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Asked in GUJCET 2025 · Reactors and conditions for fission and fusion
Given: each deuteron has radius r_d=2 fm=2×10⁻¹⁵ m and charge +e.
Idea: head on, the deuterons come no closer than surface to surface, so their centres are d=2r_d=4×10⁻¹⁵ m apart. The top of the barrier is the electrostatic potential energy at that separation, all of it supplied by the kinetic energy the deuterons started with.
U=1/(4πε₀)(e²)/d=(9×10⁹×(1.6×10⁻¹⁹)²)/(4×10⁻¹⁵).
(1.6×10⁻¹⁹)²=2.56×10⁻³⁸, so U=(2.304×10⁻²⁸)/(4×10⁻¹⁵)=5.76×10⁻¹⁴ J.
That is 360 keV, shared between the two deuterons, so each needs about 180 keV — the reason fusion needs temperatures of order 10⁹ K.
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