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NUC-05

Fission energy (Coulomb snapshot)

E ≈ 1.44 Z₁ Z₂ / (A₁^{1/3}+A₂^{1/3}) MeV. Coulomb energy of two fragments at contact.

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NuclearFission

Governing equation

E≈1.44 Z1Z2A11/3+A21/3 MeVE\approx\dfrac{1.44\,Z_1 Z_2}{A_1^{1/3}+A_2^{1/3}}\,\mathrm{MeV}

Z1∈[20,60]Z_1 \in \left[20, 60\right]

where

Z_1
Z light (—)
A_1
A light (—)
Z_2
Z heavy (—)
A_2
A heavy (—)
E
Fission energy (MeV)

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NUC-05 · reactor
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①

Governing law — red

E≈1.44 Z1Z2A11/3+A21/3 MeVE\approx\dfrac{1.44\,Z_1 Z_2}{A_1^{1/3}+A_2^{1/3}}\,\mathrm{MeV}

Z1∈[20,60]Z_1 \in \left[20, 60\right]

Z1=20Z_1 = 20E=165.5E = 165.5

②

Domain of the variable — blue

Z_1_{\min} = 20,\quad Z_1_{\max} = 60

Z1∈[20,60]Z_1 \in \left[20, 60\right]

Z1=20Z_1 = 20Z1=20Z_1 = 20

③

Response curve — green

E≈1.44 Z1Z2A11/3+A21/3 MeVE\approx\dfrac{1.44\,Z_1 Z_2}{A_1^{1/3}+A_2^{1/3}}\,\mathrm{MeV}

Z1∈[20,60]Z_1 \in \left[20, 60\right]

Z1=20Z_1 = 20E=165.5E = 165.5

④

Rate of change — amber

dEdZ1\frac{\mathrm{d}E}{\mathrm{d}Z_1}

Z1∈[20,60]Z_1 \in \left[20, 60\right]

Z1=20.25Z_1 = 20.25E′=8.277E^{\prime} = 8.277

Lecture brief

Historical brief

Rutherford, Chadwick, the semi-empirical mass formula, fission (Hahn–Strassmann 1938) and fusion Q-values, then Compton and Bethe–Bloch stopping, are the nuclear toolkit. Decay, binding and dose start here. This sheet (NUC-05 — Fission energy (Coulomb snapshot)) is the form associated with Fission. Working symbols: Z1Z_1, A1A_1, Z2Z_2, A2A_2 →\rightarrow EE. A pedagogical stand-in for the ~200 MeV of ²³⁵U. Most of the Q is Coulomb, not surface.

Purpose

Purpose: compute EE from Z1Z_1, A1A_1, Z2Z_2, A2A_2 in Nuclear physics via E≈1.44 Z1Z2A11/3+A21/3 MeVE\approx\dfrac{1.44\,Z_1 Z_2}{A_1^{1/3}+A_2^{1/3}}\,\mathrm{MeV} E ≈ 1.44 Z₁ Z₂ / (A₁^{1/3}+A₂^{1/3}) MeV. Coulomb energy of two fragments at contact. Use it when a real nuclear physics question must be answered in SI before a code check.

Live realistic example

In symbols

Live case. Given Z1=36.000 —Z_1 = 36.000\,\mathrm{—}, A1=95.000 —A_1 = 95.000\,\mathrm{—}, Z2=56.000 —Z_2 = 56.000\,\mathrm{—}, A2=139.000 —A_2 = 139.000\,\mathrm{—}, the governing relation E≈1.44 Z1Z2A11/3+A21/3 MeVE\approx\dfrac{1.44\,Z_1 Z_2}{A_1^{1/3}+A_2^{1/3}}\,\mathrm{MeV} yields E=298.0 MeVE = 298.0\,\mathrm{MeV}. A drop that splits, two charged fragments. Move a slider: the numbers are this situation, not a canned story.

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Outputs

  • Fission energy E298.0 MeV
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NUC-05 · reactor
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Narration of this film

A drop that splits, two charged fragments.

A pedagogical stand-in for the ~200 MeV of ²³⁵U. Most of the Q is Coulomb, not surface.

Reading speed

Watch on YouTube