Nuclear Energy

The principle behind all nuclear energy

• The $E_{bn}$ curve has a long flat middle region between $A = 30$ and $A = 170$, where $E_{bn}$ is nearly constant at about $8.0$ MeV. Below $A = 30$ and above $A = 170$, $E_{bn}$ is less than $8.0$ MeV.
• The greater the binding energy of a bound system, the LESS its total mass. So if nuclei with less total binding energy transform into nuclei with greater total binding energy, there is a net energy release.
• Two routes to move towards the peak: a heavy nucleus decaying into two or more intermediate-mass fragments (FISSION), or light nuclei joining into a heavier nucleus (FUSION).
• Scale comparison to quote: exothermic chemical reactions (coal, petroleum) involve energies of a few eV; nuclear reactions involve MeV. For the same quantity of matter, a nuclear source yields about a million times more energy. Fission of 1 kg of uranium generates about $10^{14}$ J; burning 1 kg of coal gives about $10^{7}$ J.

Radioactivity — the natural transformations

• Radioactivity was discovered by A. H. Becquerel in 1896, purely by accident: pieces of uranium–potassium sulphate, illuminated with visible light and then wrapped in black paper and separated from a photographic plate by a piece of silver, still blackened the plate after several hours. Something emitted by the compound had penetrated both the paper and the silver.
• Later experiments showed that radioactivity is a NUCLEAR phenomenon: an unstable nucleus undergoes a decay, called radioactive decay.
• Three types occur in nature:
• $\alpha$-decay: a helium nucleus ${}^{4}_{2}\text{He}$ is emitted.
• $\beta$-decay: electrons, or positrons (particles of the same mass as the electron but exactly opposite charge), are emitted.
• $\gamma$-decay: high-energy photons (hundreds of keV or more) are emitted; $\gamma$-rays are electromagnetic radiation of wavelength shorter than X-rays.
• Radioactivity indicates nuclear instability. Stability requires a neutron-to-proton ratio of about $1:1$ for light nuclei, rising to about $3:2$ for heavy nuclei, since extra neutrons are needed to offset the growing proton–proton repulsion. Nuclei with an excess of neutrons or of protons relative to this ratio are unstable; only about $10\%$ of known isotopes are stable.
• Electron capture is a related process the board has set: ${}^{22}_{11}\text{Na} + e^{-} \to {}^{22}_{10}\text{Ne} + \nu_e$, in which the nucleus absorbs an atomic electron.
• An electron and a positron form a particle–antiparticle pair: identical masses, equal and opposite charges. When they meet they annihilate, releasing $\gamma$-ray photons ($e^{+} + e^{-} \to \gamma + \gamma + 1.02$ MeV).

The Q-value of a nuclear reaction

• For a reaction $A + b \to C + d$:

$$Q = \left[m_A + m_b - m_C - m_d\right]c^2 = \left[\sum m_{\text{initial}} - \sum m_{\text{final}}\right]_{\text{u}}\times931.5\ \text{MeV}$$

• Equivalently, by conservation of mass-energy, $Q =$ (final kinetic energy) $-$ (initial kinetic energy).
• $Q > 0$: energy released, reaction EXOTHERMIC and energetically possible. $Q < 0$: energy must be supplied, reaction ENDOTHERMIC and not possible spontaneously.
• When binding energies are given instead of masses:

$$Q = \left(\text{total B.E. of products}\right) - \left(\text{total B.E. of reactants}\right), \qquad \text{total B.E.} = A\times E_{bn}$$

• Nuclear reactions are not "balanced" in the chemical sense — elements are transmuted, so the number of atoms of each element need not be conserved. What IS separately conserved is the number of protons and the number of neutrons, i.e. the total charge number and the total mass number. (At very high energies the strictly conserved quantities are total charge and total baryon number, but that is beyond the syllabus.)

Nuclear fission

• Definition: fission is the process in which a heavy nucleus, on being bombarded by a neutron, splits into two intermediate-mass nuclei of comparable size, releasing energy and two to four neutrons.
• Standard equations from NCERT:

$${}^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \to {}^{236}_{92}\text{U} \to {}^{144}_{56}\text{Ba} + {}^{89}_{36}\text{Kr} + 3\,{}^{1}_{0}\text{n}$$

$${}^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \to {}^{236}_{92}\text{U} \to {}^{133}_{51}\text{Sb} + {}^{99}_{41}\text{Nb} + 4\,{}^{1}_{0}\text{n}$$

$${}^{1}_{0}\text{n} + {}^{235}_{92}\text{U} \to {}^{140}_{54}\text{Xe} + {}^{94}_{38}\text{Sr} + 2\,{}^{1}_{0}\text{n}$$

• The fragment products are themselves radioactive; they emit $\beta$ particles in succession until stable end products are reached.
• Energy released per fissioning nucleus is of the order of $200$ MeV. NCERT's estimate: a nucleus of $A = 240$ has $E_{bn} \approx 7.6$ MeV; the two $A = 120$ fragments have $E_{bn} \approx 8.5$ MeV; the gain is $0.9$ MeV per nucleon, so the total gain is $240\times0.9 = 216$ MeV.
• The disintegration energy first appears as kinetic energy of the fragments and of the emitted neutrons, and is eventually transferred to the surrounding matter as heat.
• Controlled fission in a nuclear reactor is the source of nuclear electricity; UNCONTROLLED fission is what releases the enormous energy of an atom bomb.
• Why heavy nuclei fission spontaneously (1-mark): a heavy nucleus contains a very large number of protons packed close together, so the total Coulomb repulsion between them is very large, while the short-range attractive nuclear force saturates and cannot grow in proportion. The heavy nucleus is therefore energetically unstable, and splits into two smaller, more tightly bound nuclei to lower its total energy.

Nuclear fusion

• Definition: fusion is the process in which two light nuclei combine to form a single heavier nucleus, releasing energy, because the heavier product is more tightly bound (higher $E_{bn}$) than the reactants.
• Energy-liberating examples from NCERT:

$${}^{1}_{1}\text{H} + {}^{1}_{1}\text{H} \to {}^{2}_{1}\text{H} + e^{+} + \nu + 0.42\ \text{MeV}$$

$${}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \to {}^{3}_{2}\text{He} + n + 3.27\ \text{MeV}$$

$${}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \to {}^{3}_{1}\text{H} + {}^{1}_{1}\text{H} + 4.03\ \text{MeV}$$

• In the first, two protons form a deuteron and a positron; in the second, two deuterons form the light isotope of helium; in the third, two deuterons form a triton and a proton.
• The Coulomb barrier: for fusion the nuclei must come close enough for the short-range attractive nuclear force to act, but both carry positive charge and so repel. They need enough energy to overcome this Coulomb barrier, whose height depends on the charges and radii of the two nuclei. For two protons it is about $400$ keV, and it is higher for nuclei of larger charge.
• Estimating the required temperature: setting the average thermal kinetic energy equal to the barrier,

$$\dfrac32 kT = K \approx 400\ \text{keV} \quad\Rightarrow\quad T \sim 3\times10^{9}\ \text{K}$$

• When fusion is achieved by raising the temperature so that the particles have enough kinetic energy to overcome the Coulomb repulsion, it is called THERMONUCLEAR FUSION.
• Worked (NCERT 13.9): the barrier height for a head-on collision of two deuterons, treated as hard spheres of radius $2.0$ fm, is the Coulomb energy at a centre-to-centre separation of $4.0$ fm:

$$U = \dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r} = \dfrac{9\times10^{9}\times(1.6\times10^{-19})^2}{4.0\times10^{-15}} = 5.76\times10^{-14}\ \text{J} = 360\ \text{keV}$$

Energy generation in stars — the proton–proton cycle

• Thermonuclear fusion is the source of the energy output in the interior of stars.
• The Sun's core is at $1.5\times10^{7}$ K, considerably less than the $\sim3\times10^{9}$ K needed for particles of AVERAGE energy to fuse. Fusion in the Sun therefore involves the protons whose energies are far above the average.
• The fusion reaction in the Sun is a multi-step process in which hydrogen is burned into helium; the fuel is the hydrogen in the core. The proton–proton (p, p) cycle is:

$${}^{1}_{1}\text{H} + {}^{1}_{1}\text{H} \to {}^{2}_{1}\text{H} + e^{+} + \nu + 0.42\ \text{MeV} \quad \text{(i)}$$

$$e^{+} + e^{-} \to \gamma + \gamma + 1.02\ \text{MeV} \quad \text{(ii)}$$

$${}^{2}_{1}\text{H} + {}^{1}_{1}\text{H} \to {}^{3}_{2}\text{He} + \gamma + 5.49\ \text{MeV} \quad \text{(iii)}$$

$${}^{3}_{2}\text{He} + {}^{3}_{2}\text{He} \to {}^{4}_{2}\text{He} + {}^{1}_{1}\text{H} + {}^{1}_{1}\text{H} + 12.86\ \text{MeV} \quad \text{(iv)}$$

• For reaction (iv) to occur, the first three must occur TWICE, so that two ${}^{3}_{2}\text{He}$ nuclei are available to unite. Taking $2\times$(i) $+\ 2\times$(ii) $+\ 2\times$(iii) $+$ (iv), the net effect is

$$4\,{}^{1}_{1}\text{H} + 2e^{-} \to {}^{4}_{2}\text{He} + 2\nu + 6\gamma + 26.7\ \text{MeV}$$

• In words: four hydrogen nuclei (with two electrons) combine to form one helium-4 nucleus, releasing about $26.7$ MeV.
• Beyond hydrogen: as the core hydrogen is depleted and becomes helium, the core cools, the star collapses under gravity and the core temperature rises. At about $10^{8}$ K helium nuclei fuse into carbon, and successively heavier elements can be built. But elements more massive than those near the PEAK of the $E_{bn}$ curve cannot be produced this way.
• The Sun is about $5\times10^{9}$ years old and has enough hydrogen for roughly another 5 billion years. After that, hydrogen burning stops, the Sun cools and collapses under gravity, its core temperature rises, its outer envelope expands, and it becomes a RED GIANT.

Controlled thermonuclear fusion

• A thermonuclear fusion device replicates on Earth the process natural to stars.
• The aim in a controlled fusion reactor is steady power generation, by heating the nuclear fuel to a temperature in the range of $10^{8}$ K.
• At such temperatures the fuel is a PLASMA — a mixture of positive ions and electrons.
• The central challenge is CONFINEMENT of this plasma, since no material container can withstand such a temperature. Several countries, India among them, are developing the techniques. If successful, fusion reactors would supply almost unlimited power.

Fission compared with fusion

Worked numericals of every type CBSE sets

From binding energies per nucleon

• A nucleus $A = 240$ with $E_{bn} = 7.6$ MeV breaks into two fragments of $A = 120$ with $E_{bn} = 8.5$ MeV. Initial total B.E. $= 240\times7.6 = 1824$ MeV; final $= 2\times120\times8.5 = 2040$ MeV; energy released $= 2040 - 1824 = 216$ MeV.
• A heavy nucleus $P$, $A = 240$, $E_{bn} = 7.6$ MeV, splits into $Q$ ($A = 110$, $E_{bn} = 8.5$ MeV) and $R$ ($A = 130$, $E_{bn} = 8.4$ MeV). Initial $= 1824$ MeV; final $= 110\times8.5 + 130\times8.4 = 935 + 1092 = 2027$ MeV; energy released $= 2027 - 1824 = 203$ MeV.
• ${}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \to {}^{3}_{2}\text{He} + n$, with B.E.$( {}^{2}_{1}\text{H}) = 2.23$ MeV and B.E.$( {}^{3}_{2}\text{He}) = 7.73$ MeV. Energy released $= 7.73 - (2.23 + 2.23) = 3.27$ MeV. (The neutron is a free nucleon and has no binding energy.)

From masses

• ${}^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \to {}^{4}_{2}\text{He} + n$, with $m( {}^{2}_{1}\text{H}) = 2.014102$ u, $m( {}^{3}_{1}\text{H}) = 3.016049$ u, $m( {}^{4}_{2}\text{He}) = 4.002603$ u, $m_n = 1.008665$ u.
• Reactants: $2.014102 + 3.016049 = 5.030151$ u. Products: $4.002603 + 1.008665 = 5.011268$ u.
• $\Delta m = 0.018883$ u, so energy released $= 0.018883\times931.5 \approx 17.59$ MeV.
• Fission of iron into two aluminium nuclei (a favourite trap): $Q = \left[m( {}^{56}_{26}\text{Fe}) - 2m( {}^{28}_{13}\text{Al})\right]\times931.5 = [55.93494 - 55.96382]\times931.5 = (-0.02888)\times931.5 \approx -26.9$ MeV. The $Q$-value is NEGATIVE, so this fission is NOT energetically possible: energy would have to be supplied, and ${}^{56}_{26}\text{Fe}$ does not spontaneously split this way.
• ${}^{1}_{1}\text{H} + {}^{3}_{1}\text{H} \to {}^{2}_{1}\text{H} + {}^{2}_{1}\text{H}$: $Q = [1.007825 + 3.016049 - 2(2.014102)]\times931.5 = -4.03$ MeV, endothermic.
• ${}^{12}_{6}\text{C} + {}^{12}_{6}\text{C} \to {}^{20}_{10}\text{Ne} + {}^{4}_{2}\text{He}$: $Q = [24.000000 - 19.992439 - 4.002603]\times931.5 = +4.62$ MeV, exothermic.

"How long will the lamp glow" — the counting chain

• Step 1: number of nuclei in the sample, $N = \dfrac{m(\text{g})}{M(\text{g mol}^{-1})}\times N_A$. For deuterium take $M = 2$ g mol$^{-1}$.
• Step 2: energy released per nucleus. Since ${}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \to {}^{3}_{2}\text{He} + n + 3.27$ MeV consumes TWO deuterium nuclei, the energy per deuterium nucleus is $\dfrac{3.27}{2} = 1.635$ MeV $= 2.616\times10^{-13}$ J.
• Step 3: total energy $E = N\times E_{\text{per nucleus}}$.
• Step 4: time $t = \dfrac{E}{P}$, then divide by $3.154\times10^{7}$ (or $365\times24\times3600$) s to get years.
• Worked, 2.0 kg of deuterium and a 100 W lamp: $N = \dfrac{2000}{2}\times6.023\times10^{23} = 6.023\times10^{26}$; $E = 6.023\times10^{26}\times2.616\times10^{-13} \approx 1.58\times10^{14}$ J; $t = \dfrac{1.58\times10^{14}}{100} = 1.58\times10^{12}$ s $\approx 5\times10^{4}$ years.
• Worked, 2.0 kg of deuterium and an 800 W lamp: $E \approx 1.575\times10^{14}$ J; $t = \dfrac{1.575\times10^{14}}{800} = 1.97\times10^{11}$ s $\approx 6.2\times10^{3}$ years.
• Worked, 100 g of deuterium and a 500 W bulb: $N = \dfrac{100}{2}\times6.023\times10^{23} = 3.011\times10^{25}$; $E = 3.011\times10^{25}\times2.616\times10^{-13} \approx 7.88\times10^{12}$ J; $t = \dfrac{7.88\times10^{12}}{500} = 1.58\times10^{10}$ s $\approx 500$ years.
• Worked, 1 kg of ${}^{239}_{94}\text{Pu}$ with $180$ MeV per fission: $N = \dfrac{1000}{239}\times6.023\times10^{23} = 2.52\times10^{24}$; energy $= 2.52\times10^{24}\times180 \approx 4.5\times10^{26}$ MeV.

Nucleon number conserved yet energy released

• In ${}^{2}_{1}\text{H} + {}^{2}_{1}\text{H} \to {}^{3}_{2}\text{He} + {}^{1}_{0}\text{n} + 3.27$ MeV the nucleon count is 4 on both sides, yet energy is released.
• Explanation to write: the sum of the masses of the two deuterons is slightly GREATER than the sum of the masses of the helium-3 nucleus and the neutron, even though the number of nucleons is unchanged. By conservation of mass-energy this small mass difference (the mass defect $\Delta m$) is converted into energy, which appears as the $3.27$ MeV released. This is the $Q$-value of the reaction, $Q = \Delta m(\text{u})\times931.5$ MeV. Underlying it is the change in total BINDING energy: the product is more tightly bound than the reactants.

Standard results

• $Q = (\sum m_{\text{initial}} - \sum m_{\text{final}})\times931.5$ MeV $=$ (final KE) $-$ (initial KE) $=$ (B.E. of products) $-$ (B.E. of reactants).
• $Q > 0$ exothermic and possible; $Q < 0$ endothermic and not possible.
• Fission of ${}^{235}_{92}\text{U}$: about $200$ MeV per nucleus; $10^{14}$ J per kg of uranium against $10^{7}$ J per kg of coal.
• Fusion of two deuterons: $3.27$ MeV per event, hence $1.635$ MeV per deuterium nucleus.
• Coulomb barrier for two protons $\approx 400$ keV; for two deuterons touching at $4$ fm, $360$ keV; thermonuclear temperature $\sim 3\times10^{9}$ K; Sun's core $1.5\times10^{7}$ K; controlled fusion aims at $10^{8}$ K with the fuel as a plasma, confinement being the main difficulty.
• Net p–p cycle: $4\,{}^{1}_{1}\text{H} + 2e^{-} \to {}^{4}_{2}\text{He} + 2\nu + 6\gamma + 26.7$ MeV.
• Radioactivity: a nuclear phenomenon, discovered by Becquerel (1896); three natural decays, $\alpha$ (${}^{4}_{2}\text{He}$ emitted), $\beta$ (electrons or positrons) and $\gamma$ (high-energy photons).
• In every nuclear reaction, mass number and charge number are each conserved; the number of atoms of an element need not be.