What is the distance of closest approach(in x10-14 m) when a 5.0 MeV proton approaches a gold nucleus?

At the distance r0 of closest approach,KE of a proton = PE of proton and the gold nucleusK=12mv2=14πε0⋅Ze⋅er0q1=Ze,q2=eor  r0=14πε0⋅Ze2KBut K=5.0MeV=5.0×1.6×10−13J For gold, Z=79∴ r0=9×109×79×1.6×10−1925.0×1.6×10−13 =2.28×10−14m≃2.3×10−14m