The Workbook

\(R = 1.2\,A^{1/3}\) fm. The cube roots are \(2\), \(3\), \(4\) and \(5\).

\(A = 8\): \(2.4\) fm. \(A = 27\): \(3.6\) fm. \(A = 64\): \(4.8\) fm. \(A = 125\): \(6.0\) fm.

1. \(\dfrac{R_{216}}{R_{27}} = \left(\dfrac{216}{27}\right)^{1/3} = 8^{1/3} = 2\). Eight times the nucleons, twice the radius.

2. \(V \propto R^3 \propto A\). \(\dfrac{V_8}{V_{64}} = \dfrac{8}{64} = \dfrac{1}{8}\). The volume is proportional to the number of nucleons: each nucleon takes the same room, like marbles in a bag.

1. \(\rho = \dfrac{Am}{\frac{4}{3}\pi R_0^3 A} = \dfrac{3m}{4\pi R_0^3}\). The \(A\) cancels, so every nucleus has the same density.

2. \(\rho = \dfrac{3 \times 1.67 \times 10^{-27}}{4\pi \times (1.2 \times 10^{-15})^3} = \dfrac{5.0 \times 10^{-27}}{2.17 \times 10^{-44}} \approx 2.3 \times 10^{17}\) kg/m\(^3\).

   This is \(2.3 \times 10^{14}\) times the density of water. A teaspoon of nuclear matter would weigh about a billion tonnes.

1. \(R = R_0 A^{1/3}\) is of the form \(y = mx\) with \(x = A^{1/3}\): a straight line through the origin of slope \(R_0\).

2. At \(A^{1/3} = 5\), \(R = 6.0\) fm. Slope \(R_0 = \dfrac{6.0}{5} = 1.2\) fm.

3. \(A = 64\) gives \(A^{1/3} = 4\). The line is at \(R = 4.8\) fm there.

\(A^{1/3} = \dfrac{R}{R_0} = \dfrac{6.0}{1.2} = 5\), so \(A = 125\).

Mass \(\approx 125 \times 1.67 \times 10^{-27} = 2.1 \times 10^{-25}\) kg.

\(M = \rho \times \dfrac{4}{3}\pi R^3\), so \(R^3 = \dfrac{3M}{4\pi\rho} = \dfrac{3 \times 2 \times 10^{30}}{4\pi \times 2.3 \times 10^{17}} = 2.1 \times 10^{12}\) m\(^3\).

\(R = (2.1 \times 10^{12})^{1/3} \approx 1.3 \times 10^{4}\) m \(= 13\) km. The Sun, \(7 \times 10^{5}\) km across, would shrink to the size of a city.

\(E = mc^2 = 1.6605 \times 10^{-27} \times (2.998 \times 10^{8})^2 = 1.6605 \times 10^{-27} \times 8.988 \times 10^{16} = 1.492 \times 10^{-10}\) J.

In eV: \(\dfrac{1.492 \times 10^{-10}}{1.602 \times 10^{-19}} = 9.316 \times 10^{8}\) eV \(= 931.6\) MeV, which rounds to the usual \(931.5\) MeV.

1. \(\Delta m = (1.00728 + 1.00866) - 2.01355 = 2.01594 - 2.01355 = 0.00239\) u.

2. \(B = 0.00239 \times 931.5 \approx 2.23\) MeV. Per nucleon: \(\dfrac{2.23}{2} \approx 1.11\) MeV. The deuteron is loosely bound.

1. Sum of parts: \(2 \times 1.00728 + 2 \times 1.00866 = 4.03188\) u.

   \(\Delta m = 4.03188 - 4.00151 = 0.03037\) u. \(B = 0.03037 \times 931.5 \approx 28.3\) MeV.

2. \(\dfrac{B}{A} = \dfrac{28.3}{4} \approx 7.07\) MeV, far more than the deuteron's \(1.1\) MeV. Helium-4 is very tightly bound.

\(B = 8.8 \times 56 \approx 493\) MeV.

\(\Delta m = \dfrac{493}{931.5} \approx 0.529\) u, about half a nucleon mass.

Removing one neutron turns \(^4\)He into \(^3\)He. Energy needed \(= B(^4\text{He}) - B(^3\text{He}) = 28.3 - 7.7 = 20.6\) MeV.

The average \(B/A\) is \(7.1\) MeV, but the last nucleon is held by all three others in a closed, very stable arrangement. Breaking it costs much more than the average.

1. \(^2\)H: \(1.1\) MeV. \(^4\)He: \(7.1\) MeV. \(^{56}\)Fe: \(8.8\) MeV. \(^{235}\)U: \(7.6\) MeV.

2. \(^{56}\)Fe, at the peak of the curve. Its nucleons are the hardest to pull apart.

3. Fission releases energy when a heavy nucleus (\(A > 170\), on the falling part of the curve) splits into pieces with higher \(B/A\). Fusion releases energy when light nuclei (\(A < 30\), on the rising part) join into a nucleus with higher \(B/A\).

Binding energy before: \(235 \times 7.6 = 1786\) MeV. After: \(235 \times 8.5 \approx 1998\) MeV.

The nucleons are bound more tightly by about \(1998 - 1786 \approx 210\) MeV. This energy is released, mostly as kinetic energy of the fragments. The usual figure is \(200\) MeV.

Before: two deuterons, \(4\) nucleons at \(1.1\) MeV each, \(B = 4.4\) MeV. After: helium-4, \(4 \times 7.1 = 28.4\) MeV.

Energy released \(= 28.4 - 4.4 = 24\) MeV. Per nucleon this is far more than fission gives, which is why fusion powers the stars.

1. The nuclear force has a very short range. Each nucleon binds only to its nearest neighbours, so adding more nucleons adds binding in proportion to \(A\). The force is saturated, and \(B/A\) stays nearly constant.

2. Every proton repels every other proton electrically, and this force has a long range. In a heavy nucleus the repulsion grows faster than the short-range attraction, so \(B/A\) falls.

3. In a small nucleus most nucleons are on the surface and have few neighbours to bind to. So the binding per nucleon is low.

1. \(2p + 2n\): \(4.03188\) u. \(^4\)He: \(4.00151\) u. \(\Delta m = 0.03037\) u.

2. \(E = 0.03037 \times 931.5 \approx 28.3\) MeV.

3. The energy is carried away as gamma-ray photons and as kinetic energy of the products. The nucleus is left lighter by exactly this energy.

X: \(\dfrac{B}{A} = \dfrac{500}{60} \approx 8.3\) MeV. Y: \(\dfrac{B}{A} = \dfrac{1800}{240} = 7.5\) MeV. X is more stable, although its total \(B\) is smaller.

To break up a nucleus you must free each nucleon. The energy needed per nucleon, \(B/A\), tells how firmly the average nucleon is held. A big nucleus has a large \(B\) only because it has many nucleons.

1. \(216^{1/3} = 6\), so \(R_0 = \dfrac{7.2}{6} = 1.2\) fm.

2. \(27^{1/3} = 3\), so \(R = 1.2 \times 3 = 3.6\) fm. The ratio \(3.6 : 7.2 = 1 : 2\) matches the drawing.

3. Neither. Nuclear density is the same for every nucleus, about \(2.3 \times 10^{17}\) kg/m\(^3\).

1. \(28.3\) MeV \(= 28.3 \times 10^{6} \times 1.6 \times 10^{-19} = 4.5 \times 10^{-12}\) J.

2. \(6.02 \times 10^{23} \times 4.5 \times 10^{-12} \approx 2.7 \times 10^{12}\) J, for only \(4\) g of helium. Nuclear energies are millions of times chemical energies.

1. It is very short-ranged: strong up to about \(1\) fm, and practically zero beyond a few fm. The electric force falls off slowly as \(1/r^2\).

2. It is the same for proton-proton, proton-neutron and neutron-neutron pairs: it does not depend on charge.

3. It saturates: a nucleon binds only to its nearest neighbours. It is also about \(100\) times stronger than the electric force at \(1\) fm.

1. \(N = \dfrac{1000}{235} \times 6.02 \times 10^{23} \approx 2.56 \times 10^{24}\) nuclei.

2. \(200\) MeV \(= 200 \times 1.6 \times 10^{-13} = 3.2 \times 10^{-11}\) J. \(E = 2.56 \times 10^{24} \times 3.2 \times 10^{-11} \approx 8.2 \times 10^{13}\) J. Burning coal gives about \(3 \times 10^{7}\) J per kg, nearly three million times less.

1. The order follows \(B/A\). Most stable first: \(^{56}\)Fe, \(^{235}\)U, \(^4\)He, \(^2\)H.

2. Splitting releases energy only if the pieces have higher \(B/A\): uranium. Fusing releases energy if the product has higher \(B/A\): deuterium and helium.

3. Before: \(56 \times 8.8 = 492.8\) MeV. After: \(56 \times 8.4 = 470.4\) MeV. The products are less bound, so \(22\) MeV must be supplied. Iron cannot be a nuclear fuel either way.

Mass of parts \(= 938.3 + 939.6 = 1877.9\) MeV/\(c^2\). The binding energy is missing from the mass: \(m_d = 1877.9 - 2.2 = 1875.7\) MeV/\(c^2\).

To pull the nucleons apart you must do work \(B\) against the nuclear force. So the separated nucleons have energy \(B\) more than the nucleus, and by \(E = mc^2\) they have mass \(B/c^2\) more.

The "missing" mass did not vanish. When the nucleus formed, this energy left as radiation and kinetic energy. Mass and energy together are conserved.

1. \(A\): \(1 + 235 = 141 + 92 + x\), so \(x = 3\) neutrons. \(Z\): \(92 = 56 + 36\). Correct.

2. \(A\): \(236 = 140 + 94 + x\), so \(x = 2\) neutrons. \(Z\): \(92 = 54 + 38\). Correct.

Before: \(235.0439 + 1.0087 = 236.0526\) u. After: \(140.9144 + 91.9262 + 3 \times 1.0087 = 235.8667\) u.

\(\Delta m = 0.1859\) u. \(E = 0.1859 \times 931.5 \approx 173\) MeV. The fragments are also radioactive, and their later decays bring the total to about \(200\) MeV.

Each of the \(235\) nucleons is bound by about \(7.6\) MeV in uranium and about \(8.5\) MeV in the fragments. Every nucleon becomes more tightly bound by about \(0.9\) MeV.

Energy released \(\approx 0.9 \times 235 \approx 210\) MeV, close to \(200\) MeV. The energy comes from the fall in mass as binding increases.

1. The humps are at \(A \approx 95\) and \(A \approx 140\).

2. \(95 + 140 = 235\). The compound nucleus has \(A = 236\), but two or three neutrons are set free, so the two fragments share only about \(233\) to \(234\). (The humps are broad; the sum of the most common pair is roughly this.)

3. Unequal parts. Equal halves (\(A = 118\)) are near the dip in the middle: they are rare.

1. Rate \(= \dfrac{P}{E} = \dfrac{10^{9}}{3.2 \times 10^{-11}} \approx 3.1 \times 10^{19}\) fissions per second.

2. Per day: \(3.1 \times 10^{19} \times 86400 = 2.7 \times 10^{24}\) nuclei.

   Mass \(= \dfrac{2.7 \times 10^{24}}{6.02 \times 10^{23}} \times 235 \approx 1050\) g \(\approx 1.05\) kg. About one kilogram of uranium a day for a large power station.

1. \(^{235}\)U captures slow (thermal) neutrons far more readily than fast ones. Fast neutrons mostly escape or are caught by \(^{238}\)U without fission.

2. A moderator is a material in which neutrons lose energy by elastic collisions until they are thermal. Ordinary water, heavy water and graphite are used.

3. In an elastic collision the energy transferred is greatest when the two masses are equal. A neutron bouncing off a proton can lose all its energy; bouncing off a heavy nucleus it loses almost none.

1. \(\dfrac{4 \times 1}{2^2} = 1\). All of it, in one head-on collision.

2. \(\dfrac{48}{169} \approx 0.28\).

3. \(\dfrac{952}{239^2} = \dfrac{952}{57121} \approx 0.017\). Only \(1.7\%\); uranium itself is useless as a moderator.

\(k\) is the ratio of the number of fissions in one generation to the number in the previous generation.

\(k < 1\): subcritical, the reaction dies out. \(k = 1\): critical, a steady rate. \(k > 1\): supercritical, the rate grows without limit (a bomb). A working reactor is held at \(k = 1\) by its control rods.

1. Each fission leads to two more, so \(k = 2\). Generation \(3\) has \(2^3 = 8\) fissions (the eight dots on the right).

2. Generation \(10\): \(2^{10} = 1024\) fissions.

3. \(1024 \times 200 = 2.05 \times 10^{5}\) MeV \(= 2.05 \times 10^{5} \times 1.6 \times 10^{-13}\) J \(\approx 3.3 \times 10^{-8}\) J. Still tiny; the danger is that generations take only about a microsecond each.

1. \(\dfrac{N}{N_0} = (1.01)^{100} \approx 2.7\). Even \(1\%\) above critical, the rate nearly triples in \(100\) generations.

2. \(k^n = 1000\), so \(n = \dfrac{\ln 1000}{\ln 1.01} = \dfrac{6.9}{0.01} \approx 690\) generations.

1. Neutrons are born throughout the volume but escape through the surface. In a small lump the surface is large compared with the volume, so most neutrons leave before they meet a \(^{235}\)U nucleus. \(k\) stays below \(1\).

2. For a given mass a sphere has the smallest surface area, so the fewest neutrons escape. A thin sheet lets nearly every neutron out through its two faces. The critical mass is the least mass for which \(k\) reaches \(1\).

Fuel rods: hold the fissile material (enriched uranium oxide) where neutrons can reach it.

Moderator: slows the neutrons to thermal speeds (water, heavy water, graphite).

Control rods: absorb neutrons to hold \(k\) at \(1\) or shut the reactor down (cadmium, boron steel).

Coolant: carries heat from the core to the heat exchanger and boiler (water, liquid sodium, CO\(_2\)).

Shielding: stops neutrons and gamma rays leaving (thick concrete, steel, lead).

P: the thick outer wall is the shielding (concrete), stopping radiation.

Q: the dark rod that can be pushed into the core is a control rod (cadmium or boron); it absorbs neutrons.

R: the liquid around the fuel rods is the moderator, and here also the coolant (water).

S: the whole block is the reactor core, fuel rods in the moderator.

T: the box in the loop is the heat exchanger. Hot coolant (red) gives its heat to water, making steam for the turbine, and returns cold (blue).

\(\Delta m = 4 \times 1.00783 - 4.00260 = 4.03132 - 4.00260 = 0.02872\) u.

\(E = 0.02872 \times 931.5 \approx 26.7\) MeV per helium nucleus, of which the neutrinos carry away a little.

Step 1: \({}^1\)H \(+ {}^1\)H \(\to {}^2\)H \(+ e^+ + \nu\). Step 2: \({}^2\)H \(+ {}^1\)H \(\to {}^3\)He \(+ \gamma\). Step 3: \({}^3\)He \(+ {}^3\)He \(\to {}^4\)He \(+ 2\,{}^1\)H. Steps 1 and 2 happen twice for each step 3.

In the carbon-nitrogen cycle a \(^{12}\)C nucleus captures protons one by one and passes through nitrogen and oxygen isotopes. At the end it gives back a \(^{12}\)C and a \(^4\)He. The carbon is not used up; it is a catalyst.

1. Rate \(= \dfrac{3.9 \times 10^{26}}{4.27 \times 10^{-12}} \approx 9.1 \times 10^{37}\) helium nuclei per second.

2. \(E = mc^2\), so \(\dfrac{dm}{dt} = \dfrac{P}{c^2} = \dfrac{3.9 \times 10^{26}}{9 \times 10^{16}} \approx 4.3 \times 10^{9}\) kg/s. Four million tonnes a second, yet the Sun can go on for billions of years.

\(U = \dfrac{ke^2}{r} = \dfrac{9 \times 10^{9} \times (1.6 \times 10^{-19})^2}{2 \times 10^{-15}} = \dfrac{2.3 \times 10^{-28}}{2 \times 10^{-15}} = 1.15 \times 10^{-13}\) J.

In MeV: \(\dfrac{1.15 \times 10^{-13}}{1.6 \times 10^{-13}} \approx 0.72\) MeV.

1. \(\dfrac32 kT = 1.15 \times 10^{-13}\) J, so \(T = \dfrac{2 \times 1.15 \times 10^{-13}}{3 \times 1.38 \times 10^{-23}} \approx 5.6 \times 10^{9}\) K.

2. First, \(\frac32 kT\) is only the average; a small fraction of protons in the high-energy tail have far more. Second, quantum tunnelling lets protons pass through the barrier even when \(K\) is below its top. Both together make fusion slow but steady at \(1.5 \times 10^{7}\) K.

1. Reactions per kg \(= \dfrac{1}{8.3 \times 10^{-27}} = 1.2 \times 10^{26}\).

   Energy \(= 1.2 \times 10^{26} \times 17.6 \times 1.6 \times 10^{-13} \approx 3.4 \times 10^{14}\) J.

2. \(\dfrac{3.4 \times 10^{14}}{8.2 \times 10^{13}} \approx 4\). Fusion fuel gives about four times more energy per kilogram than fission fuel.

1. The hill outside \(r = R\), falling as \(1/r\), is the electric repulsion. The deep flat well inside \(R\) is the strong nuclear attraction.

2. No. Classically a particle cannot enter a region where \(U > K\); its kinetic energy would be negative. The protons would stop and turn back at the point where the red line meets the curve.

3. Tunnelling. The proton's wave has a small but non-zero amplitude beyond the barrier, so there is a small probability of appearing inside the well. With enormous numbers of protons, this small probability gives a steady fusion rate.

1. The reactants are at rest, so the total momentum is zero. After the reaction the two products must have equal and opposite momenta.

2. With equal momenta \(p\), \(K = \dfrac{p^2}{2m}\), so \(\dfrac{K_n}{K_\alpha} = \dfrac{m_\alpha}{m_n} = \dfrac{4}{1}\).

   \(K_n + K_\alpha = 17.6\) MeV: \(K_n = \dfrac45 \times 17.6 = 14.1\) MeV and \(K_\alpha = 3.5\) MeV. The light neutron takes most of the energy.

Harder: the fuel must be heated to over \(10^{8}\) K to cross the Coulomb barrier, and no solid wall can hold such a plasma. It must be confined by magnetic fields or by inertia. The confinement must last long enough, at high enough density, to give back more energy than was put in.

Worth it: deuterium is plentiful in sea water and the fuel is cheap. The products (helium) are not radioactive, there is no long-lived waste, and a runaway chain reaction is impossible.