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.)
In case of collision of two deuterons, the distance between their centres d is given as
∵ d = Radius of first atom + Radius of 2nd atom
radus of deuteron = 2 fm = 2 × 10-15 m
Hence, d = 2 × 2 × 10-15 = 4 × 10-15 m
∵ Potential energy of this two deuteron system will be V = 
where Є0 = permittivity of free space
And 
= 9 × 109 m2C-2
J
∴ 
∴ V = 360KeV
Hence, the height of barrier potential is 360 keV.
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