Derivation

Nuclear Binding Energy and the Energetics of Fission and Fusion

D-426 Home PU-304 Threads energy · matter · force Depends on Semi-Empirical Mass Formula from the Liquid-Drop Model, Mass-Energy Equivalence E = mc^2

Statement

The nuclear binding energy of a nucleus with \(Z\) protons and \(N\) neutrons is the energy required to disassemble it into free nucleons, given by \(B = \left[ Z\,m_p + N\,m_n - M_{\text{nuc}} \right]c^2\) in terms of nuclear masses, or equivalently \(B = \left[ Z\,m(^1\mathrm{H}) + N\,m_n - M_{\text{atom}} \right]c^2\) in terms of atomic masses. The \(Q\)-value of a nuclear reaction is the difference between the total binding energies of the products and reactants, \(Q = \left(\sum m_{\text{initial}} - \sum m_{\text{final}}\right)c^2\), and is positive when the final system is more tightly bound than the initial one.

Why it matters

The binding energy per nucleon curve is the single most important map in nuclear physics. It explains why fusion of light nuclei and fission of heavy nuclei both release energy, while intermediate-mass nuclei are energetically stable. The peak near iron-56 and nickel-62 at roughly 8.8 MeV per nucleon marks the boundary: reactions that move nucleons toward this peak are exothermic, those moving away are endothermic.

Quantitatively, the energy released per kilogram of fuel in nuclear reactions exceeds that of chemical combustion by a factor of about \(10^7\). Understanding the \(Q\)-value, the Coulomb barrier, and the role of quantum tunnelling is essential for both reactor physics and astrophysical nucleosynthesis.

Assumptions

Masses are rest masses in the ground state.Excited states of the nucleus carry additional energy; using an excited-state mass without accounting for the excitation energy shifts the \(Q\)-value by that amount. For example, if a fission fragment is left in a 2 MeV excited state, that energy is not available as kinetic energy of the fragments.

Atomic masses are used consistently with the electron bookkeeping.When using atomic masses, the electron masses on both sides cancel only if the total number of electrons is the same (i.e., the reaction is charge-conserving in the atomic sense). Mixing nuclear masses on one side and atomic masses on the other introduces an error of \(0.511\) MeV per mismatched electron.

The \(Q\)-value is shared as kinetic energy of the products (plus any \(\gamma\), neutrino, or antineutrino energy).If the reaction produces neutrinos (as in beta decay), a portion of the \(Q\)-value escapes as neutrino energy and is not available as recoverable kinetic energy. Ignoring this overestimates the usable energy.

The nuclear radius for Coulomb-barrier estimates uses a hard-sphere model with \(r_0 \approx 1.2\) fm.The actual nuclear surface is diffuse (Woods–Saxon profile), and the barrier is not a sharp step. The hard-sphere estimate gives the order of magnitude correctly but can be off by a factor of two in the peak barrier height. For precision, one must integrate the true potential and solve the Schrödinger equation for the transmission coefficient.

Spontaneous fission is treated as a two-body liquid-drop process.The liquid-drop model captures the competition between surface tension (favouring a single drop) and Coulomb repulsion (favouring splitting), but it neglects shell effects, pairing, and the detailed fission barrier shape. The fissility parameter \(Z^2/A\) is a necessary but not sufficient condition; actinides with \(Z^2/A \approx 35\) can fission, while some nuclei with similar \(Z^2/A\) do not, due to shell stabilisation.

Derivation

We build the result in three parts: (i) the binding energy from the mass defect, (ii) the \(Q\)-value of a general reaction, and (iii) the Coulomb barrier and tunnelling condition for fusion.

\[ B = \left( Z\,m_p + N\,m_n - M_{\text{nuc}} \right)c^2 \]

By definition, the binding energy is the energy needed to separate all nucleons. The mass defect \(\Delta m = Z\,m_p + N\,m_n - M_{\text{nuc}}\) is positive because the bound nucleus is lighter than its constituents. A

\[ M_{\text{atom}} = M_{\text{nuc}} + Z\,m_e - B_e/c^2 \]

The atomic mass includes \(Z\) electrons and their binding energy \(B_e\) (of order keV for heavy atoms, negligible compared to MeV nuclear energies). Rearranging: \(M_{\text{nuc}} = M_{\text{atom}} - Z\,m_e + B_e/c^2\). A

\[ B = \left[ Z\,m_p + N\,m_n - M_{\text{atom}} + Z\,m_e \right]c^2 \]

Substitute Step 2 into Step 1, dropping \(B_e\) as negligible (\(\ll\) MeV). A

\[ m_p + m_e = m(^1\mathrm{H}) \]

The hydrogen-1 atomic mass equals the proton mass plus the electron mass (minus the 13.6 eV hydrogen binding energy, negligible at the MeV scale). Thus \(Z(m_p + m_e) = Z\,m(^1\mathrm{H})\). B

\[ B = \left[ Z\,m(^1\mathrm{H}) + N\,m_n - M_{\text{atom}} \right]c^2 \]

Substitute Step 4 into Step 3. This is the practical formula using tabulated atomic masses and the neutron mass. A

\[ Q = \left( \sum_{\text{initial}} m_i - \sum_{\text{final}} m_f \right)c^2 \]

Energy conservation: the rest-mass energy lost by the system appears as kinetic energy (and radiation) of the products. A positive \(Q\) means the final masses are smaller. A

\[ Q = \sum_{\text{final}} B_f - \sum_{\text{initial}} B_i \]

Expand each mass as \(M = Z\,m_p + N\,m_n - B/c^2\). The total nucleon number is conserved, so all \(m_p\) and \(m_n\) terms cancel between initial and final sums, leaving only the binding-energy difference. C

\[ \frac{B}{A} \approx 7.6\ \mathrm{MeV} \quad (\text{U-235}), \qquad \frac{B}{A} \approx 8.5\ \mathrm{MeV} \quad (\text{fission fragments}) \]

Reading the binding-energy-per-nucleon curve: heavy actinides sit at about 7.6 MeV/nucleon, while the mid-mass fission products (A \(\approx\) 90–140) sit near 8.5 MeV/nucleon. A

\[ Q_{\text{fission}} \approx A\left(\frac{B_f}{A} - \frac{B_i}{A}\right) \approx 235 \times (8.5 - 7.6)\ \mathrm{MeV} \approx 210\ \mathrm{MeV} \]

The total \(Q\)-value is the gain in binding energy times the number of nucleons. The accepted value for \(^{235}\mathrm{U} + n \to\) fragments \(+ 2\text{–}3\,n\) is about 200 MeV; the estimate is within 5%. B

\[ V_C = \frac{1}{4\pi\varepsilon_0}\frac{Z_1 Z_2 e^2}{r} \approx \frac{(1.44\ \mathrm{MeV\cdot fm})\,Z_1 Z_2}{r\ \mathrm{[fm]}} \]

Coulomb potential energy at separation \(r\) between two nuclei of charges \(Z_1 e\) and \(Z_2 e\). The constant \(e^2/(4\pi\varepsilon_0) = 1.44\) MeV·fm is the standard nuclear-physics conversion. B

\[ r \approx r_0(A_1^{1/3} + A_2^{1/3}), \qquad r_0 \approx 1.2\ \mathrm{fm} \]

The two nuclei just touch when their radii add; each radius is \(r_0 A^{1/3}\). For D–T, \(r \approx 1.2(2^{1/3}+3^{1/3}) \approx 1.2(1.26+1.44) \approx 3.2\) fm. A

\[ V_C(\text{D–T}) \approx \frac{1.44 \times 1 \times 1}{3.2}\ \mathrm{MeV} \approx 0.45\ \mathrm{MeV} \approx 450\ \mathrm{keV} \]

Both D and T have \(Z=1\). The barrier peak is a few hundred keV, far above the thermal energy \(k_BT \approx 10\) keV in a fusion plasma. B

\[ P_{\text{tunnel}} \sim \exp\!\left(-2\int_{r_1}^{r_2}\sqrt{\frac{2\mu(V_C(r)-E)}{\hbar^2}}\,dr\right) = \exp\!\left(-\frac{2\pi Z_1 Z_2 e^2}{4\pi\varepsilon_0\,\hbar\,v}\right) \]

The Gamow factor: the exponential suppression of the tunnelling probability through the Coulomb barrier. The argument is the Gamow parameter \(G = 2\pi Z_1 Z_2 \alpha c/v\), where \(\alpha \approx 1/137\). At \(E \approx 10\) keV for D–T (\(\mu \approx 1.2\ \mathrm{u}\), \(v \approx 1.3\times10^6\ \mathrm{m/s}\)), \(G \approx 11\), so \(P \sim e^{-11} \sim 2\times10^{-5}\); the fast tail of the Maxwell distribution and the enormous collision rate in a hot plasma do the rest. C

\[ Q_{\text{D–T}} = \left[m_D + m_T - m_{\mathrm{He4}} - m_n\right]c^2 \]

Apply the \(Q\)-value formula (Step 6) to the specific reaction \(D + T \to {}^4\mathrm{He} + n\). A

Result

\[ B = \left[ Z\,m(^1\mathrm{H}) + N\,m_n - M_{\text{atom}} \right]c^2, \qquad Q = \left(\sum m_i - \sum m_f\right)c^2 = \sum B_f - \sum B_i \]

Reading. The binding energy is the mass defect converted to energy; the \(Q\)-value of any reaction is the increase in total binding energy. Fission of heavy nuclei and fusion of light nuclei both move the system toward the peak of the binding-energy-per-nucleon curve (near \(A \approx 56\), \(B/A \approx 8.8\) MeV), releasing energy. Fusion of D–T releases 17.6 MeV per reaction; fission of \(^{235}\)U releases about 200 MeV per fission.

Units check. Masses in u, \(c^2\) in \(\mathrm{MeV/u}\): \(1\ \mathrm{u} = 931.494\ \mathrm{MeV}/c^2\). Thus \([\Delta m \cdot c^2] = \mathrm{u \cdot (MeV/u) = MeV}\). Consistent. In SI: \([\mathrm{kg \cdot (m/s)^2}] = \mathrm{J}\). Consistent.

Limiting cases

Deuterium (A = 2). \(B/A \approx 1.1\) MeV, far below the peak. Fusion of two deuterons releases energy because the products (He-3 or T) are more tightly bound.
Iron-56 / Nickel-62. \(B/A \approx 8.8\) MeV, the maximum. Neither fusing them further nor splitting them releases energy; they are the endpoint of stellar burning.
Uranium-235 (A = 235). \(B/A \approx 7.6\) MeV. Fission into two fragments near \(A \approx 118\) (each with \(B/A \approx 8.5\) MeV) releases \(\sim 200\) MeV.
Very light or very heavy hypothetical nuclei. For \(A \to 1\) (a free nucleon), \(B = 0\). For ever heavier nuclei in the liquid-drop model, the Coulomb energy \(a_c Z^2/A^{1/3}\) grows faster than the volume term, \(B/A\) keeps falling, and the drop becomes unstable to fission: deforming it costs no energy once the Coulomb energy exceeds twice the surface energy, \(a_c Z^2/A^{1/3} \gt 2a_s A^{2/3}\), i.e. \(Z^2/A \gtrsim 2a_s/a_c \approx 50\).

Breaks when

Mixing nuclear and atomic masses. Using the nuclear mass of one species and the atomic mass of another in the same \(Q\)-value expression introduces an error of \(Z_e \times 0.511\) MeV per mismatched electron. For a reaction with \(Z\) electrons on one side and \(Z'\) on the other, the error is \((Z - Z') \times 0.511\) MeV.
Ignoring neutrino energy in beta-decay channels. In reactions that produce \(\nu\) or \(\bar{\nu}\) (e.g., \(^{14}\mathrm{N} + p \to {}^{15}\mathrm{O} + \gamma\) followed by \(\beta^+\) decay), a significant fraction of the \(Q\)-value (up to the full endpoint energy) escapes as neutrino kinetic energy and is not recoverable as heat or radiation in a detector.
Assuming all \(Q\)-value appears as fragment kinetic energy. In fission, roughly 10 MeV of the 200 MeV (about 5%) leaves as antineutrinos from the beta decays of the fission products and is lost for good; the prompt and delayed \(\gamma\)-rays, by contrast, are absorbed in the reactor. The recoverable heat is about 190–195 MeV per fission.

Failure modes

Using the wrong mass table. Looking up the mass of \(^{235}\)U as a bare nucleus when the table lists atomic masses (or vice versa). Always check whether the tabulated mass includes electrons and use the formula that matches.
Forgetting that \(Q\) is in MeV, not MeV per nucleon. The \(Q\)-value of a reaction is the total energy release. Dividing by \(A\) gives the energy per nucleon, but the total energy available is the full \(Q\). Confusing the two underestimates the energy by a factor of \(A\).
Assuming the Coulomb barrier is the minimum kinetic energy required. The barrier peak for D–T is about 450 keV, but fusion occurs at 10 keV plasma temperatures because of quantum tunnelling. The Gamow factor suppresses the rate but does not make it zero.
Adding binding energies instead of taking the difference. \(Q = B_{\text{final}} - B_{\text{initial}}\), not \(B_{\text{final}} + B_{\text{initial}}\). The initial binding energy is already “locked in”; only the increase in binding is released.

Discussion

The binding-energy-per-nucleon curve is a fingerprint of the nuclear force. The volume term (proportional to \(A\)) reflects the saturation of the strong force: each nucleon interacts only with its nearest neighbours, so the total binding grows linearly with \(A\). The surface term (proportional to \(A^{2/3}\)) penalises nucleons at the surface that have fewer neighbours. The Coulomb term (proportional to \(Z^2/A^{1/3}\)) penalises the electrostatic repulsion among protons, which grows faster than the volume term as \(Z\) increases. The asymmetry term (proportional to \((N-Z)^2/A\)) penalises an imbalance between neutrons and protons, a consequence of the Pauli principle. The interplay of these four terms in the semi-empirical mass formula produces the characteristic curve: rising from \(A=1\), peaking near \(A \approx 56\), and slowly declining for heavy nuclei.

Fission and fusion are two sides of the same coin. Both move the system from a lower \(B/A\) to a higher \(B/A\), releasing the difference. Fission splits a heavy nucleus (low \(B/A\)) into two lighter fragments (higher \(B/A\)); fusion combines two light nuclei (low \(B/A\)) into a heavier product (higher \(B/A\)). The peak at iron/nickel is the thermodynamic sink: no further nuclear reaction can extract energy from nuclei at or beyond this point. This is why massive stars end their lives in iron-core collapse rather than continued fusion.

The practical challenge of fusion is the Coulomb barrier. Two positively charged nuclei repel each other with a potential that peaks at hundreds of keV, yet the thermal energy in a 10 keV plasma is two orders of magnitude smaller. The resolution is quantum mechanical: the wavefunction of the incoming nuclei has a non-zero amplitude on the far side of the barrier, and the tunnelling probability, though exponentially small (\(\sim 10^{-5}\) for D–T at 10 keV, and far smaller for higher charges), is multiplied by the enormous collision rate in a dense, hot plasma to yield a measurable reaction rate. This is the Gamow peak in astrophysics and the basis of the Lawson criterion in reactor design.

At a deeper level, the binding energy is a consequence of the strong interaction described by quantum chromodynamics (QCD). The semi-empirical mass formula is a phenomenological parameterisation; the underlying physics is the confinement of quarks and gluons into nucleons and the residual strong force (mediated by pion exchange at nuclear distances) that binds nucleons together. The \(Q\)-value of any reaction is ultimately determined by the difference in the QCD ground-state energies of the initial and final configurations. Perturbative QCD does not apply at the \(\sim 1\) fm scale; instead, lattice QCD calculations and effective field theories (chiral perturbation theory, pionless EFT) are the modern tools for computing nuclear binding energies from first principles.

Common misconceptions. (1) “Mass is converted to energy” is an oversimplification: the rest mass of the system decreases, and the difference appears as kinetic energy and radiation; the total energy (mass-energy) is conserved. (2) Fusion and fission rearrange nucleons, not quarks: the protons and neutrons survive intact, and only their binding changes. (3) The energy released in fission is not from “splitting the atom” in a mechanical sense; it is the difference in nuclear binding energy between the initial heavy nucleus and the final fragments. (4) A positive \(Q\)-value does not mean the reaction will occur at a measurable rate; the Coulomb barrier and angular-momentum selection rules can make the rate vanishingly small even when \(Q > 0\).

Worked examples

Example 1. Compute the \(Q\)-value of the D–T fusion reaction \(D + T \to {}^4\mathrm{He} + n\) using atomic masses: \(m_D = 2.014102\ \mathrm{u}\), \(m_T = 3.016049\ \mathrm{u}\), \(m_{\mathrm{He4}} = 4.002603\ \mathrm{u}\), \(m_n = 1.008665\ \mathrm{u}\), and \(1\ \mathrm{u} = 931.494\ \mathrm{MeV}/c^2\).

\[ \Delta m = (m_D + m_T) - (m_{\mathrm{He4}} + m_n) \]

The mass defect is the difference between the total initial and final masses. A

\[ \Delta m = (2.014102 + 3.016049) - (4.002603 + 1.008665) = 5.030151 - 5.011268 = 0.018883\ \mathrm{u} \]

Arithmetic: sum the initial masses, sum the final masses, subtract. A

\[ Q = \Delta m \times 931.494\ \mathrm{MeV/u} = 0.018883 \times 931.494 \] \[ = 17.59\ \mathrm{MeV} \approx 17.6\ \mathrm{MeV} \]

Convert the mass defect to energy using the standard conversion factor. A

\[ Q_{\text{D–T}} = 17.6\ \mathrm{MeV} \]

Reading. Each D–T fusion reaction releases 17.6 MeV, shared as kinetic energy of the alpha particle (3.5 MeV) and the neutron (14.1 MeV) in the centre-of-mass frame.

Units check. \([\Delta m] = \mathrm{u}\), \([931.494] = \mathrm{MeV/u}\), so \([Q] = \mathrm{u \cdot MeV/u = MeV}\). Correct.

Example 2. A 1.0 kg sample of D–T fuel (equal masses of deuterium and tritium) is fully fused. Estimate the total energy released in joules and compare it with the energy from burning 1.0 kg of gasoline (\(\approx 4.6 \times 10^7\) J).

\[ N_{\text{pairs}} = \frac{m_{\text{fuel}}}{m_D + m_T} = \frac{1.0\ \mathrm{kg}}{(2.014102 + 3.016049)\ \mathrm{u} \times 1.66054 \times 10^{-27}\ \mathrm{kg/u}} \]

The number of D–T pairs is the total mass divided by the mass of one pair. Convert u to kg using \(1\ \mathrm{u} = 1.66054 \times 10^{-27}\) kg. A

\[ m_{\text{pair}} = 5.030151 \times 1.66054 \times 10^{-27} = 8.353 \times 10^{-27}\ \mathrm{kg} \] \[ N_{\text{pairs}} = \frac{1.0}{8.353 \times 10^{-27}} = 1.197 \times 10^{26} \]

Compute the pair mass in kg, then the number of pairs. A

\[ E_{\text{total}} = N_{\text{pairs}} \times Q = 1.197 \times 10^{26} \times 17.6\ \mathrm{MeV} \times 1.602 \times 10^{-13}\ \mathrm{J/MeV} \]

Multiply the number of reactions by the energy per reaction, converting MeV to joules. A

\[ E_{\text{total}} = 1.197 \times 10^{26} \times 17.6 \times 1.602 \times 10^{-13} \] \[ = 1.197 \times 10^{26} \times 2.820 \times 10^{-12} = 3.37 \times 10^{14}\ \mathrm{J} \]

Arithmetic: \(17.6 \times 1.602 \times 10^{-13} = 2.820 \times 10^{-12}\) J; multiply by \(1.197 \times 10^{26}\). A

\[ \frac{E_{\text{D–T}}}{E_{\text{gasoline}}} = \frac{3.37 \times 10^{14}}{4.6 \times 10^{7}} \approx 7.3 \times 10^{6} \]

Compare with the chemical energy of 1 kg of gasoline. A

\[ E_{\text{D–T}}(1\ \mathrm{kg}) \approx 3.4 \times 10^{14}\ \mathrm{J} \approx 7 \times 10^{6} \times E_{\text{gasoline}}(1\ \mathrm{kg}) \]

Reading. One kilogram of D–T fuel releases about seven million times more energy than one kilogram of gasoline. This is the fundamental energy-density advantage of nuclear over chemical fuels.

Units check. \([\mathrm{kg / (u \cdot kg/u)}] = \) dimensionless (number of pairs). \([\text{pairs} \times \mathrm{MeV} \times \mathrm{J/MeV}] = \mathrm{J}\). Correct.

Problems

Compute the binding energy per nucleon of \(^{56}\mathrm{Fe}\) using atomic masses: \(m(^{56}\mathrm{Fe}) = 55.934937\ \mathrm{u}\), \(m(^1\mathrm{H}) = 1.007825\ \mathrm{u}\), \(m_n = 1.008665\ \mathrm{u}\), \(Z = 26\), \(N = 30\), \(1\ \mathrm{u} = 931.494\ \mathrm{MeV}/c^2\).
Solution
\[ B = [26 \times 1.007825 + 30 \times 1.008665 - 55.934937] \times 931.494 \] \[ = [26.203450 + 30.259950 - 55.934937] \times 931.494 \] \[ = [56.463400 - 55.934937] \times 931.494 = 0.528463 \times 931.494 = 492.25\ \mathrm{MeV} \] \[ B/A = 492.25 / 56 = 8.79\ \mathrm{MeV/nucleon} \]
A \(^{235}\)U nucleus absorbs a neutron and fissions into \(^{141}\)Ba and \(^{92}\)Kr plus 3 neutrons. Using atomic masses \(m(^{235}\mathrm{U}) = 235.043930\ \mathrm{u}\), \(m(^{141}\mathrm{Ba}) = 140.914406\ \mathrm{u}\), \(m(^{92}\mathrm{Kr}) = 91.926156\ \mathrm{u}\), \(m_n = 1.008665\ \mathrm{u}\), compute the \(Q\)-value.
Solution
Initial: \(m_i = m(^{235}\mathrm{U}) + m_n = 235.043930 + 1.008665 = 236.052595\ \mathrm{u}\). Final: \(m_f = m(^{141}\mathrm{Ba}) + m(^{92}\mathrm{Kr}) + 3\,m_n = 140.914406 + 91.926156 + 3.025995 = 235.866557\ \mathrm{u}\). \[ \Delta m = 236.052595 - 235.866557 = 0.186038\ \mathrm{u} \] \[ Q = 0.186038 \times 931.494 = 173.3\ \mathrm{MeV} \] This is the prompt energy of this particular split. Both fragments are neutron-rich and later beta-decay towards stability, releasing roughly another 20 MeV (part of it carried off by antineutrinos), which brings the total close to the usual figure of about 200 MeV per fission.
Estimate the Coulomb barrier height for the D–T reaction, assuming the nuclei just touch at \(r = r_0(A_1^{1/3} + A_2^{1/3})\) with \(r_0 = 1.2\) fm.
Solution
\[ r = 1.2 \times (2^{1/3} + 3^{1/3}) = 1.2 \times (1.260 + 1.442) = 1.2 \times 2.702 = 3.24\ \mathrm{fm} \] \[ V_C = \frac{1.44\ \mathrm{MeV \cdot fm} \times Z_1 Z_2}{r} = \frac{1.44 \times 1 \times 1}{3.24} = 0.444\ \mathrm{MeV} \approx 444\ \mathrm{keV} \]
A 5.0 g sample of \(^{235}\)U undergoes complete fission, releasing 200 MeV per fission. Compute the total energy in joules and in kilotonnes of TNT (1 kt TNT \(= 4.184 \times 10^{12}\) J).
Solution
Number of \(^{235}\)U atoms: \[ N = \frac{5.0\ \mathrm{g}}{235.044\ \mathrm{g/mol}} \times 6.022 \times 10^{23}\ \mathrm{mol^{-1}} = 0.02127 \times 6.022 \times 10^{23} = 1.281 \times 10^{22} \] Total energy: \[ E = 1.281 \times 10^{22} \times 200\ \mathrm{MeV} \times 1.602 \times 10^{-13}\ \mathrm{J/MeV} \] \[ = 1.281 \times 10^{22} \times 3.204 \times 10^{-11} = 4.10 \times 10^{11}\ \mathrm{J} \] In kilotonnes of TNT: \[ \frac{4.10 \times 10^{11}}{4.184 \times 10^{12}} = 0.098\ \mathrm{kt} \approx 98\ \mathrm{tonnes\ of\ TNT} \]
The reaction \(^3\mathrm{He} + D \to {}^4\mathrm{He} + p\) has \(Q = 18.3\) MeV. If the \(^3\)He and D nuclei approach each other with a centre-of-mass energy of 10 keV, compute the Gamow parameter \(G = 2\pi Z_1 Z_2 \alpha c / v\) and the approximate tunnelling probability \(P \sim e^{-G}\). Take \(Z_1 = 2\), \(Z_2 = 1\), \(\alpha = 1/137\), and compute \(v\) from \(E_{\text{cm}} = \tfrac{1}{2}\mu v^2\) with reduced mass \(\mu = m_D m_{^3\mathrm{He}}/(m_D + m_{^3\mathrm{He}})\).
Solution
Reduced mass: \[ \mu = \frac{2.014102 \times 3.016029}{2.014102 + 3.016029}\ \mathrm{u} = \frac{6.074}{5.030}\ \mathrm{u} = 1.208\ \mathrm{u} = 1.208 \times 1.66054 \times 10^{-27} = 2.006 \times 10^{-27}\ \mathrm{kg} \] Velocity from \(E_{\text{cm}} = 10\ \mathrm{keV} = 1.602 \times 10^{-15}\ \mathrm{J}\): \[ v = \sqrt{\frac{2 E_{\text{cm}}}{\mu}} = \sqrt{\frac{2 \times 1.602 \times 10^{-15}}{2.006 \times 10^{-27}}} = \sqrt{1.598 \times 10^{12}} = 1.264 \times 10^{6}\ \mathrm{m/s} \] Gamow parameter: \[ G = \frac{2\pi Z_1 Z_2 \alpha c}{v} = \frac{2\pi \times 2 \times 1 \times (1/137) \times 3.0 \times 10^{8}}{1.264 \times 10^{6}} \] \[ = \frac{2\pi \times 2 \times 0.00730 \times 3.0 \times 10^{8}}{1.264 \times 10^{6}} = \frac{2.757 \times 10^{7}}{1.264 \times 10^{6}} = 21.8 \] Tunnelling probability: \[ P \sim e^{-G} = e^{-21.8} \approx 3.5 \times 10^{-10} \] This is the probability per collision; the enormous number of collisions in a hot plasma compensates for this small probability.