In an experimental fusion reactor, deuterium and tritium nuclei fuse according to the equation: ^21H+ ^31H→ ^42He+ ^10n The nuclear masses involved are m(^21H)=2.013553 u, m(^31H)=3.015500 u, m(^42He)=4.001506 u and m(^10n)=1.008665 u. Take 1 u=1.66×10^-27 kg=931.5 MeV and c=3.0…

Answer: ((a)) Helium-4 has a higher binding energy per nucleon than deuterium or tritium, so fusing them into a nucleus closer to the peak of the curve releases the difference in binding energy as kinetic energy of the products ((b)) m = 0.0189 u (3 s.f.) ((c)) E = 17.6 MeV (3 s.f.) ((d)) E = 2.82×10^-12 J (3 s.f.). Part (a): Why fusion releases energy The graph of binding energy per nucleon against nucleon number rises steeply for light nuclei, peaking around iron before falling slowly for heavy nuclei. Deuterium and tritium, both very light nuclides…

A Level Physics 9702: Energy Released in Deuterium-Tritium Fusion

In an experimental fusion reactor, deuterium and tritium nuclei fuse according to the equation: $^2_1\text{H}+\,^3_1\text{H}\rightarrow\,^4_2\text{He}+\,^1_0\text{n}$

The nuclear masses involved are $m(^2_1\text{H})=2.013553\text{ u}$ , $m(^3_1\text{H})=3.015500\text{ u}$ , $m(^4_2\text{He})=4.001506\text{ u}$ and $m(^1_0\text{n})=1.008665\text{ u}$ .

Take $1\text{ u}=1.66\times10^{-27}\text{ kg}=931.5\text{ MeV}$ and $c=3.00\times10^8\text{ m s}^{-1}$ .

(a) With reference to the graph of binding energy per nucleon against nucleon number, explain why this fusion reaction releases energy. [2]

(b) Calculate the total mass before the reaction and the total mass after the reaction, and hence find the mass defect $\Delta m$ for this reaction, in u. [2]

(d) Show that this energy is equivalent to about $2.82\times10^{-12}\text{ J}$ . [2]

Part (a): Why fusion releases energy

The graph of binding energy per nucleon against nucleon number rises steeply for light nuclei, peaking around iron before falling slowly for heavy nuclei. Deuterium and tritium, both very light nuclides, sit low on the left-hand, steeply-rising part of this curve, so each of their nucleons is relatively weakly bound.

Helium-4 lies much closer to the peak of the curve, so each of its nucleons is much more strongly bound. When deuterium and tritium fuse into helium-4 (plus a neutron), the nucleons end up more tightly bound overall. This increase in total binding energy corresponds, by $E=c^2\Delta m$ , to a decrease in total mass. The “missing” mass is released as the kinetic energy of the helium nucleus and the neutron.

Part (b): Mass defect

Total mass of the reactants (deuterium $+$ tritium): $m_{before}=2.013553+3.015500=5.029053\text{ u}$

Total mass of the products (helium-4 $+$ neutron): $m_{after}=4.001506+1.008665=5.010171\text{ u}$

Mass defect: $\Delta m=m_{before}-m_{after}=5.029053-5.010171=0.018882\text{ u}$

Recompute as a check, combining differently: $(2.013553-1.008665)+(3.015500-4.001506)=1.004888+(-0.986006)=0.018882\text{ u}$ , the same result.

$\Delta m=\boxed{0.0189}\text{ u (3 s.f.)}$

Part (c): Energy released, in MeV

Using the given conversion $1\text{ u}=931.5\text{ MeV}$ directly on the mass defect from (b): $E=\Delta m\times931.5=0.018882\times931.5$

$E=0.018882\times900+0.018882\times31.5=16.99+0.595=17.59\text{ MeV}$

Recompute as a check by multiplying directly: $0.0189\times931.5=17.6\text{ MeV}$ , consistent with the split-sum method.

$E=\boxed{17.6}\text{ MeV (3 s.f.)}$

Part (d): Show that this is about 2.82 × 10⁻¹² J

Convert the mass defect to kilograms using $1\text{ u}=1.66\times10^{-27}\text{ kg}$ : $\Delta m=0.018882\times1.66\times10^{-27}=3.134\times10^{-29}\text{ kg}$

Apply $E=c^2\Delta m$ : $E=(3.00\times10^8)^2\times3.134\times10^{-29}=9.00\times10^{16}\times3.134\times10^{-29}$

$E=2.821\times10^{-12}\text{ J}$

Recompute as a check, keeping powers of ten separate: $9.00\times3.134=28.21$ , and $10^{16}\times10^{-29}=10^{-13}$ , giving $E=28.21\times10^{-13}=2.821\times10^{-12}\text{ J}$ , consistent with the value above, confirming $E\approx2.82\times10^{-12}\text{ J}$ as required to show.

Final answers

(a) Fusion moves the nucleons from a low binding energy per nucleon (deuterium, tritium) to a higher binding energy per nucleon (helium-4, near the peak of the curve); the increase in binding energy is released as kinetic energy of the products
(b) $\Delta m=\boxed{0.0189}\text{ u}$
(c) $E=\boxed{17.6}\text{ MeV}$
(d) $E\approx\boxed{2.82\times10^{-12}}\text{ J}$