Calculate the height of the potential barrier for a head on collision of two deuterons.

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Solution

For a head on collision, distance between centers of two deuterons= $2 \times r a d i u s = r = 2 \times 2 f e r m i = 4 \times 10^{- 15} m$
Charge of each deuteron= $1.6 \times 10^{- 19} C$
Potential energy= $\frac{e^{2}}{4 π ϵ_{0} r} = 360 k e V$
Potential energy= $2 \times$ Kinetic Energy of Deuteron
$= 360 k e V$
This is the measure of height of Coulomb Barrier Potential.
Thus Kinetic energy of each deuteron= $180 k e V$ .

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