CHAPTER 13 The Structure of Atoms - Dartmouth Collegechem81/pdfs/13.pdf · 299 CHAPTER 13 The Structure of Atoms SECTION 13.1 13.1 We m ight expect the configurations of the atoms

299

CHAPTER 13

The Structure of Atoms

SECTION 13.1

13.1 We might expect the configurations of the atoms from Sc–Zn to follow a regularpattern of filling the 3d subshell that characterizes this first row of transitionmetals: from 1s22s22p63s23p64s23d1 for Sc to 1s22s22p63s23p64s23d10 forZn. Instead, one finds that the trend is followed from Sc through V, but Cr,which one might expect to be 1s22s22p63s23p64s23d4, is found instead to be1s22s22p63s23p64s13d5. Likewise, Cu is not 1s22s22p63s23p64s23d9, but1s22s22p63s23p64s13d10 instead. These two anomalies are explained byinvoking the extra stability of filled and half-filled subshells. Chromium lowersits total energy by moving an expected 4s electron into the 3d subshell, givingeach a half-filled configuration. Similarly, Cu moves a 4s electron into 3d tomake this subshell full with 10 electrons. Our task in this chapter is to under-stand why these changes lead to extra stability.

13.2 Elements that comprise the first “g block” (l = 4) would number 2(2l + 1) in all,or, for l = 4, 18 in all. The factor of 2 accounts for the two electrons that can bespin paired into any one orbital, and 2l + 1 is the number of orbitals (m values)that are associated with any l value.

SECTION 13.2

13.3 If y = x1 + x2 – x3, the results of applying P12, P13, and P23 to y are

P12 y = x2 + x1 – x3 = (+1)y

P13 y = x3 + x2 – x1 ≠ (±1)y

P23 y = x1 + x3 – x2 ≠ (±1)y .

Only the P12 operator gives an eigenvalue relation, with P12 y = (+1)y.

300

13.4 If y1(x1) = sin x1 and y2(x2) = cos x2, we can construct a function f+(x1, x2)that is symmetric with respect to exchange (i.e., that has an eigenvalue of +1with respect to the P12 operator):

f+(x1, x2) = y1(x1)y2(x2) + y1(x2)y2(x1) = sin x1 cos x2 + sin x2 cos x1

P12 f+(x1, x2) = sin x2 cos x1 + sin x1 cos x2 = +f+(x1, x2) .

Similarly, an antisymmetric function f–(x1, x2) is

f–(x1, x2) = y1(x1)y2(x2) – y1(x2)y2(x1) = sin x1 cos x2 – sin x2 cos x1

P12 f–(x1, x2) = sin x2 cos x1 – sin x1 cos x2 = –f–(x1, x2) .

13.5 If the two particles are identical spinless bosons that do not interact, the totalwavefunction has no spin function factors and it must be symmetric withrespect to interchange of the two particle’s coordinates. These requirements aremet by the function Ψ+ of Eq. (13.4). For two identical particles in a box oflength L, this equation is

Ψ+(x1, x2) = 1

2 ψ1(x1)ψ2(x2) + ψ1(x2)ψ2(x1)

= 2

2L sin

πx1

L sin

2πx2

L + sin

πx2

L sin

2πx1

L .

Note that the technique used to arrive at a symmetric total function composed oftwo individual particle functions is identical to that used in the previous problem(with the addition here of the normalization factor 1/ 2). If the particles areidentical spin 1/2 fermions, the total wavefunction must be antisymmetric withrespect to coordinate interchange. The Ψ– function of Eq. (13.4) satisfies thisrequirement, but it is incomplete in that it does not include the spin wavefunc-tions of the two particles. If we multiply this function by a spin wavefunctionthat is symmetric with respect to exchange, we still preserve the antisymmetriccharacter of the entire wavefunction. There are three ways we can do this:

Ψ1(x1, x2) = 1

2 ψ1(x1)ψ2(x2) – ψ1(x2)ψ2(x1) α(1)α(2)

Ψ2(x1, x2) = 1

2 ψ1(x1)ψ2(x2) – ψ1(x2)ψ2(x1) β(1)β(2)

Ψ3(x1, x2) = 1

2 ψ1(x1)ψ2(x2) – ψ1(x2)ψ2(x1)

1

2 α(1)β(2) + α(2)β(1) .

301

These three functions include the three spin function combinations we can writethat are symmetric, but there is a fourth acceptable function: one with a sym-metric spatial part multiplying an antisymmetric spin part:

Ψ4(x1, x2) = 1

2 ψ1(x1)ψ2(x2) + ψ1(x2)ψ2(x1)

1

2 α(1)β(2) – α(2)β(1) .

These four functions are quite general, and we will encounter them again asEquations (13.23a)–(13.23d) where they are used to describe an excited state ofhelium.

13.6 The symmetric (Ψ+) and antisymmetric (Ψ–) total spatial wavefunctions can bewritten as in Problem 13.5:

Ψ±(φ 1, φ 2) = 1

2 ψ1(φ 1)ψ2(φ 2) ± ψ1(φ 2)ψ2(φ 1)

= 1

2

12π

eiφ1 e–iφ2 ± eiφ2 e–iφ1

= 1

2

12π

ei(φ1 – φ2) ± e–i(φ1 – φ2)

= 1

2

12π

ei∆φ ± e–i∆φ .

Using the identity e±i∆φ = cos ∆φ ± i sin ∆φ, we can write

Ψ+(φ 1, φ 2) = Ψ+(∆φ ) = 2

2π cos ∆φ and Ψ–(φ 1, φ 2) = Ψ–(∆φ ) =

i 22π

sin ∆φ .

The first is symmetric, since particle interchange turns ∆φ into –∆φ and cos ∆φ= cos(–∆φ), and the second is antisymmetric because sin ∆φ = –sin(–∆φ). If theparticles are bosons, Ψ+ is the correct wavefunction (times a symmetric spinpart if the bosons are not spinless). If they are fermions, we can either multiplyΨ+ by an antisymmetric spin wavefunction, as in Ψ4 of the previous problem,or we can multiply Ψ– by a symmetric spin wavefunction, as in Ψ1–Ψ3 in theprevious problem. If both particles have quantum number m = 0, the spatialwavefunction for either one is simply ψ0(φ) = 1/ 2π; there is no particle coor-dinate dependence. All spatial functions we construct from this ψ0 must besymmetric with respect to coordinate interchange, because there are no spatialcoordinates!

302

13.7 Since the ground state energy of a one-electron atom varies like Z2E1 where Z isthe nuclear charge and E1 is the H atom ground state energy, the independentelectron energy for a two-electron atom is just 2Z2E1. The electron–electronrepulsion term that is neglected in the independent electron model is juste2/(4πε0)r12 where r12 is the electron–electron separation. It is reasonable toexpect this distance to be comparable to the atomic radius, which must be on theorder of that for hydrogen. The general expression for <r> on page 431 in thetext reduces to <r> = 3a0/2Z for the ground state of a one-electron atom ofnuclear charge Z. If we take r12 ≅ 3a0/2Z, we can expect the neglected term toraise the independent electron energy by

e2

(4πε0)r12

= 2Ze2

(4πε0)3a0

= 4Z3

e2

2(4πε0)a0

= – 4ZE1

3 .

The simpler estimate of the magnitude of this effect given for He (Z = 2) onpage 456 of the text, –2E1 = 27.2 eV, compares reasonably with this estimate,which is –8E1/3 = 36.3 eV. Adding 36.3 eV to the independent electron energyfor He, –108.78 eV, gives –72.5 eV, in reasonable agreement with the experi-mental –79.0 eV.

SECTION 13.3

13.8 Our approach here is to substitute our trial wavefunction φ(x) = N exp(–βx2)and the harmonic oscillator Hamiltonian H = –(¨ 2

/2µ)(d 2/dx2) + kx2/2 intoEq. (13.9), the variation expression for the total energy <E>, and to find thatvalue of the variation parameter β that minimizes <E>. We start with

<E> =

φHφ dx–∞

∞

φ2 dx

–∞

∞ =

N 2 e–βx2 – ¨ 2

2µ d2

dx2 + kx2

2 e–βx2 dx

–∞

∞

N 2 e–2βx2 dx–∞

∞ .

We evaluate this expression integral by integral (and note that the normalizationconstant N cancels in the expression for <E>). The integral in the denominatoris one we have seen before, in both Example 12.4 and in Problem 11.22. It isof the form

–∞∞

e–αx 2 dx = π/α with α = 2β so that

303

e–2βx2 dx

–∞

∞ =

π2β

.

Next, we operate on the trial wavefunction with the Hamiltonian:

– ¨ 2

2µ d2

dx2 + kx2

2 e–βx2 = –

¨ 2

2µ 4β2

x2 – 2β e–βx2 + k2

x2 e–βx2 .

We see that the numerator in the expression for <E> is a sum of two integrals:

e–βx2 – ¨ 2

2µ d2

dx2 + kx2

2 e–βx2 dx

–∞

∞

= k2

– 4β2¨ 2

2µ x 2 e–2βx2 dx

–∞

∞ +

2β¨ 2

2µ e–2βx2 dx

–∞

∞ .

The first integral is also known: –∞∞

x 2e–αx 2 dx = α–3/2 π /2, so that we have

e–βx2 – ¨ 2

2µ d2

dx2 + kx2

2 e–βx2 dx

–∞

∞

= k2

– 4β2¨ 2

2µ

π

4 2β3/2 +

βπ¨ 2

2µ

= π

4β2¨ 2

µ + k

8 2β3/2 .

We divide this expression by the denominator integral and write <E> as

<E> = π

4β2¨ 2

µ + k

π/2β 8 2β3/2 =

β¨ 2

2µ + k

8β .

304

Next, we minimize <E> with respect to β:

d<E>dβ

= ¨ 2

2µ – k

8β2 = 0 or β = 1

2

kµ

¨ 2 .

As shown on page 404 in the text, our best variational wavefunction is the exactground-state wavefunction:

φ(x) = N e–βx2 = N exp – 12

kµ

¨ x2 .

13.9 For the trial wavefunction φ(x) = N sin2 (πx/L), the variational energy expres-sion is

<E> =

N 2 sin2 πxL

– ¨ 2

2m d2

dx2 sin2

πxL

dx

0

L

N 2 sin4 πxL

dx0

L .

As is always the case, the normalization constant cancels from the expressionfor <E>. The integral in the denominator is given in the problem and equals3L/8. The integral in the numerator is

sin2 πxL

– ¨ 2

2m d2

dx2 sin2 πx

Ldx

0

L

= – §¨ 2

π2

mL2 sin2 πx

L cos2 πx

L – sin2 πx

Ldx

0

L

= – ¨ 2

π2

mL2 sin2 πx

L – 2sin4 πx

Ldx

0

L

= – ¨ 2

π2

mL2 L

2 – 2 3L

8 =

¨ 2π2

4mL

where we have used the identity cos2 (πx/L) = 1 – sin2 (πx/L) to go from thesecond to third step and the integrals in the problem to go from the third to thefourth. Dividing this by the denominator, 3L/8, gives <E> = 2π2¨2/3mL2.The exact energy is π2¨2/2mL2, which is three-fourths the variation energy.

305

13.10 We can follow the logic of Problem 13. 8 here with only a change in the poten-tial energy portion of the Hamiltonian from kx2/2 to bx4 where b is a constant.We will also need the definite integral

–∞∞

x 4e–αx 2 dx = 3α–5/2 π /4. We

write

<E> =

φHφ dx–∞

∞

φ2 dx

–∞

∞ =

N 2 e–αx 2 – ¨ 2

2µ d2

dx2 + bx4 e–αx 2 dx

–∞

∞

N 2 e–2αx 2 dx–∞

∞ .

The integral in the denominator equals (from Problem 13.8) π/2α , and the in-tegral in the numerator is

e–αx 2 – ¨ 2

2µ d2

dx2 + bx4 e–αx 2 dx

–∞

∞

= α¨ 2

µ e–2αx 2 dx

–∞

∞ –

α2¨ 2

2µx 2e–2αx 2 dx

–∞

∞ + b x 4e–2αx 2 dx

–∞

∞

= α¨ 2

µ

π2α

– α2¨ 2

2µ

π2(2α)3/2

+ b 3 π

4(2α)5/2 .

Thus, <E> is

<E> =

α¨ 2

µ

π2α

– α2¨ 2

2µ

π2(2α)3/2

+ b 3 π

4(2α)5/2

π2α

= 3b16α2

+ α¨ 2

2µ .

We minimize <E> with respect to our parameter α and find

ddα

3b16α2

+ α¨ 2

2µ = – 3b

8α3 +

¨ 2

2µ = 0 or α =

3bµ

4¨ 2

1/3 .

306

This gives the following variation theory estimate of the ground state energy:

<E> = 3b16α2

+ α¨ 2

2µ = 3b

16 3bµ

4¨ 2

2/3 +

3bµ

4¨ 2

1/3¨ 2

2µ = 3

4 ¨

3¨b

4µ2

1/3

.

13.11 The first-order energy correction is the expectation value of the perturbation cal-culated using the zeroth-order wavefunctions, Eq. (13.16):

En(1) = ψn

(0)* H′ ψn(0) dτ = Hnn

′ .

For potential (a), the perturbation is zero from x = 0 to L/2 and constant at V0from x = L/2 to L. The energy expression in this case is

En(1) = ψn

(0)* 0 ψn(0) dτ

0

L/2

+ ψn(0)* V0 ψn

(0) dτL/2

L

= V0 ψn(0) 2

dτL/2

L

= V0

2

where the final integral must equal 1/2, since it represents the probability thatthe unperturbed system is in the right-hand half of the box and this probabilityequals 1/2 by symmetry. For potential (b), the perturbation is zero from x = 0to L/4 and from x = 3L/4 to L. These regions contribute nothing to the energycorrection. The region from x = L/4 to 3L/4 represents the constant V0 pertur-bation and an energy correction equal to (using the unperturbed wavefunctionsfor the particle-in-a-box from Chapter 12, page 396 in the text)

En(1) = V0 ψn

(0) 2 dτ

L/4

3L/4

= V0 2L

sin2 nπx

L dτ

L/4

3L/4

.

The integral given in the problem lets us write this as

En(1) =

V0

2 +

V0

2nπ sin

nπ2

– sin 3nπ

2 .

307

(Note that we cannot jump to the conclusion that our integral here equals 1/2even though it covers half the box. The particle-in-a-box wavefunctions aresymmetric about the middle of the box, but the center half of them is not equiva-lent to the sum of the outer two quarters.) Note that as n → ∞, both potentials,(a) and (b), have energy corrections that equal V0/2. This is expected, since n→ ∞ recovers the classical limit in which the particle simply experiences, onaverage, an energy change of V0/2 no matter what state it is in and no matterwhere the potential step is placed. For finite n, however, the potential (b)expression takes on one of three values, depending on n. If we writesin(3nπ/2) = sin(nπ/2 + nπ), then the trigonometric identity sin(a + b) = cos asin b + sin a cos b lets us write

En(1) =

V0

2 +

V0

2nπ sin

nπ2

– sin nπ2

+ nπ

= V0

2 +

V0

2nπ sin

nπ2

– cos nπ2

sin nπ – cos nπ sin nπ2

= V0

2 +

V0

2nπ sin

nπ2

1 – cos nπ ,

since sin nπ = 0 if n is an integer, as it is here. The factor in square brackets inthe last expression has three distinct values: for n = any even integer, it is zero;for n = 1, 5, 9, …, it equals +2; for n = 3, 7, 11, …, it equals –2. Thus, wecan summarize the first-order energy correction for potential (b) as

En(0) =

V0

2 , n = 2, 4, 6, …

V0

2 +

V0

nπ , n = 1, 5, 9, …

V0

2 –

V0

nπ , n = 3, 7, 11, … .

Note that the n = even integer result is the same as for potential (a), since thewavefunctions for these states are symmetric about not only the middle of thebox but also about the steps in the potential at x = L/4 and 3L/4. The diagramon the next page shows the square of the unperturbed wavefunctions for n = 1,2, and 3, which are representative of each of these three perturbation energies.

308

ψ2

xL3L/4L/2L/40

V0V(x)

n = 1

n = 2

n = 3

13.12 Following Example 13.4 and exploiting the expressions in the text for the per-turbation Hamiltonian matrix elements of a quartic perturbation, we can write,with E0

(0) – Em(0) = –m¨ω,

ψ0(1) =

H0m′

E0(0) – Em

(0)∑

m ≠ 0

= H02

′

–2¨ω ψ2

(0) + H04

′

–4¨ω ψ4

(0) .

The matrix elements are given in the text on page 464. Substituting them intothis expression gives

ψ0(1) = 3b

–2 2¨ω ¨ω

k

2

ψ2(0) + 3b

–4 6¨ω ¨ω

k

2

ψ4(0)

= – 3b¨ω2 2k 2

ψ2(0) + 1

2 3 ψ4

(0) .

The general expressions in the summary to Chapter 12 show us how to writethe unperturbed ground-state wavefunction and the two unperturbed excitedstate wavefunctions used above:

309

ψ0(0) = k

π¨ω

1/4 e–kx2/2¨ω

ψ2(0) = k

π¨ω 122·2

1/2 4kx2

¨ω – 2 e–kx2/2¨ω

= 2 kx2

¨ω – 1

2 k

π¨ω

1/4 e–kx2/2¨ω

= 2 kx2

¨ω – 1

2 ψ0

(0)

ψ4(0) = k

π¨ω 124·24

1/2 16 k 2x4

¨ 2ω2

– 48 kx2

¨ω + 12 e–kx2/2¨ω

= 6 k 2x4

3¨ 2ω2

– k x2

¨ω + 1

4 k

π¨ω

1/4 e–kx2/2¨ω

= 6 k 2x4

3¨ 2ω2

– k x2

¨ω + 1

4 ψ0

(0) .

The first-order corrected ground-state wavefunction is thus, after some algebra,

ψ0 = ψ0(0) + ψ0

(1)

= ψ0(0) –

3b¨ω2 2k 2

ψ2(0) + 1

2 3 ψ4

(0)

= ψ0(0) 1 +

b¨ωk 2

916

– 13

kx2

¨ω – 1

4 k 2x4

¨ 2ω2

.

13.13 Using the notation of the problem and the second-order energy expressions forthe two-level system discussed on page 464 of the text, we can write

E0(2) =

H01′

2

–∆E(0) and E1

(2) = H01

′ 2

∆E(0) .

Thus, the perturbed energies E0 and E1 are

310

E0 = E0(0) + E0

(2) = E0(0) –

H01′

2

∆E(0) and E1 = E1

(0) + E1(2) = E1

(0) + H01

′ 2

∆E(0) ,

and the perturbed energy difference is

∆E = ∆E(0) + 2 H01

′ 2

∆E(0) = ∆E(0) 1 +

2 H01′

2

∆E(0) 2 .

If ∆E = 2∆E(0), this expression shows us that

2 = 1 + 2 H01

′ 2

∆E(0) 2 or H01

′ = ∆E(0)

2 .

13.14 As is discussed in more detail in Chapter 15, this problem is related to the sys-tem’s dipole moment. We will see there that the dipole moment of a chargedsystem (which is what we have here—the net charge is q) depends on where weplace the coordinate origin in our system. The first-order energy correction is

Hnn′ = En

(1) = –qF0 ψn(0)xψn

(0) dx = –qF0<x>n .

This is independent of n. If the coordinate system is placed so that the boxextends from x = 0 to x = L, as we did in Chapter 12, <x> = L/2 for all n, but ifthe coordinate system is placed so that the origin is at the middle of the box,<x> = 0 and the first-order correction vanishes! This seems to be a paradox:how can the energy of the system depend on the location of an abstract coordi-nate origin? The resolution to this dilemma (and it recurs in the next problem) isthe recognition that we can measure only energy differences. Any constantenergy added to all the energy levels of the system amounts to no more than ashift of the (arbitrary) definition of the zero of total energy.

13.15 The first-order energy correction is

Hnn′ = En

(1) = –qF0 ψn(0)xψn

(0) dx = –qF0<x>n ,

311

exactly as in the previous problem, but here, the only sensible choice for acoordinate origin is at the minimum of the harmonic potential well. With thischoice, <x> = 0 for all quantum numbers n due to the symmetry of the poten-tial. Since this result, En

(1) = 0, is a constant, any coordinate origin must alsogive En

(1) = a constant, and since only energy differences can be measured, thisterm in the perturbation is not observable. To see if higher order terms areobservable, we use the matrix element expressions given in the problem (andthat were also derived in Problem 12.13 in terms of the dimensionless harmonicoscillator coordinate q rather than the physical distance coordinate x used here—hence, Problem 12.13 does not include the factor 1/α) along with the generalsecond-order energy expression:

En(2) =

Hnm′ 2

En(0) – Em

(0)∑

n ≠ m

= q2F02

xn, n + 12

–¨ω +

xn, n – 12

¨ω

= q2F0

2

¨ωα2 – n + 1

2 + n

2 = –

q2F02

2k .

Since this term is also a constant for every state, the second-order energy cor-rection is also not observable.

13.16 Now we consider an exact solution to the system discussed in the previousproblem. We are using a simple mathematical trick—completing the square—torewrite the potential energy. We write, as suggested in the problem, the follow-ing potential energy function:

V(x) = kx2

2 – qF0x + E′ – E′ = a(x – x′)2 – E′ = ax2 – 2axx′ + ax′2 – E′ .

Equating coefficients of each power of x in the last expression to those in theoriginal expression for V(x) shows that

k2

= a , 2ax′ = kx′ = qF0 , x′ = qF0

k , E′ = ax′2 =

q2F02

2k .

These equalities let us write the potential energy as

312

V(x) = 12

k(x – x′)2 – q2F0

2

2k ,

which is just an ordinary harmonic oscillator (define y = x – x′ and recognizethe pure harmonic term ky2/2) with the zero of energy shifted down by theconstant amount q2F0

2/2k (which must be positive so that the shift is guaranteedto be down). Thus, the energy expression must be

E = ¨ω v + 12

– q2F0

2

2k

where v is the usual harmonic oscillator quantum number and ω = k/m wherem is the particle mass.

13.17 The unperturbed wavefunctions and energies for the particle-on-a-ring systemare derived in Example 12.6:

ψm(0) = e

imφ

2π , Em

(0) = m2¨ 2

2MR2 , m = 0, ±1, ±2, …

and the perturbation from the electric field is

H ′ = –qRF0 cos φ = –qRF0 eiφ + e–iφ

2 .

The general matrix element for this perturbation is

Hmn′ = –

qRF0

4π e–imφ eiφ + e–iφ einφ dφ

0

2π

= – qRF0

4πei(n + 1 – m)φ dφ

0

2π + ei(n – 1 – m)φ dφ

0

2π .

These integrals are both of the type 0

2πeikφ dφ where k is an integer. If k = 0,

the integral equals 2π, but if k is any other integer, the integral is zero:

313

eikφ dφ0

2π =

e2πik

ik –

1ik

= 0 ,

since e2πik = 1 for k any nonzero integer. Thus, Hmn′ = 0 unless n + 1 – m = 0

(so that the first integral in the final expression above for Hmn′ equals 2π) or n –

1 – m = 0 (so that the second integral equals 2π). Therefore

Hmn′ = –

qRF0

2 , n = m + 1 or n = m – 1

0 otherwise .

This result tells us immediately that the first-order perturbation theory correctionto the energy is zero for every state, since En

(1) = Hnn′ and Hnn

′ = 0 for all n.The second-order energy expression is given by the sum in Eq. (13.18), whichhere has only two terms due to the general expression for the perturbationmatrix elements we just found:

En(2) =

Hnm′ Hmn

′

En(0) – Em

(0)∑

m ≠ n

= Hn, n + 1

′ 2

En(0) – En + 1

(0) +

Hn, n – 1′ 2

En(0) – En – 1

(0) .

Substituting the expressions for the unperturbed energies and the perturbationmatrix elements into this equation gives

En(2) =

–qRF0/2 2

n2 – (n + 1)2 (¨ 2/2MR2)

+ –qRF0/2 2

n2 – (n – 1)2 (¨ 2/2MR2)

= (qRF0)2

4(¨ 2/2MR2)

1–2n – 1

+ 12n – 1

= q2R 4F0

2M

¨ 2 14n2 – 1

.

The total energy, correct through this second-order expression, is thus

En = En(0) + En

(2) = n2¨ 2

2MR2 +

q2R 4F02M

¨ 2 14n2 – 1

.

314

If we measure energy in multiples of ¨2/2MR2 and measure the strength of theperturbation through the dimensionless ratio ε = qRF0/(¨2/2MR2), we canwrite the total energy in dimensionless form as

En

¨ 2/2MR2

= n2 + ε2

2 4n2 – 1 .

This expression lets us see clearly that our perturbation term has the propertieswe expect: for the ground state, n = 0, this term is negative—the second-ordercorrection always lowers the ground state energy—but for all other states, theperturbation is positive and decreasing with increasing n. A comparison of theunperturbed and perturbed energies for the lowest few states is shown below ina diagram drawn to scale for the case ε = 1 (which is an unreasonably largevalue for a “perturbation”).

0

2

4

–1

1

3

5

WithoutPerturbation

WithPerturbation

n = 0

n = ±1

n = ±2

En/

(h2 /

2MR

2 )

SECTION 13.4

13.18 We start with the expression in the problem, written to expose the one-electronatom potential energy operator, the factor in parentheses below, written for aone-electron atom of nuclear charge Ze:

– (2 – Ze) e2

(4πε0)r = –

2 – Ze

Ze

Zee2

(4πε0)r =

2 – Ze

Ze –

Zee2

(4πε0)r .

The Virial Theorem for a one-electron atom tells us that <E> = <V>/2 where theaverage expressed by the angle brackets is taken over the ground state here.Performing this average over the expression above gives us

315

– (2 – Ze) e2

(4πε0)r =

2 – Ze

Ze –

Zee2

(4πε0)r =

2 – Ze

Ze <V> =

2 – Ze

Ze (2E1)

where E1 represents the ground-state energy for a one-electron atom of nuclearcharge Ze. Since

E1 = – Ze

2e2

2(4πε0)a0

from Eq. (12.46b), we have

– (2 – Ze) e2

(4πε0)r =

2 – Ze

Ze (2E1) = –

2 – Ze

Ze

Ze2e2

(4πε0)a0

= Ze2 – 2Ze

Ze2e2

(4πε0)a0

,

which is the expression we seek.

13.19 For H–, the real nuclear charge is Z = 1 rather than the He value Z = 2. Thischanges the expression for <E> from that on page 470 in the text just beforeEq. (13.20) into

<E> = e2

(4πε0)a0

Ze

2

2 – Ze

2 + Ze2 – Ze +

Ze2

2 – Ze

2 + Ze2 – Ze +

58

Ze

= e2

(4πε0)a0

Ze2 –

118

Ze = (27.196 eV) Ze2 –

118

Ze .

As in the text, we next minimize <E> with respect to Ze and find

d<E>dZe

= (27.196 eV) 2Ze – 118

= 0 or Ze = 1116

.

Thus, the variational energy is

<E> = (27.196 eV) 1116

2 –

118

1116

= –12.854 eV .

316

Note that this number is greater than the energy of H + e–, which is just theground-state H atom energy, –13.598 eV. The experimental H– energy is thenegative of the first and second ionization energies: –(13.598 eV + 0.7542 eV)= –14.352 eV. Not only does the variational calculation predict a significantlygreater energy for H– than is observed, it predicts an energy greater than that ofH + e–, which means it does not predict H– to be stable.

13.20 Figure 12.14c represents the true H atom 3s function with its two characteristicradial nodes. Slater orbitals do not have radial nodes. The diagrams belowcompare the true 3s wavefunction and radial distribution function to their Slaterapproximations.

10 20r/a0

0

ψ

300

3s

Slater

10 20r/a0300

Rad

ial D

istr

ibut

ion

Func

tion Slater

3s

13.21 If we consider Al, with Z = 13 and a 1s22s22p63s23p1 configuration, perfectscreening would predict that the 3p Ze would be 1 (the other 12 electrons screen12 of the nuclear proton charges perfectly), the 3s Ze would be 3 (10 otherelectrons screening perfectly), the 2p Ze would be 9 (4 other electrons screeningperfectly), the 2s Ze would be 11 (perfect screening from the two 1s electrons),

317

and the 1s Ze would be 13 (completely unscreened innermost orbital). Thesepredictions and the true values are compared in the table below.

1s 2s 2p 3s 3pPerfect screening 13 11 9 3 1Actual screening 12.59 8.21 8.96 4.12 4.07

Comparison shows that penetration is significantly different for s and p orbitals,and that 2p orbitals are almost perfectly screened (in this model), but 3p orbitalsare not.

13.22 Before we attack this problem head-on and quickly fill a page or two with alge-bra, it is helpful to write the Hamiltonian in a slightly simpler way that keeps usfrom writing too many fundamental constants again and again. We recognizethat

– ¨ 2

2µ = a0

2E1 and – e2

(4πε0) = 2a0E1 so that

H = E1 a02 1

r2 ddr

r2 ddr

+ 2a0r .

Here, E1 is the H-atom ground state energy and a0 is the Bohr radius. This letsus write the variation integral for <E> as

<E> = φ H φ dτ

φ2 dτ

=

E1 e–αr2

a02

r2 ddr

r2 ddr

+ 2a0

r e–αr2

r2 dr

0

∞

e–2αr2 r2 dr

0

∞ .

The integral in the denominator equals π/(2α)3/4, and the numerator is a sumof three integrals whose values are known:

318

E1 e–αr2

a02

r2 ddr

r2 ddr

+ 2a0

r e–αr2

r2 dr

0

∞

= 2E1a0 e–2αr2 r dr

0

∞ – 3a0α e–2αr2

r2 dr0

∞ + 2a0α2

e–2αr2 r4 dr

0

∞

= E1a0 1

2α –

3a0 π

4 2α +

3a0 2π

16 α .

Thus, <E> is, after some simplifying algebra,

<E> = E1a0 4 2α

π – 3a0α .

The best value for α is the one that minimizes this energy:

d<E>

dα = E1a0

2 2

πα – 3a0 = 0 or α =

8

9a02π

.

Substituting this into the energy expression gives the best energy:

<E> = E1a0

4 28

9a02π

π – 3a0

8

9a02π

= 8E1

3π = 0.8489E1 .

Note that, since E1 is less than zero, <E> here is greater than E1, as we shouldexpect from a variational calculation. If we plot on the same graph this bestGTO (with N equal to 16 3/9a0

3/2π, a value we can find from the normalizationintegral expression and the value of the integral in the denominator for <E>)along with the 1s STO, we find the figure shown at the top of the next page,which looks very different from the graph on page 473 in the text. That figurewas drawn with α chosen to represent the STO in a least-squares sense, and itsα value, 0.967/a0

2, predicts <E> = 0.2374E1, a much higher energy than thevariational result.

319

1 2r/a0

ψ

30

GTO

STO

13.23 For the 1s12p1 configuration of He, the electron spins are free to take on eithervalue, α or β, without violating the Pauli Principle (since they are inequivalentelectrons). In the abbreviated Slater determinant notation, the twelve possibili-ties are:

|| 1sα(1) 1pxα(2) || || 1sα(1) 1pyα(2) || || 1sα(1) 1pzα(2) ||

|| 1sβ(1) 1pxβ(2) || || 1sβ(1) 1pyβ(2) || || 1sβ(1) 1pzβ(2) ||

|| 1sα(1) 1pxβ(2) || || 1sα(1) 1pyβ(2) || || 1sα(1) 1pzβ(2) ||

|| 1sβ(1) 1pxα(2) || || 1sβ(1) 1pyα(2) || || 1sβ(1) 1pzα(2) ||

Note that the first three have both spins α, the next three have both β, the nexthave an αβ assignment, and the last three have a βα assignment. The fullSlater determinant for the last possibility, for example, is

|| 1sβ(1) 1pzα(2) || = 1

2

1sβ(1) 1pzα(1)

1sβ(2) 1pzα(2) .

The full Slater determinant for the other eleven possibilities follow in analogousfashion.

13.24 If we assume a 1s3 configuration for Li, then we have the following Slaterdeterminant:

1

6

1sα(1) 1sβ(1) 1sα(1)

1sα(2) 1sβ(2) 1sα(2)

1sα(3) 1sβ(3) 1sα(3)

320

where the final column could as easily have been written with β spin as with α.Note that either possible determinant has one column duplicated; here, the firstand third are the same. If we apply the general expansion of a 3 × 3 determi-nant given in the problem to one with two identical columns, we find that it isidentically zero:

a b ad e dg h g

= aeg + bdg + adh – aeg – bdg – adh = 0 .

13.25 Since φ1 = φ2 = 1s for the He ground state, the definitions of the J and K inte-grals (following Eq. (13.26) on page 477 of the text) show us that what wehave here is Jii = J11 = J = Kii = K11 = K.

SECTION 13.5

13.26 The spin-orbit energy expression, Eq. (13.31), is

Eso(1) =

¨ 2hso

2 j(j + 1) – l(l + 1) – s(s + 1) ,

and to find the spin-orbit coupling constant, hso, from energy differences(intervals) between terms split by spin-orbit interaction, we must calculate thisenergy for each level of the interval and subtract them to express the interval interms of hso. For a 2P3/2 level, s = 1/2, l = 1, and j = 3/2. For a 2P1/2 level, s= 1/2, l = 1, and j = 1/2. This gives us

Eso(1)( P3/2

2 ) = ¨ 2

hso2

32

52

– 2 – 12

32

= ¨ 2

hso2

Eso(1)( P1/2

2 ) = ¨ 2

hso2

12

32

– 2 – 12

32

= –¨ 2hso

so that the interval is ¨2hso/2 – (–¨2hso) = 3¨2hso/2. The halogen intervals inExample 13.5 are expressed in cm–1 units, and the units for hso in our expres-sions above are (energy)/(angular momentum)2 = J/(J s)2 = kg–1 m–2 in SIunits. Thus, we convert the interval energies from cm–1 to J units before divid-ing them by 3¨2/2. This gives us the table shown at the top of the next page,which reproduces the data in the text along with the results of our calculationshere.

321

Halogen F Cl Br I

Interval/cm–1 404 881 3685 7603

Interval/10–21 J 8.025 17.50 73.20 151.0

hso/1047 kg–1 m–2 4.811 10.49 43.88 90.54

The units we have here for hso are perfectly acceptable, but somewhat awk-ward. Often, spin-orbit coupling constants are expressed in cm–1 units directly(as if ¨ was not in the energy expression and with energy expressed in cm–1

units). With this convention, we would express hso simply as two-thirds theinterval. Finally, note that we have the correct magnitude of hso here, as theproblem requested, but in fact, hso is negative for the halogens, since their twolowest energy levels are said to be inverted. The higher J term, 2P3/2, is theground state. By convention, the “noninverted” energy level ordering is takento be in increasing order of J.

13.27 Table 13.2 tells us that the d2, d8 configurations have 1S, 1D, 1G, 3P, and 3Fterms. We assign possible J values to each as shown in the table below.

Term S L Possible J values Complete term symbols1S 0 0 0 1S01D 0 2 2 1D21G 0 4 4 1G43P 1 1 0, 1, 2 3P0, 3P1, 3P23F 1 3 2, 3, 4 3F2, 3F3, 3F4

The S values come from the relationship between the term symbol multiplicity(the pre-superscript) and S: multiplicity = 2S + 1; the L value is coded in thecentral letter in the usual way: S, P, D, F, G → L = 0, 1, 2, 3, 4; and the pos-sible J values range from L + S down to |L – S| in unit steps. According toHund’s rules, the triplets are lower in energy than the singlets; of the triplets,the term of greatest L is lowest (L = 3, or 3F), and of the 3F possibilities, theminimum J is lowest for the less-than-half-filled d2 configuration (3F2), but themaximum J is lowest for the greater-than-half-filled d8 configuration (3F4). Anexample of a 3F2 ground state is d2 Ti, and Ni is the example of a d8 3F4ground state.

13.28 The ground-state valence configuration of the alkaline earths is ns2. The firstexcited configuration is therefore ns1np1. For this configuration, the s electronhas l = 0 and the p electron has l = 1 so that L must equal 1, a P term. The

322

electron spins are independent and can pair in an S = 0 (singlet) or in an S = 1(triplet) fashion, leading to 1P and 3P terms. The 1P term can only have J = 1(a 1P1 term symbol), but the 3P can have J = 0, 1, or 2 for 3P0, 3P1, and 3P2term symbols. The triplet with minimum J is lowest, according to Hund’s rulesfor a less-than-half-filled configuration: 3P0 is the lowest energy level of thosederived from the ns1np1 configuration.

13.29 The elements half-way across any l block in the Periodic Table have half-filledsubshells: H–Fr are ns1, N–Bi are np3, Mn–Re are nd5, and Eu–Am are nf7.According to Hund’s rules, the maximum spin multiplicity state will be lowestin energy, and for half-filled subshells, the maximum multiplicity comes fromall spins aligned. This accounts for the increasing 2S + 1 multiplicities:

ns1 → S = 1/2 → doublet multiplicity

np3 → S = 3/2 → quartet multiplicity

nd5 → S = 5/2 → sextet multiplicity

nf7 → S = 7/2 → octet multiplicity.

Since all of these electrons have the same l value (s, p, d, or f), the only way toalign all these spins and obey the Pauli Principle is to assign each to a unique mlvalue. For example, the np3 configuration with all spins aligned must have ml= –1 for one electron, 0 for another, and +1 for the third. With this assign-ment, the total orbital angular momentum adds to L = 0, an S state.

13.30 The table suggested in the problem is shown below:

m = +1 ↑↓ ↑ ↑ ↑ ↑ ↓ ↓ ↓ ↓m = 0 ↑↓ ↓ ↑ ↓ ↑ ↑ ↑ ↓ ↓

m = –1 ↑↓ ↓ ↑ ↓ ↑ ↑ ↓ ↓ ↑mL 2 0 –2 1 1 0 0 1 1 0 0 –1 –1 –1 –1

mS 0 0 0 0 1 0 1 –1 0 –1 0 1 0 –1 0

Note that there are 15 columns in the table, and we interpret the mL and mSvalues in the last two rows as coordinates for nonzero entries in a Slater dia-gram. The entire diagram, given in the problem, is reproduced at the top of thenext page with the subdiagrams for each term symbol outlined by a rectangle orsquare.

323

0 1 1 1 0

2 3 2 11

1 1 1 00

–2 –1 0 +1 +2

+1

0

–1

mS

mL

1D

1S

3P

The diagram for any one term symbol has a “1” in each circle within the rectan-gle outlining it above, and the total Slater diagram is simply the sum of all the1’s in all the overlapping subdiagrams.

13.31 For three equivalent p electrons, Table 13.2 tells us we should find 2P, 2D, and4S terms. The table of possible spin combinations for the p3 configuration has20 columns, as shown below:

m = +1 ↑↓ ↑↓ ↑↓ ↑↓ ↑ ↓ ↑ ↓m = 0 ↑ ↓ ↑↓ ↑↓ ↑↓ ↑↓

m = –1 ↑ ↓ ↑ ↓ ↑↓ ↑↓mL 2 2 1 1 1 1 –1 –1 –1 –1

mS +1/2 –1/2 +1/2 –1/2 +1/2 –1/2 +1/2 –1/2 +1/2 –1/2

m = +1 ↑ ↑ ↑ ↑ ↓ ↓ ↓ ↓m = 0 ↑ ↓ ↑ ↑ ↓ ↓ ↑ ↑ ↓ ↓

m = –1 ↑↓ ↑↓ ↑ ↓ ↑ ↓ ↑ ↓ ↑ ↓mL –2 –2 0 0 0 0 0 0 0 0

mS +1/2 –1/2 +3/2 +1/2 +1/2 –1/2 +1/2 –1/2 –1/2 –3/2

The Slater diagram corresponding to this table, with the subdiagrams for eachterm symbol, is shown at the top of the next page. Note that this diagram isfive columns across, as in the previous problem, but it is four rows tall due tothe half-integral values for mS that are due to the odd number of electrons here.Note also the general rule that any subdiagram must extend as far to the right asit does to the left of the diagram’s center, and that each must also extend as farabove as below that center. Moreover, the subdiagram’s spin multiplicity tellsus its height in rows and its L value tells us its width in columns.

324

0 0 1 0 0

2 3 2 11

2 3 2 11

–2 –1 0 +1 +2

+3/2

+1/2

–1/2

mS

mL

2P

4S

2D

0 1 0 00–3/2

13.32 In the Ce ground-state configuration, 4f15d16s2, we see that the 6s subshell isfull and can be ignored for the purposes of determining term symbols. The felectron has l = 3, and the d electron has l = 2 so that L is at most 3 + 2 = 5 andat least |3 – 2| = 1 with all integral values in between allowed. This gives P, D,F, G, and H terms. Since the electrons are independent, they can be spin-paired for S = 0, a singlet term, or unpaired for S = 1, a triplet. These spinmultiplicities are allowed for each L value so that the possible term symbolsneglecting J are 1P, 3P, 1D, 3D, 1F, 3F, 1G, 3G, 1H, and 3H. Including Jexpands this list to read 1P1, 3P0,1,2, 1D2, 3D1,2,3, 1F3, 3F2,3,4, 1G4, 3G3,4,5,1H5, and 3H4,5,6. That this rather innocuous-looking configuration gives riseto 20 different energy states might help you understand why the lanthanide ele-ments have very complicated energy level patterns and atomic spectra.

13.33 Not all of the singly-charged anions of the elements from H through Ne arestable, but of those that are, we can use isoelectronic analogies to write theirground-state term symbols. For example, H– is isoelectronic to He and thushas a 1S0 term symbol. Continuing, He– is unstable, Li– is like Be: 1S0, Be– isunstable, B– is like C: 3P0, C– is like N: 4S3/2, N– is unstable, O– is like F:2P3/2, F– is like Ne: 1S0, and Ne– is unstable. For the singly-charged cations,H+ has no electrons and thus needs no terms symbol, He+ is 2S1/2, Li+ is 1S0,Be+ is 2S1/2, B+ is 1S0, C+ is 2P1/2, N+ is 3P0, O+ is 4S3/2, F+ is 3P2, andNe+ is 2P3/2.

13.34 From Problem 13.27, we know that the possible term symbols for Ti in the3d24s2 configuration are 1S0, 1D2, 1G4, 3P0, 3P1, 3P2, 3F2, 3F3, and 3F4 withthe 3F2 term the ground state. The next two levels are quite nearby and must

325

correspond to 3F3 and 3F4, respectively. We expect the other set of triplet lev-els, 3P0, 3P1, and 3P2, to come next and in this order. We then assign the stateat 8602 cm–1 to 1G4, since it is the singlet state of highest L. The final ob-served state at 12 118 cm–1 must be the 1D2 state, since the 1S0 state is as yetunobserved.

13.35 To couple the 8S and 3P terms that represent the f and sp electrons, respec-tively, in the 4f76s16p1 configuration of Eu, we first decompose them into theircorresponding S and L values: 8S → S = 7/2 and L = 0; 3P → S = 1, L = 1.The L values can only couple to give L = 1 so that all states must be P states.The S values can couple to a resultant S that has any of the values 5/2, 7/2, or9/2 so that the multiplicity possibilities are 6, 8, and 10. The nine completeterm symbols including J are 6P3/2,5/2,7/2, 8P5/2,7/2,9/2, and 10P7/2,9/2,11/2.

GENERAL PROBLEMS

13.36 The trial wavefunction is

φ(x) = c1ψ1(x) + c2ψ2(x) = 2L

1/2 c1 sin

πxL

+ c2 sin 2πx

L ,

and the perturbation is zero from x = 0 to x = L/2 but constant at V0 = E1 =¨2π2/2mL2 from x = L/2 to x = L. The variational energy expression is givenby Eq. (13.13):

<E> =

∑i = 1

2

ci*cjHij∑

j = 1

2

∑i = 1

2

ci*cjSij∑

j = 1

2 ,

and it indicates that our first task is to find the matrix elements Sij and Hij. TheS matrix is particularly easy to find, since our trial wavefunction contains onlythe orthonormal functions ψ1 and ψ2:

Sii = ψi2 dx

0

L

= 1 , Sij = ψiψ j dx0

L

= 0 , S = 1 00 1

.

326

The Hamiltonian matrix is also easy to calculate. The complete Hamiltonianoperator H is just the sum of the usual particle-in-a-box kinetic energy term Tplus the potential energy term that equals zero over one half of the box and E1over the other. For the H11 matrix element, we write

H11 = ψ1Hψ1 dx0

L

= 2L

sin πxL

H sin πxL

dx0

L

= 2L

sin πxL

T sin πxL

dx0

L

+ E1 2L

sin2 πxL

dxL/2

L

= E1 + E1

2 =

32

E1

where the integral containing T must equal E1 because T is the entire Hamilto-nian for the ordinary particle-in-a-box problem, for which ψ1 is the eigenfunc-tion with eigenenergy E1. Likewise, the integral from L/2 to L represents theonly nonzero part of the potential energy, and by symmetry, this integral equals1/2, the probability that the ordinary system is found in this region. We find theH22 matrix element similarly, taking advantage of the fact that E2, the secondenergy of the ordinary system, equals 22E1 = 4E1. We have

H22 = ψ2Hψ2 dx0

L

= 2L

sin 2πx

L H sin

2πxL

dx0

L

= 2L

sin 2πx

L T sin

2πxL

dx0

L

+ E1 2L

sin2 2πx

L dx

L/2

L

= E2 + E1

2 = 4E1 +

E1

2 =

92

E1 .

The so-called “off-diagonal” Hamiltonian matrix elements H12 and H21 areequal by symmetry. We find them to be

327

H12 = ψ1Hψ2 dx0

L

= 2L

sin πxL

H sin 2πx

L dx

0

L

= 2L

sin πxL

T sin 2πx

L dx

0

L

+ E1 2L

sin πxL

sin 2πx

L dx

L/2

L

= E2 2L

sin πxL

sin 2πx

L dx

0

L

+ E1 2L

sin πxL

sin 2πx

L dx

L/2

L

= 0 + E1 – 4

3π = –

4E1

3π

where the zero value follows from the orthogonality of ψ1 and ψ2 and the finalintegral is given in the problem. Thus, the entire Hamiltonian matrix is

H = 3E1

2 –

4E1

3π

– 4E1

3π

9E1

2

= E1

32

– 4

3π

– 4

3π

92

.

The best energy is the solution to the quadratic expression shown at the bottomof page 460 in the text:

0 = (H11 – <E>S11) (H22 – <E>S22) – (H12 – <E>S12)2

= 32

E1 – <E> 92

E1 – <E> – E12

16

9π2 .

Two solutions emerge:

<E> = E1 3 ± 12

4 16

9π2

+ 9 =

1.4411E1

4.5589E1

and we choose the smaller of the two as the best approximation to the trueground-state energy.

13.37 We continue the previous problem here with a look at the variational wavefunc-tion itself. We now must use the best energy found there, 1.4411E1, to find the

328

coefficients c1 and c2. These are found from their normalization condition c12 +

c22 = 1 and the two equations shown on page 460 in the text, which, given our

H and S matrices, read

c1 32

E1 – 1.4411E1 – c2E14

3π = 0

–c1E14

3π + c2

92

E1 – 1.4411E1 = 0 .

Simultaneous solution of these equations gives c1 = ±0.9905 and c2 = ±0.1374with an ambiguity in sign due to the quadratic nature of the equations. We candecide which signs to choose because we know the best energy from the previ-ous problem. We write

<E>E1

= 1.4411 = c1ψ1 + c2ψ2 HE1

c1ψ1 + c2ψ2 dx0

L

= c1ψ1 + c2ψ2 TE1

c1ψ1 + c2ψ2 + VE1

c1ψ1 + c2ψ2 dx0

L

= c1ψ1 + c2ψ2 c1ψ1 + 4c2ψ2 dx 0

L

+ c1ψ1 + c2ψ22 dx

L/2

L

= c12 + 4c2

2 + c1

2

2 + 2c1c2 –

43π

+ c2

2

2 = 3c2

2 + 32

– 8c1c2

3π

where we have used the integral given in the previous problem as well as theorthonormality and symmetry of ψ1 and ψ2. This expression shows us that c1and c2 must be both positive or both negative; the choice of opposite signs doesnot satisfy this equation. Thus, we have, choosing the positive signs,

φ(x) = c1ψ1 + c2ψ2 = 0.9905ψ1 + 0.1374ψ2 .

This wavefunction is graphed on the next page. Over the potential step, from x= L/2 to L, the particle has a smaller kinetic energy than it has from 0 to L/2.This means it has a greater de Broglie wavelength over the step, and this forcesthe maximum in φ to lie between 0 and L/2. One can show that the mean posi-

329

tion of the particle, <x>, is 0.4510L. This is somewhat surprising, since theclassical expectation would be that the particle spends more time over the poten-tial step, where it moves slower, so that <x> would be >L/2.

0 L/2 L

ψ

13.38 For the general case of a two-electron atom of nuclear charge Z, we can easilywrite the expression for <E> if we generalize the expression on page 470 in thetext. The two factors of 2 in that expression become Z, and we introduce theground-state H-atom energy, E1, to simplify notation. We find

<E> = –2E1 Ze

2

2 – Ze

2 + Ze2 – ZZe +

Ze2

2 – Ze

2 + Ze2 – ZZe +

58

Ze

= –2E1 Ze2 – 2ZZe +

58

Ze .

We minimize this expression with respect to Ze:

d<E>dZe

= –2E1 2Ze – 2Z + 58

= 0 or Ze = Z – 516

.

Substituting this value for Ze into the general expression for <E> gives

<E> = E1 2Z2 – 54

Z + 25128

.

This energy is the energy of the two-electron atom with respect to a zero of en-ergy that represents the nucleus and the two electrons infinitely far apart. Theenergy of the one-electron atom with the second electron infinitely far away isZ2E1 with respect to this energy zero so that the first ionization energy of thetwo-electron atom is Z2E1 – <E>:

330

IP = Z2E1 – <E> = –E1 Z2 – 54

Z + 25128

.

With E1 = –13.595 eV, this expression becomes IP/eV = 13.595Z2 – 16.994Z+ 2.6553. We can use this expression to compare with the experimental IPs asshown in the table below (where IP is expressed in eV units).

H– He Li+ Be2+ B3+ C4+ N5+ O6+ F7+

Z 1 2 3 4 5 6 7 8 9

Expt 0.7542 24.596 75.619 153.85 259.298 391.986 551.925 739.114 953.6

Calc –0.7535 23.05 74.03 152.2 257.6 390.1 549.9 736.8 950.9

Note that the calculated H– IP is negative, which means this level of theorypredicts H– is unstable. (This is the result we found in Problem 13.19.) Theabsolute error is remarkably constant, varying smoothly from about 1.4 eV toabout 2.7 eV, but the relative error is decreasing with increasing Z.

13.39 Triplet Ps emits three photons so that two have off-setting angular momenta andthe third carries the angular momentum inherent in the triplet. One photon de-cay could conserve the triplet’s angular momentum, but in the Ps atom’s refer-ence frame, there is no linear momentum before the decay and thus there mustbe none after the decay. One photon decay would produce linear momentum(that of the photon) where none had existed before, in violation of linear mo-mentum conservation. Three photons can be emitted with linear momenta thatpoint in different directions so that they add, vectorially, to zero.

13.40 It is sufficient here to consider, for simplicity in notation and algebra, only the n= 1 ground state and an abbreviated expansion of the true wavefunction in theform

ψ1 = ψ1(0)

+ λc11ψ1(0)

+ λc12ψ2(0)

where we have expanded the first-order corrected wavefunction through onlythe n = 2 state. The normalization integral is

331

ψ1*ψ1 dτ = ψ1

(0)* + λc11

* ψ1(0)*

+ λc12* ψ2

(0)* ψ1(0)

+ λc11ψ1(0)

+ λc12ψ2(0)

dτ

= ψ1(0)*ψ1

(0) + λc11ψ1

(0)*ψ1(0)

+ λc12ψ1(0)*ψ2

(0) + λc11

* ψ1(0)*ψ1

(0)

+ λ2c11

* c11ψ1(0)*ψ1

(0) + λ2

c11* c12ψ1

(0)*ψ2(0)

+ λc12* ψ2

(0)*ψ1(0)

+ λ2c12

* c11ψ2(0)*ψ1

(0) + λ2

c12* c12ψ2

(0)*ψ2(0)

dτ

= 1 + λc11 + 0 + λc11* + λ 2

c11* c11 + 0 + 0 + 0 + λ2

c12* c12

= 1 + λ c11 + c11* + λ2

where we have exploited the orthonormality properties of ψ1(0)

and ψ2(0)

to eval-uate the many integrals. If we take c11 to be real, then c11 + c11

* = 2c11. Fi-nally, to be consistent with the idea that we are deriving an expression validthrough first order only, we drop the second-order term λ2 and end with a nor-malization expression

ψ1*ψ1 dτ = 1 + 2λc11 .

In order to make this integral equal 1, we must have c11 = 0.