For head on collision, distance between centres of two deuterons = r = 2 × radius ⇒ r = 4 fm = 4 × $10^{-15}$ m charge of each deuteron, e = 1.6 × $10^{-19}$ C Potential energy = $\frac{e^2}{4\pi\epsilon_0 r}$ = $\frac{9 \times 10^9 (1.6 \times 10^{-19})^2}{4 \times 10^{-15}}$ joule = $\frac{9 \times 1.6 \times 1.6 \times 10^{-14}}{4 \times 1.6 \times 10^{-16}}$ keV P.E. = 360 keV or P.E. = 2 × K.E. of each deuteron = 360 keV ∴ K.E. of each deuteron = $\frac{360}{2}$ = 180 keV. This is a measure of height of Coulomb barrier.