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 the head on collision of two deuterons, the two nuclei come close up to a distance equal to the sum of their radii r=2+2=4fm=4×10−15m The charge on a deuteron e= charge on a proton =1.6×10−19c Potential energy of the neuterons during head-on collision U=4π∈0−1×re2=−9×109×4×10−15(1.6×10−19)2J=−9×109×4×10−15×1.6×10−19(1.6×10−19)2=−360KeV A deuteron will require a Kinetic energy of 360keV to surmount the potential barrier
