Calculate the height of the potential barrier for a head-on collision of two deuterons. (Hint: The height of the potential barrier is given by the Coulomb repulsion between the two deuterons when they just touch each other. Assume that they can be taken as hard spheres of radius $2.0 \; fm$.)

(D) When two deuterons collide head-on,the distance between their centres,$d$,is the sum of their radii.
Radius of a deuteron nucleus $= 2.0 \; fm = 2.0 \times 10^{-15} \; m$.
Thus,$d = 2.0 \times 10^{-15} + 2.0 \times 10^{-15} = 4.0 \times 10^{-15} \; m$.
The charge on each deuteron nucleus is equal to the elementary charge $e = 1.6 \times 10^{-19} \; C$.
The potential energy $V$ of the two-deuteron system is given by the Coulomb potential: $V = \frac{1}{4 \pi \epsilon_0} \frac{e^2}{d}$.
Using $\frac{1}{4 \pi \epsilon_0} = 9.0 \times 10^9 \; N \cdot m^2/C^2$:
$V = \frac{9.0 \times 10^9 \times (1.6 \times 10^{-19})^2}{4.0 \times 10^{-15}} \; J$.
To convert to $eV$,divide by $e = 1.6 \times 10^{-19} \; C$:
$V = \frac{9.0 \times 10^9 \times 1.6 \times 10^{-19}}{4.0 \times 10^{-15}} \; eV = 3.6 \times 10^5 \; eV = 360 \; keV$.
Hence,the height of the potential barrier is $360 \; keV$.