Chapter 13: Nuclei

The nuclear density is the ratio of the mass of the nucleus to the volume it occupies. We are given the mass of the iron nucleus and its mass number $$A$$.

Step 1 — Mass of the nucleus in kg.

$$m = 55.85\,\mathrm{u} = 55.85 \times 1.6605 \times 10^{-27}\,\mathrm{kg} = 9.27 \times 10^{-26}\,\mathrm{kg}$$

Step 2 — Volume of the nucleus. A nucleus of mass number $$A$$ has radius $$R = R_0 A^{1/3}$$ with $$R_0 = 1.2 \times 10^{-15}\,\mathrm{m}$$. Treating the nucleus as a sphere,

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi R_0^3\,A$$

Step 3 — Nuclear density.

$$\rho = \frac{\text{mass}}{\text{volume}} = \frac{m}{\frac{4}{3}\pi R_0^3\,A} = \frac{9.27 \times 10^{-26}}{\frac{4}{3}\pi\,(1.2 \times 10^{-15})^3 \times 56}$$

$$\rho = 2.29 \times 10^{17}\,\mathrm{kg\,m^{-3}}$$

This is enormously larger than the density of ordinary matter (water is only $$10^3\,\mathrm{kg\,m^{-3}}$$), because almost the entire mass of an atom is packed into the tiny nuclear volume. The density of matter in neutron stars is comparable to this value, which is why such stars resemble one big nucleus.

By Einstein's mass–energy relation, a mass $$m$$ has an energy equivalent $$E = mc^2$$.

Given: $$m = 1\,\mathrm{g} = 10^{-3}\,\mathrm{kg}$$ and $$c = 3 \times 10^8\,\mathrm{m\,s^{-1}}$$.

Energy equivalent.

$$E = mc^2 = 10^{-3} \times (3 \times 10^8)^2$$

$$E = 10^{-3} \times 9 \times 10^{16} = 9 \times 10^{13}\,\mathrm{J}$$

To appreciate how large this is, convert it to kilowatt-hours using $$1\,\mathrm{kWh} = 3.6 \times 10^6\,\mathrm{J}$$:

$$E = \frac{9 \times 10^{13}}{3.6 \times 10^6}\,\mathrm{kWh} = 2.5 \times 10^7\,\mathrm{kWh}$$

Thus the energy locked up in just $$1\,\mathrm{g}$$ of matter is enormous. This is why nuclear processes, which release even a tiny fraction of this rest-mass energy, are so powerful compared with chemical processes.

Step 1 — Energy equivalent of $$1\,\mathrm{u}$$ in joules. One atomic mass unit is $$1\,\mathrm{u} = 1.6605 \times 10^{-27}\,\mathrm{kg}$$. By $$E = mc^2$$, with $$c = 2.9979 \times 10^8\,\mathrm{m\,s^{-1}}$$:

$$E = (1.6605 \times 10^{-27}) \times (2.9979 \times 10^8)^2$$

$$E = 1.6605 \times 10^{-27} \times 8.9874 \times 10^{16} = 1.4924 \times 10^{-10}\,\mathrm{J}$$

Step 2 — Convert to MeV. Using $$1\,\mathrm{MeV} = 1.602 \times 10^{-13}\,\mathrm{J}$$,

$$E = \frac{1.4924 \times 10^{-10}}{1.602 \times 10^{-13}}\,\mathrm{MeV} = 931.5\,\mathrm{MeV}$$

Hence the standard conversion

$$1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$$

Step 3 — Mass defect of $$\mathrm{^{16}_{8}O}$$ in $$\mathrm{MeV}/c^2$$. The nucleus $$\mathrm{^{16}_{8}O}$$ has $$8$$ protons and $$8$$ neutrons; its mass defect (sum of the free-nucleon masses minus the atomic mass) was found in the chapter to be

$$\Delta M = 8\,m_H + 8\,m_n - m(\mathrm{^{16}_{8}O}) = 0.13691\,\mathrm{u}$$

Since energy and mass are related by $$E = (\Delta M)c^2$$, a mass defect can be quoted directly in $$\mathrm{MeV}/c^2$$ simply by multiplying by the conversion factor:

$$\Delta M = 0.13691\,\mathrm{u} = 0.13691 \times 931.5\,\mathrm{MeV}/c^2$$

$$\Delta M \approx 127.5\,\mathrm{MeV}/c^2$$

(Equivalently, the binding energy of $$\mathrm{^{16}_{8}O}$$ is about $$127.5\,\mathrm{MeV}$$.)

Concept. The binding energy of a nucleus is the energy equivalent of its mass defect $$\Delta M$$ — the amount by which the total mass of the separated nucleons exceeds the mass of the bound nucleus.

Step 1 — Composition. The nucleus $$\mathrm{^{14}_{7}N}$$ has $$Z = 7$$ protons and $$N = A - Z = 14 - 7 = 7$$ neutrons.

Step 2 — Mass defect. Using the atomic mass of hydrogen $$m_H = 1.007825\,\mathrm{u}$$ and the neutron mass $$m_n = 1.008665\,\mathrm{u}$$,

$$\Delta M = Z\,m_H + N\,m_n - m(\mathrm{^{14}_{7}N})$$

$$\Delta M = 7(1.007825) + 7(1.008665) - 14.00307$$

$$\Delta M = 7.054775 + 7.060655 - 14.00307$$

$$\Delta M = 14.115430 - 14.00307 = 0.112360\,\mathrm{u}$$

(Using atomic masses is consistent here: the $$7$$ electron masses contained in $$7\,m_H$$ are exactly matched by the $$7$$ electrons of the neutral nitrogen atom $$m(\mathrm{^{14}_{7}N})$$, so they cancel.)

Step 3 — Binding energy. Convert the mass defect to energy using $$1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$$:

$$E_b = \Delta M \times 931.5\,\mathrm{MeV} = 0.112360 \times 931.5$$

$$E_b \approx 104.7\,\mathrm{MeV}$$

Concept. The binding energy is $$E_b = (\Delta M)c^2$$, where the mass defect is $$\Delta M = Z\,m_H + N\,m_n - m(\text{atom})$$. We use $$m_H = 1.007825\,\mathrm{u}$$, $$m_n = 1.008665\,\mathrm{u}$$ and $$1\,\mathrm{u} = 931.5\,\mathrm{MeV}/c^2$$.

(a) Iron, $$\mathrm{^{56}_{26}Fe}$$: here $$Z = 26$$ and $$N = 56 - 26 = 30$$.

$$\Delta M = 26(1.007825) + 30(1.008665) - 55.934939$$

$$\Delta M = 26.203450 + 30.259950 - 55.934939$$

$$\Delta M = 56.463400 - 55.934939 = 0.528461\,\mathrm{u}$$

$$E_b = 0.528461 \times 931.5\,\mathrm{MeV} \approx 492.3\,\mathrm{MeV}$$

Binding energy per nucleon: $$E_{bn} = \dfrac{492.3}{56} \approx 8.79\,\mathrm{MeV}$$.

(b) Bismuth, $$\mathrm{^{209}_{83}Bi}$$: here $$Z = 83$$ and $$N = 209 - 83 = 126$$.

$$\Delta M = 83(1.007825) + 126(1.008665) - 208.980388$$

$$\Delta M = 83.649475 + 127.091790 - 208.980388$$

$$\Delta M = 210.741265 - 208.980388 = 1.760877\,\mathrm{u}$$

$$E_b = 1.760877 \times 931.5\,\mathrm{MeV} \approx 1640.3\,\mathrm{MeV}$$

Binding energy per nucleon: $$E_{bn} = \dfrac{1640.3}{209} \approx 7.85\,\mathrm{MeV}$$.

Comparison. Although bismuth has a much larger total binding energy, its binding energy per nucleon ($$7.85\,\mathrm{MeV}$$) is smaller than that of iron ($$8.79\,\mathrm{MeV}$$). Iron lies near the peak of the binding-energy curve, so each of its nucleons is more tightly bound, making it the more stable nucleus.

The energy required to separate every nucleon in the coin is just the total binding energy of all the copper nuclei it contains. We need (i) the number of atoms and (ii) the binding energy of one nucleus.

Step 1 — Number of atoms in the coin. The coin (mass $$3.0\,\mathrm{g}$$) is taken to be entirely $$\mathrm{^{63}_{29}Cu}$$, of molar mass $$\approx 63\,\mathrm{g\,mol^{-1}}$$. Hence

$$N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{3.0}{63} \times 6.022 \times 10^{23}$$

$$N = 2.868 \times 10^{22}\ \text{atoms}$$

Step 2 — Binding energy of one $$\mathrm{^{63}_{29}Cu}$$ nucleus. Here $$Z = 29$$ and $$N_n = 63 - 29 = 34$$. With $$m_H = 1.007825\,\mathrm{u}$$ and $$m_n = 1.008665\,\mathrm{u}$$,

$$\Delta M = 29\,m_H + 34\,m_n - m(\mathrm{^{63}_{29}Cu})$$

$$\Delta M = 29(1.007825) + 34(1.008665) - 62.92960$$

$$\Delta M = 29.226925 + 34.294610 - 62.92960 = 0.591935\,\mathrm{u}$$

$$E_b = 0.591935 \times 931.5\,\mathrm{MeV} \approx 551.4\,\mathrm{MeV}$$

This is the energy needed to break one copper nucleus into its free protons and neutrons.

Step 3 — Total energy for the whole coin.

$$E = N \times E_b = (2.868 \times 10^{22}) \times 551.4\,\mathrm{MeV}$$

$$E \approx 1.58 \times 10^{25}\,\mathrm{MeV}$$

Converting to joules with $$1\,\mathrm{MeV} = 1.602 \times 10^{-13}\,\mathrm{J}$$:

$$E \approx 1.58 \times 10^{25} \times 1.602 \times 10^{-13}\,\mathrm{J} \approx 2.53 \times 10^{12}\,\mathrm{J}$$

An enormous figure — about $$2.5 \times 10^{12}\,\mathrm{J}$$ stored in an ordinary coin — showing how tightly nucleons are bound in nuclei.

The radius of a nucleus of mass number $$A$$ obeys $$R = R_0 A^{1/3}$$, where $$R_0$$ is the same constant for every nucleus. Therefore the ratio of two nuclear radii depends only on the mass numbers:

$$\frac{R_{\mathrm{Au}}}{R_{\mathrm{Ag}}} = \frac{R_0\,(A_{\mathrm{Au}})^{1/3}}{R_0\,(A_{\mathrm{Ag}})^{1/3}} = \left(\frac{A_{\mathrm{Au}}}{A_{\mathrm{Ag}}}\right)^{1/3}$$

For $$\mathrm{^{197}_{79}Au}$$ the mass number is $$A_{\mathrm{Au}} = 197$$, and for $$\mathrm{^{107}_{47}Ag}$$ it is $$A_{\mathrm{Ag}} = 107$$. Hence

$$\frac{R_{\mathrm{Au}}}{R_{\mathrm{Ag}}} = \left(\frac{197}{107}\right)^{1/3} = (1.8411)^{1/3}$$

$$\frac{R_{\mathrm{Au}}}{R_{\mathrm{Ag}}} \approx 1.23$$

So the gold nucleus is only about $$1.23$$ times larger in radius than the silver nucleus, even though it has almost twice the mass — a direct consequence of the cube-root dependence of nuclear radius on mass number.

The proposed process is $$\mathrm{^{56}_{26}Fe \rightarrow ^{28}_{13}Al + ^{28}_{13}Al}$$. It conserves both mass number ($$56 = 28 + 28$$) and charge ($$26 = 13 + 13$$), so it is allowed by the conservation laws. Whether it can actually occur is decided by the sign of its $$Q$$-value.

Q-value. The electron counts balance ($$26 = 13 + 13$$), so the given atomic masses may be used directly:

$$Q = \left[\,m(\mathrm{^{56}_{26}Fe}) - 2\,m(\mathrm{^{28}_{13}Al})\,\right]c^2$$

$$Q = \left[\,55.93494 - 2(27.98191)\,\right] \times 931.5\,\mathrm{MeV}$$

$$Q = \left[\,55.93494 - 55.96382\,\right] \times 931.5\,\mathrm{MeV}$$

$$Q = (-0.02888) \times 931.5\,\mathrm{MeV} \approx -26.9\,\mathrm{MeV}$$

Conclusion. The $$Q$$-value is negative: this fission would absorb about $$26.9\,\mathrm{MeV}$$ rather than release it. Hence the fission of $$\mathrm{^{56}_{26}Fe}$$ into two $$\mathrm{^{28}_{13}Al}$$ nuclei is not energetically possible — it cannot occur spontaneously.

This is exactly what one expects: $$\mathrm{^{56}_{26}Fe}$$ sits at the peak of the binding-energy-per-nucleon curve, so it is already maximally stable. Splitting such a nucleus produces less tightly bound fragments and therefore requires an input of energy.

Step 1 — Number of atoms in $$1\,\mathrm{kg}$$ of $$\mathrm{^{239}_{94}Pu}$$. The molar mass of plutonium-239 is $$\approx 239\,\mathrm{g\,mol^{-1}}$$, so $$1\,\mathrm{kg} = 1000\,\mathrm{g}$$ contains

$$N = \frac{\text{mass}}{\text{molar mass}} \times N_A = \frac{1000}{239} \times 6.022 \times 10^{23}$$

$$N = 4.184 \times 6.022 \times 10^{23} = 2.520 \times 10^{24}\ \text{atoms}$$

Step 2 — Total energy released. Each fission releases, on average, $$180\,\mathrm{MeV}$$. If every atom undergoes fission, the total energy is

$$E = N \times 180\,\mathrm{MeV} = (2.520 \times 10^{24}) \times 180\,\mathrm{MeV}$$

$$E \approx 4.5 \times 10^{26}\,\mathrm{MeV}$$

So the complete fission of just $$1\,\mathrm{kg}$$ of $$\mathrm{^{239}_{94}Pu}$$ liberates about $$4.5 \times 10^{26}\,\mathrm{MeV}$$ — equivalent to roughly $$7.2 \times 10^{13}\,\mathrm{J}$$ of energy.

Step 1 — Number of deuterium atoms in $$2.0\,\mathrm{kg}$$. Deuterium has molar mass $$\approx 2\,\mathrm{g\,mol^{-1}}$$, so $$2.0\,\mathrm{kg} = 2000\,\mathrm{g}$$ contains

$$N = \frac{2000}{2} \times 6.022 \times 10^{23} = 6.022 \times 10^{26}\ \text{atoms}$$

Step 2 — Number of fusion reactions. Each reaction $$\mathrm{^{2}_{1}H + ^{2}_{1}H \rightarrow ^{3}_{2}He + n}$$ consumes two deuterons. Hence the number of reactions is

$$N_{\text{rxn}} = \frac{N}{2} = \frac{6.022 \times 10^{26}}{2} = 3.011 \times 10^{26}$$

Step 3 — Total energy released. Each reaction releases $$3.27\,\mathrm{MeV}$$, so

$$E = N_{\text{rxn}} \times 3.27\,\mathrm{MeV} = 3.011 \times 10^{26} \times 3.27\,\mathrm{MeV}$$

$$E = 9.846 \times 10^{26}\,\mathrm{MeV}$$

Converting to joules with $$1\,\mathrm{MeV} = 1.602 \times 10^{-13}\,\mathrm{J}$$,

$$E = 9.846 \times 10^{26} \times 1.602 \times 10^{-13}\,\mathrm{J} \approx 1.577 \times 10^{14}\,\mathrm{J}$$

Step 4 — Glowing time of the lamp. The lamp consumes power $$P = 100\,\mathrm{W} = 100\,\mathrm{J\,s^{-1}}$$, so the time for which it can be kept glowing is

$$t = \frac{E}{P} = \frac{1.577 \times 10^{14}}{100} = 1.577 \times 10^{12}\,\mathrm{s}$$

In years (with $$1\ \text{year} \approx 3.154 \times 10^7\,\mathrm{s}$$),

$$t = \frac{1.577 \times 10^{12}}{3.154 \times 10^7}\ \text{years} \approx 5.0 \times 10^4\ \text{years}$$

The lamp could glow for about $$50{,}000$$ years — a striking illustration of the energy density of nuclear fusion.

Setup. As two deuterons approach head-on, they are slowed by their mutual Coulomb repulsion. The height of the potential barrier equals the electrostatic potential energy of the pair at the instant they just touch — modelling each deuteron as a hard sphere of radius $$r = 2.0\,\mathrm{fm}$$.

Step 1 — Separation at contact. When the two spheres touch, the distance between their centres equals the sum of the radii:

$$d = r + r = 2r = 2 \times 2.0\,\mathrm{fm} = 4.0\,\mathrm{fm} = 4.0 \times 10^{-15}\,\mathrm{m}$$

Step 2 — Coulomb potential energy. Each deuteron carries charge $$+e$$. The potential energy of the two charges at separation $$d$$ is

$$U = \frac{1}{4\pi\varepsilon_0}\,\frac{e^2}{d} = \frac{(9 \times 10^9)\,(1.6 \times 10^{-19})^2}{4.0 \times 10^{-15}}$$

$$U = \frac{(9 \times 10^9)(2.56 \times 10^{-38})}{4.0 \times 10^{-15}} = \frac{2.304 \times 10^{-28}}{4.0 \times 10^{-15}}$$

$$U = 5.76 \times 10^{-14}\,\mathrm{J}$$

Step 3 — Express in electron-volts. Dividing by $$1.6 \times 10^{-19}\,\mathrm{J/eV}$$,

$$U = \frac{5.76 \times 10^{-14}}{1.6 \times 10^{-19}}\,\mathrm{eV} = 3.6 \times 10^{5}\,\mathrm{eV} = 360\,\mathrm{keV}$$

So the potential barrier is about $$360\,\mathrm{keV}$$. In a head-on collision this energy is shared equally between the two deuterons, so each must carry roughly $$180\,\mathrm{keV}$$ of kinetic energy to overcome the barrier — far above ordinary thermal energies, which is why fusion needs extremely high temperatures.

Consider a nucleus of mass number $$A$$.

Mass of the nucleus. The nucleus contains $$A$$ nucleons, each of mass approximately $$m$$ (the average nucleon mass, $$m \approx 1.66 \times 10^{-27}\,\mathrm{kg}$$). Hence its mass is

$$M \approx m A$$

Volume of the nucleus. A nucleus is very nearly spherical with radius $$R = R_0 A^{1/3}$$, so its volume is

$$V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi \left(R_0 A^{1/3}\right)^3 = \frac{4}{3}\pi R_0^3\,A$$

Nuclear matter density. Taking the ratio,

$$\rho = \frac{M}{V} = \frac{m A}{\dfrac{4}{3}\pi R_0^3\,A}$$

The mass number $$A$$ appears in both the numerator and the denominator and therefore cancels:

$$\rho = \frac{3m}{4\pi R_0^3}$$

The right-hand side contains only the constants $$m$$, $$R_0$$ and $$\pi$$ — there is no dependence on $$A$$. Hence every nucleus, light or heavy, has essentially the same matter density. (This constancy is a consequence of the short-range, saturating nature of the nuclear force: each nucleon binds only to its nearest neighbours, so nucleons pack at a fixed density regardless of nucleus size.)

Numerical estimate. With $$m \approx 1.66 \times 10^{-27}\,\mathrm{kg}$$ and $$R_0 = 1.2 \times 10^{-15}\,\mathrm{m}$$,

$$\rho = \frac{3(1.66 \times 10^{-27})}{4\pi (1.2 \times 10^{-15})^3} \approx 2.3 \times 10^{17}\,\mathrm{kg\,m^{-3}}$$