Nuclear energy

Nuclear fission is the splitting of a heavy nucleus (U-235, Pu-239) after absorbing a neutron into two medium nuclei, with 2–3 neutrons and about 200 MeV of energy per fission. Nuclear fusion is the joining of two light nuclei (isotopes of hydrogen) into a heavier nucleus with release of energy, as in the sun.

Main reaction: $^2_1D + {}^3_1T \rightarrow {}^4_2He + {}^1_0n + 17.6\ \text{MeV}$

Requirements for fusion

Nuclei are positively charged and repel each other (Coulomb barrier). To fuse they must have very high kinetic energy, which gives these requirements:

1. High temperature: about 100–150 million K for D-T (10–15 keV) so the fuel is a fully ionised plasma.
2. Sufficient density $n$ of the plasma, so that collisions are frequent.
3. Confinement time $\tau$: the plasma must be held away from walls long enough.
4. Lawson criterion: to produce more energy than is spent in heating, the product $n\tau \ge 10^{20}\ \text{s/m}^3$ for D-T at about 10 keV (triple product $nT\tau \approx 3\times10^{21}\ \text{keV s/m}^3$).
5. Fuel supply of deuterium (from seawater) and tritium (bred from lithium in the blanket), and purity of plasma.

Confinement approaches

• Magnetic confinement: the plasma is held in a magnetic field in a doughnut-shaped chamber (tokamak, stellarator). Low density, seconds-long confinement.
• Inertial confinement: a small fuel pellet is compressed and heated by lasers or ion beams to extremely high density for a very short time (about 10⁻¹¹ s), so the fusion burns before the pellet blows apart.

Mass defect of D-T fusion

Reaction: $D + T \rightarrow He + n$

$$\begin{aligned} m_{in} &= 2.014102 + 3.016049 = 5.030151\ \text{u} \\ m_{out} &= 4.002603 + 1.008665 = 5.011268\ \text{u} \\ \Delta m &= 0.018883\ \text{u} \end{aligned}$$ $$E = \Delta m \times 931.5 = 0.018883 \times 931.5 = 17.59\ \text{MeV}$$

Energy per kg of D-T fuel

One reaction uses 5.030 u of fuel, i.e. 1 kg holds $\dfrac{1000\ \text{g}}{5.030\ \text{g/mol}} \times N_A = 1.197\times10^{26}$ reactions.

$$E_{fus} = 1.197\times10^{26} \times 17.59 \times 1.602\times10^{-13} = 3.37\times10^{14}\ \text{J/kg}$$

Energy per kg of U-235

Atoms in 1 kg: $\dfrac{1000}{235}\times 6.022\times10^{23} = 2.563\times10^{24}$.

$$E_{fis} = 2.563\times10^{24} \times 200 \times 1.602\times10^{-13} = 8.21\times10^{13}\ \text{J/kg}$$

Comparison and coal equivalent

$$\frac{E_{fus}}{E_{fis}} = \frac{3.37\times10^{14}}{8.21\times10^{13}} = 4.1$$

Coal equivalent of 1 kg U-235:

$$\frac{8.21\times10^{13}}{25\times10^{6}} = 3.28\times10^{6}\ \text{kg} \approx 3280\ \text{tonnes}$$

Answer: D-T fusion releases 17.59 MeV; fusion gives $3.37\times10^{14}$ J/kg, about 4.1 times the $8.21\times10^{13}$ J/kg of fission; 1 kg of U-235 is equal to about 3.3 million kg (3280 tonnes) of coal.

(a) Health hazards

Ionising radiation (alpha, beta, gamma, X-rays, neutrons) breaks chemical bonds and damages cells.

• Acute (deterministic) effects from a high dose in short time: radiation sickness (nausea, hair loss, burns) and death above about 4–5 Sv whole-body.
• Stochastic (delayed) effects: cancer (leukaemia, thyroid, lung), shortening of life, appear after years; probability rises with dose.
• Genetic effects: mutations in reproductive cells passed to children.
• Internal exposure by inhaling or swallowing radioactive material (e.g. iodine-131 to thyroid, radon, strontium-90 to bones) is dangerous for alpha emitters.

Units: activity in becquerel (Bq); absorbed dose in gray (Gy); equivalent dose in sievert (Sv). The public limit is about 1 mSv per year and radiation workers about 20 mSv per year (averaged, ICRP).

Principles of protection

1. Time: reduce time of exposure; dose is proportional to time.
2. Distance: dose falls with the square of the distance from a point source, $I \propto 1/r^2$.
3. Shielding: place absorbers between source and person: lead and concrete for gamma and X-rays, plastic or perspex for beta, water or polythene (hydrogen-rich) and boron/cadmium for neutrons, and paper or skin for alpha. Also containment, ventilation, monitoring badges (TLD), and the ALARA principle (as low as reasonably achievable).

(b) Shielding calculation

Attenuation law: $I = I_0 e^{-\mu x}$.

Half-value layer:

$$x_{1/2} = \frac{\ln 2}{\mu} = \frac{0.693}{0.806} = 0.860\ \text{cm}$$

For 1 % transmission, $I/I_0 = 0.01$:

$$x = \frac{\ln 100}{\mu} = \frac{4.605}{0.806} = 5.71\ \text{cm}$$

Distance effect:

$$D_4 = D_1\left(\frac{1}{4}\right)^2 = \frac{4}{16} = 0.25\ \text{mSv/h}$$

Answer: HVL = 0.86 cm; thickness for 1 % = 5.7 cm of lead; dose rate at 4 m = 0.25 mSv/h.