Fusion: Climbing the Curve from the Left

When two light nuclei fuse, the product is more tightly bound (higher EbnE_{bn}, left flank of the curve) — energy is released. NCERT's three flagship reactions:

11H+11H→12H+e++ν+0.42 MeV^{1}_{1}H + {}^{1}_{1}H \to {}^{2}_{1}H + e^{+} + \nu + 0.42 \text{ MeV} 12H+12H→23He+n+3.27 MeV^{2}_{1}H + {}^{2}_{1}H \to {}^{3}_{2}He + n + 3.27 \text{ MeV} 12H+12H→13H+11H+4.03 MeV^{2}_{1}H + {}^{2}_{1}H \to {}^{3}_{1}H + {}^{1}_{1}H + 4.03 \text{ MeV}

(Proton + proton → deuteron + positron; deuteron + deuteron → helium-3 + neutron, or triton + proton.)

The Coulomb barrier

For fusion, the nuclei must come within nuclear-force range — but both are positively charged and repel. They need enough kinetic energy to climb the Coulomb barrier, whose height depends on the charges and radii involved: ~400 keV for two protons, higher for higher charges.

The temperature price

When does a gas of protons average 400 keV?

32kT=K≃400 keV⇒T∼3×109 K\frac{3}{2}kT = K \simeq 400 \text{ keV} \quad\Rightarrow\quad T \sim 3 \times 10^9 \text{ K}

Fusion achieved by sheer temperature is thermonuclear fusion. But note: the Sun's core is only 1.5×1071.5 \times 10^7 K — far below the estimate. Solar fusion runs on the high-energy tail of the proton distribution: protons with energies much above average do the burning.

Proton-proton cycle powering the sun

The Proton-Proton Cycle: The Sun's Furnace

The Sun burns hydrogen into helium in its core via a multi-step proton-proton (p, p) cycle:

(i)11H+11H→12H+e++ν+0.42 MeV\text{(i)}\quad ^{1}_{1}H + {}^{1}_{1}H \to {}^{2}_{1}H + e^{+} + \nu + 0.42 \text{ MeV} (ii)e++e−→γ+γ+1.02 MeV\text{(ii)}\quad e^{+} + e^{-} \to \gamma + \gamma + 1.02 \text{ MeV} (iii)12H+11H→23He+γ+5.49 MeV\text{(iii)}\quad ^{2}_{1}H + {}^{1}_{1}H \to {}^{3}_{2}He + \gamma + 5.49 \text{ MeV} (iv)23He+23He→24He+11H+11H+12.86 MeV\text{(iv)}\quad ^{3}_{2}He + {}^{3}_{2}He \to {}^{4}_{2}He + {}^{1}_{1}H + {}^{1}_{1}H + 12.86 \text{ MeV}

For step (iv) to run once, steps (i)-(iii) must run twice. Net effect of 2(i) + 2(ii) + 2(iii) + (iv):

4 11H+2e−→24He+2ν+6γ+26.7 MeV\boxed{4\,^{1}_{1}H + 2e^{-} \to {}^{4}_{2}He + 2\nu + 6\gamma + 26.7 \text{ MeV}}

Four hydrogen atoms burn into one helium atom, releasing 26.7 MeV.

The Sun's biography

  • Age: about 5×1095 \times 10^9 years; hydrogen enough for another ~5 billion years.
  • Then: hydrogen burning stops → core cools → gravitational collapse raises core temperature → at ~10810^8 K, helium fuses into carbon; the envelope expands — the Sun becomes a red giant.
  • Successive burnings build heavier elements — but only up to the iron peak of the binding curve; beyond it, fusion costs energy (heavier elements need other processes).

Controlled thermonuclear fusion

Replicating the star: heat fuel to ~10810^8 K, where it is a plasma (positive ions + electrons). The challenge: no container withstands such temperatures — the plasma must be confined by other means (several countries, including India, are developing techniques). Success would mean almost unlimited power.

[NEET Important] Number bank: 0.42 / 1.02 / 5.49 / 12.86 / 26.7 MeV; barrier ~400 keV; 3×1093 \times 10^9 K estimate vs 1.5×1071.5 \times 10^7 K solar core; helium→carbon at 10810^8 K. [JEE Tip] Fusion is per-KILOGRAM even richer than fission: 26.7 MeV from just 4 nucleons ≈ 6.7 MeV/nucleon versus fission's ~0.85 MeV/nucleon — hydrogen fuel outperforms uranium ~7× by mass.

Solved Examples

Example 1: The lamp on deuterium (NCERT Exercise 13.8)

How long can a 100 W lamp glow on the fusion of 2.0 kg of deuterium via 12H+12H→23He+n+3.27^{2}_{1}H + {}^{2}_{1}H \to {}^{3}_{2}He + n + 3.27 MeV?

Solution:

  1. Deuterons available: N=20002×6.023×1023=6.023×1026N = \dfrac{2000}{2} \times 6.023 \times 10^{23} = 6.023 \times 10^{26}.
  2. Reactions: each uses 2 deuterons → N2=3.01×1026\dfrac{N}{2} = 3.01 \times 10^{26} reactions, releasing 3.27 MeV each.
  3. Total energy: 3.01×1026×3.27×1.6×10−13≈1.58×10143.01 \times 10^{26} \times 3.27 \times 1.6 \times 10^{-13} \approx 1.58 \times 10^{14} J.
  4. Time: t=EP=1.58×1014100=1.58×1012t = \dfrac{E}{P} = \dfrac{1.58 \times 10^{14}}{100} = 1.58 \times 10^{12} s ≈ 5×1045 \times 10^4 years.
  5. Takeaway: two kilograms of heavy hydrogen light a bulb for fifty millennia — fusion's promise in one number.

Example 2: Coulomb barrier of two deuterons (NCERT Exercise 13.9)

Find the barrier height for a head-on collision of two deuterons (radius 2.0 fm each).

Solution:

  1. At contact: centres separated by d = 2 × 2.0 = 4.0 fm.
  2. Coulomb energy: U=e24πε0d=9×109×(1.6×10−19)24×10−15U = \dfrac{e^2}{4\pi\varepsilon_0 d} = \dfrac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15}} J.
  3. Compute: U=5.76×10−14U = 5.76 \times 10^{-14} J ≈ 360 keV.
  4. Takeaway: a few hundred keV — consistent with NCERT's ~400 keV two-proton figure; each nucleus supplies half (~180 keV) in a symmetric collision.

Example 3: The temperature estimate [Board Numerical]

Show that protons averaging 400 keV correspond to T ~ 3×1093 \times 10^9 K.

Solution:

  1. Equipartition: 32kT=K\dfrac{3}{2}kT = K.
  2. Solve: T=2K3k=2×400×103×1.6×10−193×1.38×10−23T = \dfrac{2K}{3k} = \dfrac{2 \times 400 \times 10^3 \times 1.6 \times 10^{-19}}{3 \times 1.38 \times 10^{-23}}.
  3. Compute: T≈3.1×109T \approx 3.1 \times 10^9 K.
  4. Takeaway: NCERT's estimate exactly; the Sun (at 1.5×1071.5 \times 10^7 K) burns anyway because its energetic-tail protons far exceed the average — a favourite conceptual follow-up.

Example 4: Auditing the p-p cycle [Board Numerical]

Verify that 2(i) + 2(ii) + 2(iii) + (iv) releases 26.7 MeV.

Solution:

  1. Twice (i): 2 × 0.42 = 0.84 MeV.
  2. Twice (ii): 2 × 1.02 = 2.04 MeV.
  3. Twice (iii): 2 × 5.49 = 10.98 MeV.
  4. Once (iv): 12.86 MeV.
  5. Total: 0.84 + 2.04 + 10.98 + 12.86 = 26.72 ≈ 26.7 MeV. ✔
  6. Takeaway: four protons in, one helium out, 26.7 MeV released — the Sun's balance sheet checks to the decimal.

Example 5: How much hydrogen does the Sun burn? [JEE Numerical]

The Sun radiates 3.8×10263.8 \times 10^{26} W. Estimate the mass of hydrogen converted per second (26.7 MeV per 4 protons).

Solution:

  1. Energy per proton: 26.74=6.68\dfrac{26.7}{4} = 6.68 MeV =1.07×10−12= 1.07 \times 10^{-12} J.
  2. Protons per second: 3.8×10261.07×10−12=3.6×1038\dfrac{3.8 \times 10^{26}}{1.07 \times 10^{-12}} = 3.6 \times 10^{38}.
  3. Mass rate: 3.6×1038×1.67×10−27≈6×10113.6 \times 10^{38} \times 1.67 \times 10^{-27} \approx 6 \times 10^{11} kg/s.
  4. Takeaway: the Sun burns ~600 million tonnes of hydrogen every second — and still has 5 billion years of fuel. Astronomy from arithmetic.

Example 6: Fusion vs fission per kilogram [JEE Comparison]

Compare the energy per kilogram from hydrogen fusion (26.7 MeV per 4 u) and uranium fission (200 MeV per 235 u).

Solution:

  1. Fusion: 26.74=6.7\dfrac{26.7}{4} = 6.7 MeV per nucleon (u).
  2. Fission: 200235=0.85\dfrac{200}{235} = 0.85 MeV per nucleon.
  3. Ratio: ≈ 7.8× in fusion's favour per unit mass.
  4. Takeaway: fusion fuel is nearly an order of magnitude richer — plus hydrogen is abundant and the products non-radioactive. Hence the reactor dream.

Example 7: Why fusion needs heat but fission doesn't [NEET Conceptual]

Fission is triggered by slow neutrons at room temperature; fusion needs 10710^7-10910^9 K. Why the difference?

Solution:

  1. Fission's trigger is a neutron: electrically neutral, it feels no Coulomb barrier and strolls into the uranium nucleus at any speed.
  2. Fusion's ingredients are both positive nuclei: they must overcome ~400 keV of Coulomb repulsion before the nuclear force can grab them.
  3. Only extreme temperatures give nuclei such kinetic energies — hence THERMOnuclear fusion.
  4. Takeaway: the barrier belongs to charged projectiles; neutrality is fission's skeleton key.

Example 8: The Sun's future [Board Conceptual]

Describe the sequence of events when the Sun's core hydrogen runs out.

Solution:

  1. Hydrogen burning stops → the core (now helium) cools → pressure support weakens.
  2. The star collapses under gravity, which heats the core.
  3. At ~10810^8 K, helium fuses into carbon; the outer envelope expands enormously — the Sun becomes a red giant.
  4. Successive fusion stages can build heavier elements, but only up to the binding-curve peak (iron region) — beyond that, fusion absorbs energy.
  5. Takeaway: a star's life is gravity versus fusion, negotiated stage by stage up the periodic table.

Example 9: The plasma problem [Board Conceptual]

What is the central engineering challenge of controlled fusion, and why?

Solution:

  1. Working temperature ~10810^8 K turns the fuel into plasma — a mixture of positive ions and electrons.
  2. No material container can withstand contact with matter at such temperatures.
  3. The challenge is confinement of the plasma (by non-material means); several countries including India are developing techniques.
  4. Takeaway: success promises 'almost unlimited power to humanity' — NCERT's own phrase, and a favourite short-answer quote.

Example 10: Positron's fate in the cycle [NEET Conceptual]

What happens to the positron produced in step (i) of the p-p cycle, and how much energy does that step contribute?

Solution:

  1. The positron meets one of the plasma's abundant electrons and annihilates: e++e−→γ+γe^+ + e^- \to \gamma + \gamma.
  2. Energy released: 1.02 MeV (= 2×0.5112 \times 0.511 MeV, the pair's rest-mass energy) — step (ii) of the cycle.
  3. This is why the net equation consumes 2 electrons and emits 6 gammas.
  4. Takeaway: annihilation is bookkept INSIDE the 26.7 MeV total — matter itself is part of the Sun's fuel.