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:

$V=\frac{e^{2}}{4 \pi \in_{0} d}$

Where,

$\epsilon_{0}=$ Permittivity of free space

$\frac{1}{4 \pi \in_{0}}=9 \times 10^{9} \mathrm{~N} \mathrm{~m}^{2} \mathrm{C}^{-2}$

$\therefore V=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 10^{-15}} \mathrm{~J}$

$=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{4 \times 10^{-15} \times\left(1.6 \times 10^{-19}\right)} \mathrm{eV}$

$=360 \mathrm{keV}$

Hence, the height of the potential barrier of the two-deuteron system is 360 keV.