρ ρ ndρ = n! for n = 1, 2, 3, and 4 So
Z3
32πa3 0
×
− Z2e2
4πε0a0
×
a3 0
Z3
× (2) = − Z2e2
16πε0a0
E13.10(b) The radial distribution function is defined as
P = 4πr2ψ2 so P3s = 4πr2(Y0,0 R3,0 )2,
P3s = 4πr2
1
4π
×
1 243
×
Z
a0
3
× (6 − 6ρ + ρ2)2e−ρ
whereρ ≡ 2Zr
na0 = 2Zr
3a0 here.
Trang 5But we want to find the most likely radius, so it would help to simplify the function by expressing it
in terms either ofr or ρ, but not both To find the most likely radius, we could set the derivative of
P3sequal to zero; therefore, we can collect all multiplicative constants together (including the factors
ofa0/Z needed to turn the initial r2intoρ2) since they will eventually be divided into zero
P3s = C2ρ2(6 − 6ρ + ρ2)2e−ρ
Note that not all the extrema ofP are maxima; some are minima But all the extrema of (P3s )1/2
correspond to maxima ofP3s So let us find the extrema of(P3s )1/2
d(P3s )1/2
d
dρ Cρ(6 − 6ρ + ρ2)e −ρ/2
= C[ρ(6 − 6ρ + ρ2) × (−1
2) + (6 − 12ρ + 3ρ2)]e −ρ/2
0= C(6 − 15ρ + 6ρ2−1
2ρ3)e −ρ/2 so 12− 30ρ + 12ρ2− ρ3= 0 Numerical solution of this cubic equation yields
ρ = 0.49, 2.79, and 8.72
corresponding to
r = 0.74a0/Z, 4.19a0/Z, and 13.08a0/Z
Comment If numerical methods are to be used to locate the roots of the equation which locates
the extrema, then graphical/numerical methods might as well be used to locate the maxima directly That is, the student may simply have a spreadsheet compute P3s and examine or manipulate the spreadsheet to locate the maxima
E13.11(b) Orbital angular momentum is
21/2 = ¯h(l(l + 1))1/2
There arel angular nodes and n − l − 1 radial nodes
E13.12(b) Forl > 0, j = l ± 1/2, so
E13.13(b) Use the Clebsch–Gordan series in the form
J = j1+ j2, j1+ j2− 1, , |j1− j2|
Then, withj1= 5 and j2= 3
J = 8, 7, 6, 5, 4, 3, 2
Trang 6E13.14(b) The degeneracyg of a hydrogenic atom with principal quantum number n is g = n2 The energyE
of hydrogenic atoms is
E = − hcZ2RH
n2 = −hcZ g2RH
so the degeneracy is
g = − hcZ E2RH
(a) g = − hc(2)2RH
−4hcRH = 1
(b) g = − hc(4)2RH
−1
4hcRH
= 64
(c) g = − hc(5)2RH
−hcRH = 25
E13.15(b) The letter F indicates that the total orbital angular momentum quantum numberL is 3; the superscript
3 is the multiplicity of the term, 2S + 1, related to the spin quantum number S = 1; and the subscript
4 indicates the total angular momentum quantum numberJ
E13.16(b) The radial distribution function varies as
P = 4πr2ψ2= 4
a3 0
r2e−2r/a0
The maximum value ofP occurs at r = a0since
dP
dr ∝
2r −2r2
a0
e−2r/a0 = 0 at r = a0 and Pmax= 4
a0
e−2
P falls to a fraction f of its maximum given by
f =
4r2
a3 e−2r/a0
4
a0e−2 = r2
a2 0
e2e−2r/a0
and hence we must solve forr in
f1/2
a0
e−r/a0
0.260 = r
a0
e−r/a0 solves tor = 2.08a0= 110 pm and to r = 0.380a0= 20.1 pm
0.319 = a r
0
e−r/a0 solves tor = 1.63a0= 86 pm and to r = 0.555a0= 29.4 pm
In each case the equation is solved numerically (or graphically) with readily available personal computer software The solutions above are easily checked by substitution into the equation forf
The radial distribution function is readily plotted and is shown in Fig 13.1
Trang 70.10
0.05
0.00
Figure 13.1 E13.17(b) (a) 5d → 2s is not an allowed transition, for l = −2 (.l must equal ±1).
(b) 5p → 3s is allowed , since l = −1.
(c) 5p → 3f is not allowed, for l = +2 (.l must equal ±1).
E13.18(b) For eachl, there are 2l + 1 values of m l and hence 2l + 1 orbitals—each of which can be occupied
by two electrons, so maximum occupancy is 2(2l + 1)
(a) 2s: l = 0; maximum occupancy = 2
(b) 4d: l = 2; maximum occupancy = 10
(c) 6f : l = 3; maximum occupancy = 14
(d) 6h: l = 5; maximum occupancy = 22
E13.19(b) V2+: 1s22s22p63s23p63d3= [Ar]3d3
The only unpaired electrons are those in the 3d subshell There are three S = 3
2 and32− 1 = 1
2 ForS =3
2, M S = ±1
2 and±3 2 forS =1
2, M S = ±1
2
E13.20(b) (a) Possible values ofS for four electrons in different orbitals are 2, 1, and 0 ; the multiplicity is
2S + 1, so multiplicities are 5, 3, and 1 respectively.
(b) Possible values ofS for five electrons in different orbitals are 5/2, 3/2, and 1/2 ; the multiplicity
is 2S + 1, so multiplicities are 6, 4, and 2 respectively.
E13.21(b) The coupling of ap electron (l = 1) and a d electron (l = 2) gives rise to L = 3 (F), 2 (D), and 1 (P)
terms Possible values ofS include 0 and 1 Possible values of J (using Russell–Saunders coupling)
are 3, 2, and 1 (S = 0) and 4, 3, 2, 1, and 0 (S = 1) The term symbols are
1F3;3F4,3F3,3F2;1D2;3D3,3D2,3D1;1P1;3P2,3P1,3P0 .
Hund’s rules state that the lowest energy level has maximum multiplicity Consideration of spin–orbit coupling says the lowest energy level has the lowest value ofJ (J + 1) − L(L + 1) − S(S + 1) So
the lowest energy level is 3F2
Trang 8E13.22(b) (a) 3D hasS = 1 and L = 2, so J = 3, 2, and 1 are present J = 3 has 7 states, with M J = 0,
±1, ±2, or ±3; J = 2 has 5 states, with M J = 0, ±1, or ±2; J = 1 has 3 states, with
M J = 0, or ±1
(b) 4D hasS = 3/2 and L = 2, so J = 7/2, 5/2, 3/2, and 1/2 are present J = 7/2 has 8
possible states, withM J = ±7/2, ±5/2, ±3/2 or ±1/2; J = 5/2 has 6 possible states, with
M J = ±5/2 ±3/2 or ±1/2; J = 3/2 has 4 possible states, with M J = ±3/2 or ±1/2;
J = 1/2 has 2 possible states, with M J = ±1/2.
(c) 2G hasS = 1/2 and L = 4, so J = 9/2 and 7/2 are present J = 9/2 had 10 possible
states, withM J = ±9/2, ±7/2, ±5/2, ±3/2 or ±1/2; J = 7/2 has 8 possible states, with
M J = ±7/2, ±5/2, ±3/2 or ±1/2.
E13.23(b) Closed shells and subshells do not contribute to eitherL or S and thus are ignored in what follows.
(a) Sc[Ar]3d14s2:S =1
2,L = 2; J = 5
2,3
2, so the terms are 2D5/2and2D3/2
(b) Br[Ar]3d104s24p5 We treat the missing electron in the 4p subshell as equivalent to a single
“electron” with l = 1, s = 1
2 Hence L = 1, S = 1
2, and J = 3
2,1
2, so the terms are
2P3/2and2P1/2
Solutions to problems
Solutions to numerical problems
P13.2 All lines in the hydrogen spectrum fit the Rydberg formula
1
λ = RH
1
n2 1
− 1
n2 2
13.1, with ˜ν = λ1 RH= 109 677 cm−1 Findn1from the value ofλmax, which arises from the transitionn1+ 1 → n1
1
λmaxRH = 1
n2 1
(n1+ 1)2 = 2n1+ 1
n2
1(n1+ 1)2
λmaxRH=n21(n1+ 1)2
2n1+ 1 = (656.46 × 10−9m) × (109 677 × 102m−1) = 7.20 and hencen1= 2, as determined by trial and error substitution Therefore, the transitions are given by
˜ν = λ1 = (109 677 cm−1) ×
1
4 − 1
n2 2
, n2= 3, 4, 5, 6
The next line hasn2= 7, and occurs at
˜ν = λ1 = (109 677 cm−1) ×
1
4 − 1 49
= 397.13 nm The energy required to ionize the atom is obtained by lettingn2→ ∞ Then
˜ν∞= 1
λ∞ = (109 677 cm−1) ×
1
4 − 0
= 27 419 cm−1, or 3.40 eV
Trang 9(The answer, 3.40 eV, is the ionization energy of an H atom that is already in an excited state, with
n = 2.)
Comment The series withn1= 2 is the Balmer series
P13.4 The lowest possible value ofn in 1s2nd1is 3; thus the series of2D terms correspond to 1s23d, 1s24d,
etc Figure 13.2 is a description consistent with the data in the problem statement
I
413 nm 460 nm 610 nm
Figure 13.2
If we assume that the energies of thed orbitals are hydrogenic we may write
E(1s2nd1,2D) = − hcR
n2 [n = 3, 4, 5, ]
Then for the2D→2
P transitions
˜ν = 1λ = |E(1s2hc2p1,2P)|−R
n2
.E = hν = hc λ = hc˜ν, ˜ν = .E hc
from which we can write
|E(1s22p1,2P)|
1
R
n2 =
1
610.36 × 10−7cm +R
1
460.29 × 10−7cm +R
1
413.23 × 10−7cm +R
Then
(b) − (a) solves to R= 109 886 cm−1
(a) − (c) solves to R= 109 910 cm−1
(b) − (c) solves to R= 109 963 cm−1
Mean= 109 920 cm−1 The binding energies are therefore
E(1s23d1,2D) = R
9 = −12 213 cm−1
E(1s2
2p,2
610.36 × 10−7cm− 12 213 cm−1= −28 597 cm−1
E(1s22s1,2S) = − 1
670.78 × 10−7cm − 28 597 cm−1= −43 505 cm−1
Trang 10Therefore, the ionization energy is
I (1s22s1,2S) = 43 505 cm−1, or 5.39 eV
P13.5 The 7p configuration has just one electron outside a closed subshell That electron has l = 1, s = 1/2,
andj = 1/2 or 3/2, so the atom has L = 1, S = 1/2, and J = 1/2 or 3/2 The term symbols are
2
P1/2and2P3/2, of which the former has the lower energy The 6d configuration also has just one
electron outside a closed subshell; that electron hasl = 2, s = 1/2, and j = 3/2 or 5/2, so the atom
hasL = 2, S = 1/2, and J = 3/2 or 5/2 The term symbols are 2D3/2and2D5/2, of which the former has the lower energy According to the simple treatment of spin–orbit coupling, the energy is given by
E l,s,j =1
2hcA[j (j + 1) − l(l + 1) − s(s + 1)]
whereA is the spin–orbit coupling constant So
E(2P1/2 ) = 1
2hcA[1
2(1/2 + 1) − 1(1 + 1) −1
2(1/2 + 1)] = −hcA
andE(2D3/2 ) = 1
2hcA[3
2(3/2 + 1) − 2(2 + 1) −1
2(1/2 + 1)] = −3
2hcA
This approach would predict the ground state to be 2D3/2
Comment The computational study cited above finds the2P1/2level to be lowest, but the authors caution that the error of similar calculations on Y and Lu is comparable to the computed difference between levels
P13.7 RH= kµH, RD= kµD, R = kµ [18]
whereR corresponds to an infinitely heavy nucleus, with µ = me
Sinceµ = memN
me+ mN
[N= p or d]
RH= kµH= kme
1+me
mp
1+me
mp
Likewise,RD= R
1+me
md
wherempis the mass of the proton andmdthe mass of the deuteron The two lines in question lie at
1
λH = RH
1−1 4
= 3
4RH
1
λD = RD
1−1 4
= 3
4RD and hence
RH
RD =λ λD
H = ˜ν ˜νH D Then, since
RH
RD =1+
me
md
1+me
mp
1+me
mp
RH
RD − 1