An effective charge is a fitted number that turns out to predict things it was not fitted to. That is the whole justification for the approximation the chapter is built on.

Nine problems, in three groups. 1–4 justify the machinery of §11.1a: where Eq. (11.5) comes from, what an effective charge measures, and why subshells hold what they hold. 5–7 are term-symbol exercises for §11.1b. 8–9 extend §11.3’s model to ions — and problem 9 is where the model both triumphs and quietly fails.

Where the central potential comes from

1 Eq. (11.5) derived from a charge density, by Gauss's law derivation

The $Z-1$ electrons around a nucleus of charge $Ze$ have charge density

\rho(r) = -\frac{(Z-1)e}{4\pi a^3}\,\frac{e^{-r/a}}{r/a}

(a) Show the mean charge radius of this distribution is $2a$ . (b) Find the charge inside radius $r$ . (c) Show that an electron in the field of the nucleus and this distribution has potential energy given by Eq. (11.5),

V(r) = -\frac{z(r)e^2}{4\pi\epsilon_0 r}, \qquad z(r) = (Z-1)e^{-r/a} + 1

💡 Phillips' own hint

(a) The mean charge radius is

\frac{\int_0^\infty r\,\rho(r)\,4\pi r^2\,\mathrm dr}{\int_0^\infty \rho(r)\,4\pi r^2\,\mathrm dr}

(b) The charge inside radius $r$ is $q(r) = \int_0^r \rho(r')\,4\pi r'^2\,\mathrm dr'$ .

(c) Use Gauss’s theorem for the magnitude of $\mathbf E(r)$ , then find $\phi(r)$ from $E = -\mathrm d\phi/\mathrm dr$ .

✓ Worked solution

(a) Note that $\rho\,4\pi r^2 = -\big[(Z-1)e/a^2\big]\,r\,e^{-r/a}$ — the $1/r$ in the density cancels one power of $r^2$ . Both integrals are then standard:

\int_0^\infty r\,e^{-r/a}\,\mathrm dr = a^2, \qquad \int_0^\infty r^2 e^{-r/a}\,\mathrm dr = 2a^3

so the ratio is $2a^3/a^2 = \mathbf{2a}$ . (The denominator also confirms the distribution carries total charge $-(Z-1)e$ , as it must.)

(b) Integrating to $r$ instead of infinity,

q_{\text{electrons}}(r) = -(Z-1)e\left[1 - \left(1 + \frac ra\right)e^{-r/a}\right]

and adding the nucleus gives the total enclosed charge

q(r) = e\left[1 + (Z-1)\left(1 + \frac ra\right)e^{-r/a}\right]

(c) Gauss’s theorem gives $E(r) = q(r)/4\pi\epsilon_0r^2$ , and the potential is $\phi(r) = \int_r^\infty E\,\mathrm dr'$ . The $1/r'^2$ piece integrates to $1/r$ . For the rest, integrate by parts:

\int_r^\infty \frac{e^{-r'/a}}{r'^2}\,\mathrm dr' = \frac{e^{-r/a}}{r} - \frac1a\int_r^\infty \frac{e^{-r'/a}}{r'}\,\mathrm dr'

and the surviving integral is exactly cancelled by the $(r'/a)$ term of $q(r)$ , which contributes $+\tfrac1a\int_r^\infty e^{-r'/a}/r'\,\mathrm dr'$ . What is left is clean:

\phi(r) = \frac{1}{4\pi\epsilon_0 r}\left[1 + (Z-1)e^{-r/a}\right] = \frac{z(r)}{4\pi\epsilon_0 r}

so $V(r) = -e\phi(r)$ is Eq. (11.5).

Reproduce it in NumPy

The cancellation in part (c) is the kind of result that is easy to assert and easy to get wrong, so it is checked numerically. My first run showed r·φ(r) drifting away from z(r) by an amount that grew with r — which looked like a real disagreement and was actually the integral truncating at finite radius. Adding the analytic 1/r_max tail made the two agree to six decimals.

import numpy as np
Z, a = 6, 0.5                                   # carbon, screening radius a0/2

# The charge density of the Z-1 electrons. Note rho * 4 pi r^2 is proportional to
# r e^(-r/a): the 1/r in the density cancels one power of r^2.
r    = np.linspace(1e-9, 200*a, 4_000_001)
rho  = -(Z-1)/(4*np.pi*a**3) * np.exp(-r/a)/(r/a)      # in units of e
shell = rho*4*np.pi*r**2

print("(a) normalization and the mean charge radius")
tot  = np.trapezoid(shell, r)
mean = np.trapezoid(r*shell, r)/tot
print(f"    total electron charge = {tot:+.6f} e   (want {-(Z-1)})")
print(f"    mean charge radius    = {mean/a:.6f} a   (want 2a)")

print("\n(b) charge enclosed within r, nucleus included")
q_closed = lambda R: 1 + (Z-1)*(1 + R/a)*np.exp(-R/a)
def q_numeric(R):
    m = r <= R
    return Z + np.trapezoid(shell[m], r[m])
print(f"    {'r/a':>6}{'numeric':>12}{'closed form':>14}")
for R in (0.5*a, a, 2*a, 4*a):
    print(f"    {R/a:>6.1f}{q_numeric(R):>12.5f}{q_closed(R):>14.5f}")
print("    -> q(r)/e = 1 + (Z-1)(1 + r/a) e^(-r/a).  NOTE the (1 + r/a):")
print("       this is NOT z(r), and Eq. (11.5) cannot be read off it.")

print("\n(c) the potential, where the extra factor cancels")
def phi(R):
    m = r >= R
    # beyond r_max only the bare 1/r'^2 term survives, contributing 1/r_max.
    # Dropping it leaves an R/r_max offset that grows with R and looks exactly
    # like a real disagreement -- which is what it looked like on the first run.
    return np.trapezoid(q_closed(r[m])/r[m]**2, r[m]) + 1/r[-1]
z = lambda R: (Z-1)*np.exp(-R/a) + 1
print(f"    {'r/a':>6}{'r phi(r)':>14}{'z(r), Eq. (11.5)':>20}")
for R in (0.25*a, 0.5*a, a, 2*a, 4*a):
    print(f"    {R/a:>6.2f}{R*phi(R):>14.6f}{z(R):>20.6f}")
print("""
    They agree to six decimals, so V(r) = -z(r) e^2 / 4 pi eps0 r exactly.
    The (1 + r/a) disappeared because integrating e^(-r/a)/r^2 by parts throws
    off an exponential-integral term that exactly cancels the one coming from
    the r/a piece. Neither can be done in closed form alone; the sum can.""")

prints(a) normalization and the mean charge radius total electron charge = -5.000000 e (want -5) mean charge radius = 2.000000 a (want 2a) (b) charge enclosed within r, nucleus included r/a numeric closed form 0.5 5.54906 5.54898 1.0 4.67889 4.67879 2.0 3.03010 3.03003 4.0 1.45791 1.45789 -> q(r)/e = 1 + (Z-1)(1 + r/a) e^(-r/a). NOTE the (1 + r/a): this is NOT z(r), and Eq. (11.5) cannot be read off it. (c) the potential, where the extra factor cancels r/a r phi(r) z(r), Eq. (11.5) 0.25 4.894004 4.894004 0.50 4.032653 4.032653 1.00 2.839397 2.839397 2.00 1.676676 1.676676 4.00 1.091578 1.091578 They agree to six decimals, so V(r) = -z(r) e^2 / 4 pi eps0 r exactly. The (1 + r/a) disappeared because integrating e^(-r/a)/r^2 by parts throws off an exponential-integral term that exactly cancels the one coming from the r/a piece. Neither can be done in closed form alone; the sum can.

The enclosed charge is NOT z(r), and stopping at part (b) will mislead you

Part (b) gives an enclosed charge proportional to

1 + (Z-1)\left(1 + \frac ra\right)e^{-r/a}

while Eq. (11.5) has $z(r) = (Z-1)e^{-r/a} + 1$ . They are different functions. It is tempting — and wrong — to read the screened charge straight off the enclosed charge and declare part (c) done.

The extra $(1 + r/a)$ survives in the field and disappears only in the potential, because integrating $e^{-r/a}/r^2$ by parts throws off an exponential-integral term that exactly cancels the one produced by the $r/a$ piece. Neither term can be done in closed form on its own; only the combination can.

The calculation below checks the cancellation numerically, because it is the kind of result that is easy to assert and easy to get wrong.

Reading numbers off real atoms

2 Helium's two ionization energies are very different numerical

Removing both electrons from a helium atom costs 79 eV. Show that removing just the first costs 24.6 eV.

✓ Worked solution

Removing the second electron is a problem we can solve exactly: what is left is He⁺, a hydrogen-like ion with $Z = 2$ and one electron. From ch09’s problem 14,

E = -Z^2E_R = -4 \times 13.61 = -54.4\ \mathrm{eV}

so that step costs 54.4 eV. The two steps together cost 79 eV, so the first costs

79.0 - 54.4 = \mathbf{24.6\ eV}

which is the measured first ionization energy, 24.587 eV.

Reproduce it in NumPy

One subtraction, but it works only because the intermediate state is exactly solvable: He⁺ has a single electron and is hydrogen-like, so ch09's closed form applies. The two-electron atom has no such form — which is why the problem asks about the difference rather than about either energy alone.

ER = 13.6057
print("Helium: the two ionization energies are very different")
print("=" * 62)
E_He_plus = -4*ER                       # hydrogen-like ion, Z = 2:  E = -Z^2 E_R
second = -E_He_plus
first  = 79.0 - second
print(f"  He+ is hydrogen-like with Z = 2, so E = -Z^2 E_R = {E_He_plus:.2f} eV")
print(f"  removing the SECOND electron therefore costs   {second:.1f} eV")
print(f"  both together cost 79.0 eV, so the FIRST costs {first:.1f} eV")
print(f"  measured first ionization energy of helium:    24.587 eV")
print(f"""
  The second electron costs {second/first:.2f} times as much as the first, out of the
  same atom. While both are present each screens the nucleus for the other; once
  one is gone the survivor faces the full unscreened Z = 2.""")

printsHelium: the two ionization energies are very different ============================================================== He+ is hydrogen-like with Z = 2, so E = -Z^2 E_R = -54.42 eV removing the SECOND electron therefore costs 54.4 eV both together cost 79.0 eV, so the FIRST costs 24.6 eV measured first ionization energy of helium: 24.587 eV The second electron costs 2.21 times as much as the first, out of the same atom. While both are present each screens the nucleus for the other; once one is gone the survivor faces the full unscreened Z = 2.

3 What sodium's valence electron actually sees numerical

Sodium’s other ten electrons partly screen the nuclear charge $11e$ , so the valence electron sees an effective point charge $Z^*e$ . Given binding energies 5.12 eV (3s), 2.10 eV (3p) and 1.52 eV (3d), find $Z^*$ in each case.

💡 Phillips' own hint

The effective point charges are $Z^*(3s) = 1.84$ , $Z^*(3p) = 1.18$ and $Z^*(3d) = 1.003$ . Note that the shielding by the inner electrons increases with $l$ .

✓ Worked solution

A single electron bound to a point charge $Z^*e$ has $E = -Z^{*2}E_R/n^2$ . All three states have $n = 3$ , so

Z^* = n\sqrt{\frac{E}{E_R}} = 3\sqrt{\frac{E}{13.61\ \mathrm{eV}}}

Z^*(3s) = 3\sqrt{5.12/13.61} = \mathbf{1.84} \qquad Z^*(3p) = \mathbf{1.18} \qquad Z^*(3d) = \mathbf{1.00}

Reproduce it in NumPy

No model at all here — three measured binding energies go in and three effective charges come out, using only E = −Z*²E_R/n². The trend across l is §11.1a's screening mechanism read directly off experiment, and it agrees with the same quantity computed for carbon by solving Eq. (11.6).

import numpy as np
ER = 13.6057
print("Sodium's valence electron: E = -Z*^2 E_R / n^2 with n = 3")
print("=" * 66)
print(f"  {'state':<7}{'binding (eV)':>14}{'Z* computed':>14}{'hint':>9}")
for st, E, book in (('3s', 5.12, 1.84), ('3p', 2.10, 1.18), ('3d', 1.52, 1.003)):
    print(f"  {st:<7}{E:>14.2f}{3*np.sqrt(E/ER):>14.3f}{book:>9.3f}")
print(f"""
  Sodium has ELEVEN protons and its valence electron sees between one and two.
  The trend across l is the whole of section 11.1a in three measured numbers:
  3s penetrates the neon core and glimpses the unscreened nucleus; 3d is held
  outside by the centrifugal barrier and sees Z* = 1.003 -- screened from ten of
  the eleven protons, and bound almost exactly as a hydrogen 3d electron is.""")

printsSodium's valence electron: E = -Z*^2 E_R / n^2 with n = 3 ================================================================== state binding (eV) Z* computed hint 3s 5.12 1.840 1.840 3p 2.10 1.179 1.180 3d 1.52 1.003 1.003 Sodium has ELEVEN protons and its valence electron sees between one and two. The trend across l is the whole of section 11.1a in three measured numbers: 3s penetrates the neon core and glimpses the unscreened nucleus; 3d is held outside by the centrifugal barrier and sees Z* = 1.003 -- screened from ten of the eleven protons, and bound almost exactly as a hydrogen 3d electron is.

Three numbers that contain the whole chapter

1.84, 1.18, 1.003. Sodium has eleven protons, and its valence electron sees between one and two of them.

The trend across $l$ is the point. A 3s electron penetrates the neon core and gets a glimpse of the unscreened nucleus, so it sees $Z^* = 1.84$ . A 3d electron is held out by the centrifugal barrier $l(l+1)\hbar^2/2m_er^2$ and never gets inside at all, so it sees $Z^* = 1.003$ — one unit of charge, screened from ten of the eleven protons.

That is why $E$ rises with $l$ , which is why shells do not fill in one go, which is why there is a periodic table. §11.1a computed the same quantity for carbon (4.49, 1.92, 1.24) by solving Eq. (11.6); here it is read straight off three measured numbers, with no model at all.

The 3d value is worth a second look. $Z^* = 1.003$ means sodium’s 3d electron is bound almost exactly as a hydrogen 3d electron would be. Screening is essentially perfect, and the alkali metals’ spectra look hydrogen-like for precisely this reason.

4 Why 3d holds ten and 4f holds fourteen derivation

Explain why up to 10 electrons may be assigned to 3d orbitals and 14 to 4f.

✓ Worked solution

For orbital quantum number $l$ there are $2l+1$ values of $m_l$ , running $-l, \ldots, +l$ . Each of those orbitals can hold two electrons, $m_s = \pm\tfrac12$ , and no more — Pauli forbids two electrons sharing all four of $n, l, m_l, m_s$ . So the capacity is $2(2l+1)$ :

3d: $l = 2$ , so $m_l = -2, -1, 0, 1, 2$ — five orbitals, $2 \times 5 = \mathbf{10}$ .
4f: $l = 3$ , so seven values of $m_l$ , $2 \times 7 = \mathbf{14}$ .

Note that $n$ never enters. The capacity of a subshell depends on $l$ alone, which is why 3d and 4d both hold ten.

Term symbols, three times

5 Helium: four configurations, their L, S and J derivation

For the helium configurations $(1s)^2$ , $(1s)(2s)$ , $(1s)(2p)$ and $(1s)(3d)$ , write the possible $L$ and $S$ , and for each pair the possible $J$ .

✓ Worked solution

configuration⇅	L and S⇅	terms with J⇅
(1s)²	L = 0; S = 0 only	¹S₀
(1s)(2s)	L = 0; S = 0 or 1	¹S₀, ³S₁
(1s)(2p)	L = 1; S = 0 or 1	¹P₁, ³P₀, ³P₁, ³P₂
(1s)(3d)	L = 2; S = 0 or 1	¹D₂, ³D₁, ³D₂, ³D₃

$(1s)^2$ is the only equivalent case, and it is where Pauli bites: two electrons in the same orbital must have opposite spins, so $S = 1$ is unavailable and $^3S$ does not exist. In the other three the electrons differ in $n$ , so both spin states survive.

$L$ is forced in each case because one electron has $l = 0$ : combining $l = 0$ with $l'$ gives $L = l'$ and nothing else.

6 Carbon: (2p)(3s) and (2p)(3p) derivation

For the carbon configurations $(1s)^2(2s)^2(2p)(3s)$ and $(1s)^2(2s)^2(2p)(3p)$ , write the possible $L$ , $S$ and $J$ .

✓ Worked solution

The closed $(1s)^2(2s)^2$ contributes $L = S = 0$ and can be ignored.

configuration⇅	L and S⇅	terms with J⇅
(2p)(3s)	L = 1; S = 0 or 1	¹P₁, ³P₀, ³P₁, ³P₂
(2p)(3p)	L = 2, 1, 0; S = 0 or 1	¹S₀, ³S₁, ¹P₁, ³P₀₁₂, ¹D₂, ³D₁₂₃

Both are inequivalent — the two electrons differ in $n$ — so every combination of $L$ and $S$ survives. $(2p)(3p)$ therefore gives all six terms, 36 states in total.

7 Why (2p)(3p) has a ³S state and (2p)² does not derivation

Explain why the carbon configuration $(1s)^2(2s)^2(2p)(3p)$ has a $^3S$ state but $(1s)^2(2s)^2(2p)^2$ does not.

✓ Worked solution

$^3S$ means $S = 1$ and $L = 0$ : spins parallel, orbital angular momenta cancelling.

In $(2p)(3p)$ the electrons differ in $n$ , so Pauli imposes nothing. Both may have $m_l = +1$ with both spins up, and every combination of $L$ and $S$ is available — including $L = 0$ with $S = 1$ .

In $(2p)^2$ they share $n$ and $l$ , so they must differ in $(m_l, m_s)$ . With $S = 1$ both spins are already up, so they are forced into different $m_l$ — and that is what kills $^3S$ . Building $L = 0$ requires the $m_l$ values to cancel in every way a full multiplet needs; with parallel spins the largest $M_L$ reachable is $1 + 0 = 1$ , so the only triplet available is $^3P$ .

Counting says the same thing without any of the words: $(2p)^2$ has 15 micro-states and they divide exactly as $^3P + {}^1D + {}^1S = 9 + 5 + 1$ , with no room left for a $^3S$ .

Switch the widget between (2p)² and (2p)(3p) to see problems 5–7 at once — the same two $l$ values, and the exclusion principle the only difference.

Which terms a configuration has — counted, not quoted

Inequivalent electrons — different n, so the electrons are already distinguished by a quantum number and every combination is allowed. Problem 6 and 7. Compare with (2p)²: the same two l values, but different n — and six more terms.

M_L ＼ M_S	+1	0	-1
+2	1	2	1
+1	2	4	2
0	3	6	3
-1	2	4	2
-2	1	2	1

36 micro-states. Peeling removes one whole (2L+1)(2S+1) block at a time.

peel:

term	L	S	(2S+1)(2L+1)	J = \|L−S\| … L+S
3D	2	1	3 × 5 = 15	3D1, 3D2, 3D3
1D	2	0	1 × 5 = 5	1D2
3P	1	1	3 × 3 = 9	3P0, 3P1, 3P2
1P	1	0	1 × 3 = 3	1P1
3S	0	1	3 × 1 = 3	3S1
1S	0	0	1 × 1 = 1	1S0
total			36	matches the micro-state count ✓

Every combination survives here. The two electrons already differ in n, so Pauli never engages and the table is the full outer product — 36 micro-states, and every (L, S) pair appears. Compare this with an equivalent configuration to see exactly what the exclusion principle costs.

Extending §11.3’s model to ions

8 An ion with Z−1 electrons derivation

Using §11.3’s model, write the energy of a state of spatial extent $R$ for an ion with $Z-1$ electrons around a nucleus of charge $Ze$ , and show its minimum is Eq. (11.23).

💡 Phillips' own hint

The energy of such a state is

E = (Z-1)\frac{\hbar^2}{2m_eR^2} - Z(Z-1)\frac{e^2}{4\pi\epsilon_0R} + \frac{(Z-1)(Z-2)}{2}\frac{e^2}{4\pi\epsilon_0R_{ee}}

✓ Worked solution

Count the three terms. There are $Z-1$ electrons, so $Z-1$ kinetic terms. Each is attracted by charge $Ze$ , giving $Z(Z-1)$ units of attraction. And there are $\tfrac12(Z-1)(Z-2)$ pairs repelling. Setting $R_{ee} = fR$ — with $f$ the avoidance parameter fitted once to helium — collects it into the familiar form:

E = \frac{A}{R^2} - \frac{B}{R}, \qquad A = (Z-1)A_1, \quad B = \left[Z(Z-1) - \frac{(Z-1)(Z-2)}{2f}\right]B_1

The minimum of $A/R^2 - B/R$ is $-B^2/4A$ , and with $B_1^2/4A_1 = E_R$ ,

E_{Z^-} = -\frac{\left[Z(Z-1) - \frac{(Z-1)(Z-2)}{2f}\right]^2}{Z-1}\,E_R

which is Eq. (11.23). Every step is the same minimisation as §11.3; only the coefficients change.

9 Two-electron ions, and the one case that does not work numerical

(a) Using $f = 1.67$ , show the model reproduces the measured ionization energies of the two-electron ions Li⁺ through F⁷⁺. (b) Show the model gives H⁻‘s measured binding energy of 0.75 eV if $f = 1.8$ .

✓ Worked solution

(a) For a two-electron ion of nuclear charge $Ze$ , §11.3’s construction gives $A = 2A_1$ and $B = [2Z - 1/f]B_1$ , so $E_2 = -(2Z - 1/f)^2E_R/2$ . Removing one electron leaves a hydrogen-like ion at $-Z^2E_R$ , so

E_I = \left[\frac{(2Z - \tfrac1f)^2}{2} - Z^2\right]E_R

Every one of the seven comes out within half a per cent of experiment, as the calculation below tabulates. That is a remarkable return for a model whose only empirical input is a single number fitted to helium.

(b) Here the model does not deliver what the problem claims. With $Z = 1$ and $f = 1.8$ ,

E_I = \left[\frac{(2 - 1/1.8)^2}{2} - 1\right]E_R = 0.0432\,E_R = 0.59\ \mathrm{eV}

not 0.75 eV. Reproducing 0.75 eV requires $f = 1.827$ .

Reproduce it in NumPy

Part (a) is the model at its best and part (b) at its worst, from the same three lines of arithmetic. The difference is not the physics but the conditioning: (a) predicts quantities comparable to the terms it is built from, (b) predicts a difference twenty-three times smaller than either.

import numpy as np
ER = 13.6057
# Section 11.3's model for a TWO-electron ion of nuclear charge Ze:
#   A = 2 A1,  B = (2Z - 1/f) B1,  so E2 = -(2Z - 1/f)^2 / 2  in E_R.
# Removing one electron leaves a hydrogen-like ion at -Z^2 E_R.
EI = lambda Z, f: (2*Z - 1/f)**2/2 - Z*Z

print("(a) two-electron ions, with the f = 1.67 fitted to helium and never refitted")
print("=" * 74)
data = [('Li+',3,5.57),('Be2+',4,11.32),('B3+',5,19.08),('C4+',6,28.84),
        ('N5+',7,40.61),('O6+',8,54.39),('F7+',9,70.12)]
book = {3:5.6,4:11.4,5:19.2,6:29.0,7:40.8,8:54.6,9:70.4}
print(f"  {'ion':<6}{'model':>9}{'book':>8}{'experiment':>13}{'error':>9}")
worst = 0
for ion, Z, e in data:
    m = EI(Z, 1.67); err = (m-e)/e*100; worst = max(worst, abs(err))
    print(f"  {ion:<6}{m:>9.2f}{book[Z]:>8.1f}{e:>13.2f}{err:>8.1f}%")
print(f"  worst error across all seven: {worst:.1f}%, from a model with ONE fitted number.")

print("\n(b) H-, where the same model stops working")
print("=" * 74)
for f in (1.8, 1.827):
    print(f"  f = {f:<6} binding = {EI(1,f):.5f} E_R = {EI(1,f)*ER:.3f} eV")
need = 0.75/ER
fneed = 1/(2 - np.sqrt(2*(need+1)))
print(f"  the book says f = 1.8 gives 0.75 eV. It gives {EI(1,1.8)*ER:.2f} eV.")
print(f"  reproducing 0.75 eV needs f = {fneed:.4f}.")
print(f"""
  Why so sensitive? The binding energy is a SMALL DIFFERENCE OF LARGE NUMBERS:

      E(H-) = {-(2-1/1.8)**2/2:.4f} E_R      E(H) = -1.0000 E_R
      binding = {EI(1,1.8):.4f} E_R

  Everywhere in part (a) the model predicts a quantity comparable to the numbers
  it is built from, and lands within half a per cent. Here it predicts something
  {1/EI(1,1.8):.0f} times smaller than either term, so relative errors are amplified about
  {1/EI(1,1.8):.0f}-fold -- a {abs(1.827-1.8)/1.8*100:.1f}% change in f moves the answer by {abs(EI(1,1.827)-EI(1,1.8))/EI(1,1.8)*100:.0f}%.

  So "1.8" is right to the two figures quoted AND its prediction is wrong by a
  quarter. Both at once. The lesson is conditioning, not atoms.""")

prints(a) two-electron ions, with the f = 1.67 fitted to helium and never refitted ========================================================================== ion model book experiment error Li+ 5.59 5.6 5.57 0.3% Be2+ 11.39 11.4 11.32 0.6% B3+ 19.19 19.2 19.08 0.6% C4+ 28.99 29.0 28.84 0.5% N5+ 40.80 40.8 40.61 0.5% O6+ 54.60 54.6 54.39 0.4% F7+ 70.40 70.4 70.12 0.4% worst error across all seven: 0.6%, from a model with ONE fitted number. (b) H-, where the same model stops working ========================================================================== f = 1.8 binding = 0.04321 E_R = 0.588 eV f = 1.827 binding = 0.05510 E_R = 0.750 eV the book says f = 1.8 gives 0.75 eV. It gives 0.59 eV. reproducing 0.75 eV needs f = 1.8270. Why so sensitive? The binding energy is a SMALL DIFFERENCE OF LARGE NUMBERS: E(H-) = -1.0432 E_R E(H) = -1.0000 E_R binding = 0.0432 E_R Everywhere in part (a) the model predicts a quantity comparable to the numbers it is built from, and lands within half a per cent. Here it predicts something 23 times smaller than either term, so relative errors are amplified about 23-fold -- a 1.5% change in f moves the answer by 28%. So "1.8" is right to the two figures quoted AND its prediction is wrong by a quarter. Both at once. The lesson is conditioning, not atoms.

Problem 9(b)'s f = 1.8 gives 0.59 eV, not 0.75 eV — and the reason is worth more than the error

The book asks you to show that $f = 1.8$ reproduces H⁻‘s binding energy of 0.75 eV. It gives 0.59 eV. The value that works is $f = 1.827$ .

Before calling this a misprint, look at why the answer is so sensitive:

E(\mathrm{H}^-) = -1.0432\,E_R, \qquad E(\mathrm{H}) = -1.0000\,E_R, \qquad \text{binding} = 0.0432\,E_R

The binding energy is a small difference of two large numbers. Everywhere else in problem 9 the model is predicting a quantity comparable to the numbers it is built from, and it is accurate to half a per cent. Here it is predicting something 25 times smaller than either term, so every relative error is amplified about 25-fold — and a 1.5 % change in $f$ moves the answer by 28 %.

So “1.8” is right to the two significant figures the book quotes, and its prediction is still wrong by a quarter. Both facts are true at once, and the lesson is about conditioning rather than about atoms: when a calculation subtracts two nearly equal quantities, the precision of the inputs stops being a guide to the precision of the output.

Check yourself

0 / 6 answered

Problem 1(b) gives the charge enclosed within radius $r$ as $q (r) = e [1 + (Z - 1) (1 + r / a) e^{- r / a}]$ , while Eq. (11.5) has $z (r) = (Z - 1) e^{- r / a} + 1$ .
1.Why is that not a contradiction?
2.Helium's first ionization energy is 24.6 eV and its second is 54.4 eV. Why is the second so much larger?
3.Problem 3 finds $Z^{*} = 1.84$ , $1.18$ and $1.003$ for sodium's 3s, 3p and 3d states. What does the 3d value mean physically?
4.Why does $(2 p) (3 p)$ have a $^{3} S$ term while $(2 p)^{2}$ does not?
5.Problem 9(a) reproduces seven ions' ionization energies to within half a per cent, but 9(b)'s H⁻ comes out 21% low at the stated $f = 1.8$ . What explains the contrast?
6.Problem 4 asks why 3d holds ten electrons and 4f holds fourteen. What does the answer depend on?