advanced34 min studyLesson 589 in Physics

Nuclear fission barriers and fragment distributions

Separate deformation stability, tunneling, energy partition, and the normalization of measured fragment yields.

On this page

What you’ll understand

  • Interpret the competition between surface and Coulomb deformation energies.
  • Partition the energy and momentum of a toy binary split.
  • Distinguish primary mass yields, post-emission yields, and pair correlations.

Fission is a change of a many-nucleon system into separated fragments along a complex deformation path. A positive energy release does not remove the barrier separating the initial compact shape from scission. Surface tension, electrostatic energy, shell structure, pairing, and collective inertia influence that path. The resulting fragment distribution contains dynamical and structural information that cannot be reconstructed from a single binding-energy difference.

A local curvature is not a complete fission barrier

In a liquid-drop approximation, elongating a nearly spherical nucleus increases its surface area but lowers its Coulomb energy. For a small dimensionless quadrupole deformation, the leading change is proportional to twice the spherical surface energy minus the spherical Coulomb energy. This motivates a fissility parameter comparing those two contributions. Below its liquid-drop instability value the sphere has positive curvature, but a lower-energy separated configuration can still exist beyond a finite barrier.

This local expansion does not determine the barrier maximum, scission configuration, or preferred mass split. Large deformations require more shape coordinates, and shell and pairing corrections can create multiple humps and metastable minima. A nucleus below a barrier can undergo spontaneous fission by collective tunneling, with a rate depending on the action and inertia along the chosen path. Treating a positive final Q as a rate prediction, or equating spontaneous fission with disappearance of local curvature, would conflate these distinct questions.

ΔE(ϵ)≃2Es−EC5ϵ2,xf=EC2Es\Delta E(\epsilon)\simeq\frac{2E_s-E_C}{5}\epsilon^2,\qquad x_f=\frac{E_C}{2E_s}
Es and EC are positive spherical surface and Coulomb energies in a common unit; ε and xf are dimensionless. This small-volume-preserving-quadrupole liquid-drop expansion diagnoses local stability only. It omits shell corrections, collective inertia, higher deformations, and a complete barrier path.↔ Scroll wide equations horizontally.

The Q-value feeds several energy reservoirs

Rest-mass differences determine the total energy available after all specified final states are named. At scission, much of that energy can appear as fragment kinetic energy, while primary fragments may retain intrinsic excitation. Later neutron emission, photons, beta electrons, recoil, and neutrinos distribute the remaining release along a decay chain. A ground-state Q-value is therefore not identical to the kinetic energy measured for one fragment, nor necessarily to energy deposited locally by every eventual product.

For an ideal binary split in the parent rest frame, the primary fragment momenta are equal and opposite before subsequent emission. The lighter fragment then carries more translational kinetic energy. Neutron evaporation changes fragment masses, momenta, and measured energies, so post-neutron quantities cannot be inserted uncritically into a primary two-body formula. Conservation checks should identify whether fragment mass numbers and Q-values refer to the initial compound system, the primary split, or the later products.

Q=TL+TH+EL∗+EH∗(primary binary split),TL=MHML+MHTtot,TH=MLML+MHTtotQ=T_L+T_H+E_L^*+E_H^*\quad\text{(primary binary split)},\qquad T_L=\frac{M_H}{M_L+M_H}T_{\rm tot},\quad T_H=\frac{M_L}{M_L+M_H}T_{\rm tot}
Q is the ground-state rest-energy release for the two fragment species; E* measures excitation above those ground states. All Q, T, and excitation quantities are energies. ML and MH are primary fragment inertial masses at the chosen nonrelativistic accuracy, and Ttot is their total translational kinetic energy. No particles are emitted before this specified split.↔ Scroll wide equations horizontally.

A yield curve is not a list of paired fragments

A primary single-fragment mass yield counts how often each mass appears, usually normalized to two fragments per binary event. Its integrated yield is therefore two, or two hundred percent under a common reporting convention. A distribution normalized instead as a random single-fragment probability sums to one. State the convention before interpreting peak heights. Complementary masses are correlated within an event, so drawing two masses independently from a single-fragment curve generally violates mass conservation.

Shell effects can favor asymmetric primary splits at some low excitation energies, while increasing excitation can change the pattern; neither symmetric nor asymmetric splitting is universal for all nuclei and conditions. Charge polarization, odd–even effects, excitation sharing, and neutron evaporation add further structure. A measured post-emission mass curve blends these ingredients with instrumental response. Models must be compared with observables defined at the same stage, and a successful average mass yield does not by itself validate predicted pair correlations or neutron distributions.

Local shape stability and a toy binary energy split

A toy liquid drop has Es=600 MeVE_s=600\,\mathrm{MeV} and EC=1000 MeVE_C=1000\,\mathrm{MeV}. Find its small-deformation energy at ϵ=0.10\epsilon=0.10. In a separate allowed primary split with mass ratio ML:MH=80:120M_L:M_H=80:120, let Q be 180 MeV and total fragment kinetic energy 150 MeV. Find each kinetic energy and the remaining total excitation.

  1. The curvature coefficient is (2Es−EC)/5=(1200−1000)/5=40 MeV(2E_s-E_C)/5=(1200-1000)/5=40\,\mathrm{MeV}. Multiplying by ϵ2=0.01\epsilon^2=0.01 gives a 0.40 MeV energy increase, so the toy sphere is locally stable against this small deformation.

  2. Equal opposite momenta divide the 150 MeV kinetic energy inversely with mass. The light fragment receives (120/200)(150)=90 MeV(120/200)(150)=90\,\mathrm{MeV} and the heavy fragment receives 60 MeV.

  3. Subtract total translational kinetic energy from the declared Q: EL∗+EH∗=180−150=30 MeVE_L^*+E_H^*=180-150=30\,\mathrm{MeV}. The given information does not specify how this excitation is shared between the two primary fragments.

The result

The local deformation cost is 0.40 MeV. The toy split has 90 MeV light-fragment and 60 MeV heavy-fragment kinetic energy, plus 30 MeV total excitation; local curvature does not determine the full fission probability.

Explore the model

Interactive lab

Interactive lab · function

Inspect a local deformation curvature

Represent normalized shape energy by y = a x², with x a small deformation coordinate. Reset and set both controls to a = 1 and x = 0.5; predict y = 0.25 and positive slope 1. Reset and set both to a = −1 and x = 0.5; predict y = −0.25 and slope −1. This illustrates local positive versus negative curvature only; the parabola has no fission saddle, scission, collective inertia, or fragment-yield calculation.

Inspect a local deformation curvatureThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forA local curvature test is narrower than a complete barrier or dynamical prediction.

Check your understanding

What does a positive small-deformation liquid-drop curvature establish?

Try it yourself

A toy A = 200 binary sample has random-single-fragment probabilities 0.40 at mass 80, 0.20 at mass 100, and 0.40 at mass 120, before neutron emission. Convert these into yields per fission, find the mean single-fragment mass, and explain why independent double sampling is invalid.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

There are two primary fragments per event, but their mass numbers must sum to the parent value.

Compare with the solution

Multiplying the single-fragment probabilities by two gives yields 0.80, 0.40, and 0.80 fragments per fission; they sum to two. The mean mass is 0.4(80)+0.2(100)+0.4(120)=1000.4(80)+0.2(100)+0.4(120)=100. Independent draws could produce an 80–80 pair, violating A = 200 conservation for the specified no-emission primary split. The one-fragment yield needs complementary pair correlations rather than independent sampling.

Take it with you

  • Energy release, a deformation barrier, and a fission rate are distinct.
  • Primary fragment recoil partitions kinetic energy inversely with mass.
  • Yield normalization and the stage before or after neutron emission must be explicit.

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