Nuclei: Diagram-Based Questions (4)

easy 4 min read

Question

The binding energy per nucleon curve peaks near A56A \approx 56 at about 8.8 MeV/nucleon8.8 \text{ MeV/nucleon}. Use this curve to (a) explain why fusion of light nuclei releases energy, (b) explain why fission of heavy nuclei releases energy, and (c) estimate the energy released when a 235U\,^{235}\text{U} nucleus splits into two fragments of mass numbers 141Ba\,^{141}\text{Ba} and 92Kr\,^{92}\text{Kr} (plus 2 neutrons). Use BE/nucleon values: 235U=7.6\,^{235}\text{U} = 7.6, 141Ba=8.5\,^{141}\text{Ba} = 8.5, 92Kr=8.7\,^{92}\text{Kr} = 8.7 (all in MeV).

Solution — Step by Step

The binding energy per nucleon B/AB/A tells you how tightly each nucleon is held in the nucleus. Higher B/AB/A means more stable.

  • For A<20A < 20: B/AB/A is small — light nuclei are loosely bound.
  • For A56A \approx 56 (iron region): B/AB/A peaks at 8.8 MeV\sim 8.8 \text{ MeV} — most stable.
  • For A>100A > 100: B/AB/A slowly decreases due to Coulomb repulsion.

Combining two light nuclei produces a heavier one with higher B/AB/A. The total binding energy increases, and the difference is released as kinetic energy of products. Example: 2H+3H4He+n\,^{2}\text{H} + \,^{3}\text{H} \to \,^{4}\text{He} + n jumps from B/A1B/A \approx 1-2.82.8 to 7.1 MeV7.1 \text{ MeV} — huge gain per nucleon.

Splitting a heavy nucleus produces two medium-mass nuclei with higher B/AB/A. Same logic, opposite direction on the curve.

Total binding energy before:

Ei=235×7.6=1786 MeVE_i = 235 \times 7.6 = 1786 \text{ MeV}

Total binding energy after (ignoring 2 free neutrons, which have B/A=0B/A = 0):

Ef=(141×8.5)+(92×8.7)=1198.5+800.4=1998.9 MeVE_f = (141 \times 8.5) + (92 \times 8.7) = 1198.5 + 800.4 = 1998.9 \text{ MeV}

Energy released:

ΔE=EfEi=1998.91786=212.9 MeV\Delta E = E_f - E_i = 1998.9 - 1786 = 212.9 \text{ MeV}

Why This Works

Binding energy is the energy released when nucleons come together to form a nucleus from infinity. Equivalently, it’s the energy you’d need to break the nucleus apart. So total binding energy = (number of nucleons) × (BE per nucleon), and any reaction that increases this total releases energy.

The curve shape comes from the competition between the strong nuclear force (short-range, attractive, saturated near a nucleon) and Coulomb repulsion (long-range, builds up with ZZ). Strong force dominates for light/medium nuclei; Coulomb dominates for heavy nuclei.

Alternative Method

You can compute fission energy from mass defects directly: ΔE=(mass of reactantsmass of products)×c2\Delta E = (\text{mass of reactants} - \text{mass of products}) \times c^2. This requires looking up atomic masses to several decimal places and converting via 1 u=931.5 MeV/c21 \text{ u} = 931.5 \text{ MeV}/c^2. The binding-energy method is faster for back-of-envelope estimates and is what NEET/JEE typically expects.

The actual fission energy of 235U\,^{235}\text{U} is about 200 MeV200 \text{ MeV} — our 213 MeV213 \text{ MeV} estimate is in the right ballpark. Always sanity-check against this number in PYQs. Anything above 300 MeV300 \text{ MeV} or below 150 MeV150 \text{ MeV} for a single fission is a red flag.

Common Mistake

Students conflate “high BE per nucleon = high mass” with “stable = heavier”. Iron is stable because each nucleon is tightly bound, not because it’s heavy. Uranium has more total binding energy than iron, but the per-nucleon binding is lower — that’s why uranium can split and release energy.

The other slip is forgetting the neutrons in the fission products. While free neutrons have B/A=0B/A = 0, the number of nucleons on each side must balance — here 235=141+92+2235 = 141 + 92 + 2. If your nucleon count doesn’t match, your equation is wrong.

Final answer: Energy released per fission 213 MeV\approx 213 \text{ MeV}.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next