Lesson 9 of 15

Coulomb Barrier

Coulomb Barrier

Two nuclei approaching each other experience a strong electrostatic repulsion. The Coulomb barrier is the potential energy at the point of nuclear contact — the height that must be overcome (or tunneled through) for a nuclear reaction to occur.

VC=keZ1Z2e2R1+R2V_C = \frac{k_e Z_1 Z_2 e^2}{R_1 + R_2}

where:

  • ke=8.988×109k_e = 8.988 \times 10^9 N·m²/C² (Coulomb constant)
  • e=1.602×1019e = 1.602 \times 10^{-19} C (elementary charge)
  • Ri=1.2×1015Ai1/3R_i = 1.2 \times 10^{-15} \cdot A_i^{1/3} m (nuclear radius)

Converting to MeV

VC[MeV]=VC[J]1.602×1013 J/MeVV_C [\text{MeV}] = \frac{V_C [\text{J}]}{1.602 \times 10^{-13} \text{ J/MeV}}

Contact Distance

The sum of the two nuclear radii gives the contact distance in femtometers (fm, 101510^{-15} m):

d=R1+R2=1.2(A11/3+A21/3) fmd = R_1 + R_2 = 1.2 \left(A_1^{1/3} + A_2^{1/3}\right) \text{ fm}

Why It Matters

The D+T fusion reaction has a Coulomb barrier of only ~0.44 MeV (both particles have Z=1Z=1), making it the easiest fusion reaction. Heavy-ion fusion (e.g., Pb+Pb) requires barriers of hundreds of MeV.

Your Task

Implement:

  • coulomb_barrier_J(Z1, A1, Z2, A2) — barrier in Joules
  • coulomb_barrier_MeV(Z1, A1, Z2, A2) — barrier in MeV
  • contact_distance_fm(A1, A2) — contact distance in femtometers

All physical constants must be defined inside each function.

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