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.
For head on collision distance between centres of two deuterons `= r = 2xx` radius
`r = 4 fm = 4xx10^(-15)m`
Charge of each deuteron `e = 1.6xx10^(-10)C` Potential energy
`(e^(2))/(4pi epsilon_(0)r)=(9xx10^(9)(1.6xx10^(-19))^(2))/(4xx10^(-15))` Joule
`= (9xx1.6xx1.6xx10^(-14))/(4xx1.6xx10^(-16))` KeV
PE = 360 KeV
P.E `= 2xx` K.E of each deuteron = 360 KeV
K.E of each deuteron `= (360)/(2)=180` KeV
This is a measure of height of Coulomp barrier.
