9.P — Worked Problems: Atoms and Nuclei

Twelve problems across Part 9. Every solution shows the arithmetic. Attempt each before opening it.

Problem 1 — Closest approach in Rutherford scattering

A 7.7 MeV alpha particle is fired head-on at a silver nucleus (Z = 47). Find the distance of closest approach, and compare with the nuclear radius.

Solution

All kinetic energy converts to electrostatic potential energy at the turning point:

\frac{1}{2}mv^2 = \frac{k(2e)(Ze)}{r_{\min}}

r_{\min} = \frac{2Zke^2}{E}

A useful shortcut: ke^2 = 1.44 MeV·fm.

r_{\min} = \frac{2(47)(1.44\ \text{MeV fm})}{7.7\ \text{MeV}} = \frac{135.4}{7.7} = 17.6\ \text{fm}

Nuclear radius of silver (A = 108):

R = 1.2\times(108)^{1/3} = 1.2\times4.76 = 5.7\ \text{fm}

The alpha stops three times further out than the nuclear surface.

What to notice. Because the alpha never touches the nucleus, the scattering is purely Coulombic and Rutherford's formula holds exactly. This is why the experiment worked — had the alphas been energetic enough to reach the nucleus, the strong force would have intervened and the simple analysis would have failed.

Turn it around and it becomes a measuring technique: raise the energy until the scattering departs from Rutherford's formula, and the departure point gives the nuclear radius. This is how nuclear sizes were first measured, and it is where R = 1.2A^{1/3} fm comes from.

Problem 2 — Binding energy of uranium-238

Uranium-238 has an atomic mass of 238.050788 u. Find its total binding energy and binding energy per nucleon, and compare with iron-56.

Solution

Z = 92, N = 146. Using atomic masses so electron masses cancel:

92(1.007825)+146(1.008665) = 92.71990+147.26509 = 239.98499\ \text{u}

\Delta m = 239.98499-238.05079 = 1.93420\ \text{u}

B = 1.93420\times931.494 = 1801.7\ \text{MeV}

\frac{B}{A} = \frac{1801.7}{238} = 7.57\ \text{MeV/nucleon}

Iron-56 gives 8.79 MeV/nucleon (Chapter 9.2).

Difference: 8.79-7.57 = 1.22 MeV per nucleon.

What to notice. Splitting uranium into two iron-sized fragments would release about 1.2 MeV per nucleon, and with 238 nucleons that is 285 MeV. Real fission releases about 200 MeV, because the fragments are not iron — they land around A = 95 and A = 140, where the binding is about 8.5 MeV/nucleon rather than 8.79.

Note also the fractional mass loss: 1.934/239.985 = 0.81 %. Less than one percent of the mass of a uranium nucleus is binding energy, and that fraction runs every nuclear power station.

Problem 3 — Q value and alpha energy

Radium-226 decays to radon-222 by alpha emission. Masses: 226.025410, 222.017578, 4.002603 u. Find the Q value and the kinetic energy of the alpha particle.

Solution

Q = \left[226.025410-222.017578-4.002603\right]\times931.494

= 0.005229\times931.494 = 4.871\ \text{MeV}

The energy splits by momentum conservation. The two products have equal and opposite momenta, so:

\frac{p^2}{2m_\alpha}+\frac{p^2}{2m_{\text{Rn}}} = Q

KE_\alpha = Q\times\frac{m_{\text{Rn}}}{m_\alpha+m_{\text{Rn}}} = 4.871\times\frac{222.0}{226.0} = 4.785\ \text{MeV}

And the recoiling radon nucleus gets:

KE_{\text{Rn}} = 4.871-4.785 = 0.086\ \text{MeV} = 86\ \text{keV}

What to notice. The alpha takes 98.2 % of the energy, and the measured value is 4.784 MeV — matching to four figures.

The 86 keV of recoil is small and not negligible. It is roughly 10,000 times a chemical bond energy, so the recoiling radon atom is ejected from whatever it was chemically bound to. This is why radioactive decay breaks up the crystal it happens in, and why old minerals containing uranium show radiation damage tracks that can themselves be used for dating.

Problem 4 — Radioactive dating with uranium

A rock contains uranium-238 and lead-206 in a mole ratio of 1.00 to 0.60. Given t_{1/2} = 4.468\times10^{9} years, find the rock's age. Assume all the lead came from uranium decay.

Solution

Every lead-206 atom was once a uranium-238 atom, so the original uranium was 1.00+0.60 = 1.60 units.

\frac{N}{N_0} = \frac{1.00}{1.60} = 0.625

\lambda = \frac{0.693}{4.468\times10^{9}} = 1.551\times10^{-10}\ \text{y}^{-1}

0.625 = e^{-\lambda t} \quad\Longrightarrow\quad \ln(0.625) = -\lambda t

-0.4700 = -(1.551\times10^{-10})t

t = \frac{0.4700}{1.551\times10^{-10}} = 3.03\times10^{9}\ \text{years}

About three billion years old.

What to notice. This method — with refinements — gave the Earth's age as 4.54\times10^{9} years, from Clair Patterson's 1956 measurement on meteorites. The assumption that no initial lead was present is the weak point, and Patterson's solution was to use the isochron method: measure several minerals from the same rock, plot isotope ratios against each other, and the slope gives the age while the intercept gives the initial lead, so the assumption is no longer needed.

Patterson also discovered, in the course of this work, that his samples were contaminated by industrial lead from petrol additives, and spent the rest of his career campaigning to get lead out of fuel. A geochronology problem removed lead from petrol worldwide.

Problem 5 — Activity of a source

A medical source contains 2.0 μg of technetium-99m (t_{1/2} = 6.01 h, atomic mass 99 u). Find its initial activity in becquerels, and its activity after 24 hours.

Solution

Number of atoms:

N = \frac{2.0\times10^{-6}\ \text{g}}{99\ \text{g/mol}}\times6.022\times10^{23} = (2.020\times10^{-8})(6.022\times10^{23}) = 1.216\times10^{16}

Decay constant:

\lambda = \frac{0.693}{6.01\times3600\ \text{s}} = \frac{0.693}{21636} = 3.203\times10^{-5}\ \text{s}^{-1}

Activity:

A_0 = \lambda N = (3.203\times10^{-5})(1.216\times10^{16}) = 3.90\times10^{11}\ \text{Bq} = 390\ \text{GBq}

After 24 hours — four half-lives:

A = \frac{390}{2^4} = \frac{390}{16} = 24.4\ \text{GBq}

What to notice. Two micrograms gives 390 gigabecquerels. A typical patient dose is about 0.8 GBq, so this source would supply roughly 500 doses — from a quantity of material invisible to the eye.

Technetium-99m's 6-hour half-life is the reason it dominates nuclear medicine: long enough to transport, inject and image, short enough that the patient's exposure ends within a day. And it is supplied as molybdenum-99 (t_{1/2} = 66 h), which decays into it — hospitals receive a "technetium generator" and elute fresh Tc-99m from it daily for a week. The 66-hour parent is what makes the 6-hour daughter practical.

Problem 6 — Effective nuclear charge

Use Slater's rules to compute Z_{\text{eff}} for (a) a 3p electron in sulphur (Z = 16), (b) a 3d electron in copper (Z = 29), (c) a 4s electron in copper.

Solution

(a) Sulphur, 1s^22s^22p^63s^23p^4. For a 3p electron, the same group is [3s,3p] with 6 electrons total, so 5 others:

S = 5(0.35)+8(0.85)+2(1.00) = 1.75+6.80+2.00 = 10.55

Z_{\text{eff}} = 16-10.55 = 5.45

(b) Copper 3d, [\text{Ar}]3d^{10}4s^1. For a 3d electron, the group is [3d] with 10 electrons, so 9 others, and everything to the left counts 1.00:

S = 9(0.35)+18(1.00) = 3.15+18.00 = 21.15

Z_{\text{eff}} = 29-21.15 = 7.85

(c) Copper 4s. The group is [4s,4p] with 1 electron, so none other. The n-1 shell for a 4s electron includes all the n=3 electrons (18 of them: 3s, 3p and 3d), and the n=1 and n=2 shells count 1.00:

S = 0(0.35)+18(0.85)+10(1.00) = 15.30+10.00 = 25.30

Z_{\text{eff}} = 29-25.30 = 3.70

What to notice. The 3d electron feels Z_{\text{eff}} = 7.85 and the 4s electron only 3.70 — more than double. This is why copper's outer electron is the 4s one and why it ionises to Cu⁺ by losing 4s, leaving the filled 3d^{10} intact.

It also explains why copper conducts so well. A single, weakly held s electron per atom with Z_{\text{eff}} = 3.7 delocalises readily into the metallic sea, and the filled d shell below it is compact and does not scatter the conduction electrons much. Compare with iron, whose partly filled d shell provides scattering states and whose resistivity is six times higher.

Problem 7 — Ionisation energy estimate

Estimate the first ionisation energy of potassium using Z_{\text{eff}} = 2.20 (from Chapter 9.3) and compare with the measured 4.34 eV.

Solution

IE \approx \frac{13.6\,Z_{\text{eff}}^2}{n^2}

For the 4s electron, n = 4:

IE = \frac{13.6\times(2.20)^2}{16} = \frac{13.6\times4.84}{16} = \frac{65.82}{16} = 4.11\ \text{eV}

Measured: 4.34 eV. Within 5 %.

Now try it with n = 3 to see how much the shell matters:

IE = \frac{13.6\times4.84}{9} = 7.31\ \text{eV}

Nearly 70 % too high.

What to notice. The agreement at n = 4 is better than this crude model deserves, and it works because Slater's rules were fitted with exactly this kind of estimate in mind.

The n = 3 comparison makes the point about periodic trends concrete: moving down a group changes n, and since IE \propto 1/n^2, that single change accounts for most of the drop from lithium (5.39 eV, n=2) to potassium (4.34 eV, n=4) — partly offset by Z_{\text{eff}} rising from 1.30 to 2.20. Two competing effects, and n wins.

Problem 8 — Electron configurations and ions

Write the ground-state configurations of (a) Fe, (b) Fe³⁺, (c) Br⁻, (d) Cr, and identify the number of unpaired electrons in each.

Solution

(a) Fe, Z = 26:

[\text{Ar}]3d^64s^2

The six 3d electrons in five orbitals: by Hund's rule, five go in singly with parallel spins and the sixth pairs. 4 unpaired.

(b) Fe³⁺. Remove the two 4s electrons first, then one 3d:

[\text{Ar}]3d^5

Five electrons in five orbitals, all parallel. 5 unpaired.

(c) Br⁻, Z = 35 plus one electron = 36:

[\text{Ar}]3d^{10}4s^24p^6 = [\text{Kr}]

0 unpaired — a closed shell, which is why the bromide ion is stable.

(d) Cr, Z = 24, the exception from Chapter 9.3:

[\text{Ar}]3d^54s^1

6 unpaired — five in 3d and one in 4s.

What to notice. Chromium has six unpaired electrons, the most of any element in the fourth period, which is a direct consequence of the exception. The exception exists because six unpaired electrons is such a favourable arrangement.

Fe³⁺'s 3d^5 is why it is so common and stable — a half-filled subshell with maximum exchange energy. This is why iron rusts to Fe(III) rather than stopping at Fe(II), and why Fe³⁺ compounds are so abundant in nature.

Problem 9 — Isotope abundance from atomic weight

Chlorine's atomic weight is 35.453 u, and its two stable isotopes have masses 34.96885 and 36.96590 u. Find their relative abundances.

Solution

Let x be the fraction of ³⁵Cl, so (1-x) is ³⁷Cl:

34.96885x + 36.96590(1-x) = 35.453

34.96885x + 36.96590 - 36.96590x = 35.453

-1.99705x = 35.453-36.96590 = -1.51290

x = \frac{1.51290}{1.99705} = 0.7576

75.76 % chlorine-35 and 24.24 % chlorine-37.

What to notice. The accepted values are 75.76 % and 24.24 %. The atomic weight on the periodic table is not the mass of any actual atom — no chlorine atom weighs 35.453 u — it is a weighted average of isotopes.

This is why atomic weights are not near-integers. Chlorine's 35.45 was one of the strongest arguments against Dalton's atomic theory in the nineteenth century, since Prout had proposed that all elements are multiples of hydrogen and chlorine flatly refused to fit. Soddy's discovery of isotopes in 1913 resolved a century-old objection.

The 3:1 ratio also shows up directly in mass spectrometry: any chlorine-containing molecule gives a pair of peaks two mass units apart with intensities in a 3:1 ratio, which is one of the fastest ways to spot chlorine in an unknown compound.

Problem 10 — Fission energy

Compute the energy released per fission of uranium-235 into barium-141 and krypton-92 plus three neutrons, using the binding energies: U-235 at 7.591 MeV/nucleon, Ba-141 at 8.326, Kr-92 at 8.516.

Solution

Total binding energy before:

B_i = 235\times7.591 = 1783.9\ \text{MeV}

Total binding energy after:

B_f = 141(8.326)+92(8.516) = 1173.97+783.47 = 1957.4\ \text{MeV}

(The three free neutrons have zero binding energy.)

Energy released is the increase in binding:

Q = B_f - B_i = 1957.4-1783.9 = 173.5\ \text{MeV}

What to notice. This matches the 173 MeV computed by masses in Chapter 6.4, as it must — binding energy and mass defect are the same quantity.

Adding the delayed contributions brings the total to about 200 MeV: prompt gammas (7 MeV), beta particles from the fragments (7 MeV), delayed gammas (6 MeV), and antineutrinos (10 MeV, which escape entirely and are lost).

The 10 MeV carried away by antineutrinos is unrecoverable, which is a permanent 5 % loss in every reactor — and it is also the signal used for reactor monitoring in Chapter 8.5. The neutrinos that make a reactor 5 % less efficient are what let you verify what it is doing from outside.

Problem 11 — Nuclear versus chemical energy

Compare the energy per kilogram from (a) burning carbon to CO₂ (393.5 kJ/mol), (b) fissioning uranium-235 (200 MeV per nucleus), and (c) fusing hydrogen to helium (26.7 MeV per four protons).

Solution

(a) Carbon. One mole is 12 g:

\frac{393.5\ \text{kJ}}{0.012\ \text{kg}} = 3.28\times10^{4}\ \text{kJ/kg} = 3.28\times10^{7}\ \text{J/kg}

(b) Uranium-235. Nuclei per kg:

N = \frac{1000}{235}\times6.022\times10^{23} = 2.563\times10^{24}

E = (2.563\times10^{24})(200\times10^{6}\times1.602\times10^{-19}) = (2.563\times10^{24})(3.204\times10^{-11})

= 8.21\times10^{13}\ \text{J/kg}

(c) Hydrogen fusion. Protons per kg:

N = \frac{1000}{1.008}\times6.022\times10^{23} = 5.975\times10^{26}

Each set of four releases 26.7 MeV:

E = \frac{5.975\times10^{26}}{4}\times(26.7\times10^{6}\times1.602\times10^{-19}) = (1.494\times10^{26})(4.277\times10^{-12})

= 6.39\times10^{14}\ \text{J/kg}

Ratios:

\text{Fission} : \text{Chemical} = \frac{8.21\times10^{13}}{3.28\times10^{7}} = 2.5\times10^{6}

\text{Fusion} : \text{Fission} = \frac{6.39\times10^{14}}{8.21\times10^{13}} = 7.8

What to notice. Fission is 2.5 million times better than burning coal, and fusion is nearly eight times better than fission.

The factor of a million comes directly from the energy scales: chemical bonds are electron-volts, nuclear binding is mega-electron-volts. That is the whole explanation, and it is why one kilogram of uranium can replace 2500 tonnes of coal.

For perspective, the fusion figure of 6.4\times10^{14} J/kg is 0.7 % of c^2, exactly as Chapter 6.4 said. And complete matter–antimatter annihilation would give 9\times10^{16} J/kg, another factor of 140 beyond fusion, which is the ceiling.

Problem 12 — Percent ionic character

Compute the percent ionic character of (a) HCl, (b) NaCl, (c) CsF, using Pauling's formula and the electronegativities: H 2.20, Cl 3.16, Na 0.93, Cs 0.79, F 3.98.

Solution

\%\ \text{ionic} = 100\left[1-e^{-0.25(\Delta\chi)^2}\right]

(a) HCl: \Delta\chi = 3.16-2.20 = 0.96

(\Delta\chi)^2 = 0.9216, \qquad 0.25\times0.9216 = 0.2304

\% = 100(1-e^{-0.2304}) = 100(1-0.7943) = 20.6\ \%

(b) NaCl: \Delta\chi = 3.16-0.93 = 2.23

(\Delta\chi)^2 = 4.9729, \qquad 0.25\times4.9729 = 1.2432

\% = 100(1-e^{-1.2432}) = 100(1-0.2884) = 71.2\ \%

(c) CsF: \Delta\chi = 3.98-0.79 = 3.19

(\Delta\chi)^2 = 10.176, \qquad 0.25\times10.176 = 2.544

\% = 100(1-e^{-2.544}) = 100(1-0.0785) = 92.2\ \%

What to notice. Caesium fluoride is the most ionic compound possible — the least electronegative stable element paired with the most electronegative — and it is still only 92 % ionic. There is no such thing as a purely ionic bond.

Equally, HCl at 21 % ionic is usually called covalent, and that 21 % is doing real work: it is why HCl dissolves in water and dissociates completely, while H₂ does not dissolve at all. The labels "ionic" and "covalent" name the ends of a continuum, and most real bonds are somewhere in the middle.

Note also the melting points, which track the ionic character: HCl at −114 °C, NaCl at 801 °C, CsF at 682 °C. The gas-versus-solid distinction is a direct consequence of \Delta\chi, and Chapter 10.1 explains the mechanism.