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Aishwarya pandey
In the question it is asking the closest approach of deutron to the silver nucleus. See the closest approach would be the distance at which the electrostatic repulsion energy becomes equal to the energy of deutron particle. Deutron have 1e charge and silver nucleus has atomic number 47, so charge will be 47e. where e= 1.6×10^19 = charge on one electron = charge on one proton. So as the deutron particle approaches the silver nucleus it will experience a repulsion due to electrostatic charge and at a minimum distance when it's energy becomes equal to the repulsive energy it will be repelled outwards from it's path. You can understand it by attached diagram. so in the given solution the kinetic energy of deutron has been equated to the electrostatic potential energy between the deutron and the silver nucleus and using this equation the value of r, at which repulsion occurs, has been found out. If you haven't got any point now please ask again. I'm here to clear your doubts.