When two deuterons collide head-on, the distance between their centres, d is given as:
Radius of 1st deuteron + Radius of 2nd deuteron
Radius of a deuteron nucleus = 2 fm = 2 × 10−15 m
d = 2 × 10−15 + 2 × 10−15 = 4 × 10−15 m
Charge on a deuteron nucleus = Charge on an electron = e = 1.6 × 10−19 C
Potential energy of the two-deuteron system:
\(V = \frac{e^2}{4\pi∈_od}\)
Where,
∈o = Permittivity of free space
\(\frac{1}{4\pi∈_o} = 9 \times 10^9 Nm^2C^{-2}\)
\(∴ V = \frac{9 \times 10^9 x\times (1.6 \times 10^{-19})^2}{4 \times 10^{-15}} J\)
\(V = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{4 \times 10^{-15} \times (1.6 \times 10^{-19})} eV\)
V = 360 keV
Hence, the height of the potential barrier of the two-deuteron system is 360 keV.