Repulsion

What is the magnitude of the electrostatic repulsion between two protons on opposite sides of the diameter of the nucleus?
Remember the Coulomb law: F = kq1q2/r2. Here the charges are only protons and the distance between charges is the diameter of the core!
atomic number Z = 26 and atomic mass number A = 54

Here, k = 9x10 ^ 9, q1 = q2 = 1.602 * 10 ^ -19 C

Repulsive force is, therefore, 9 * 10 ^ 9 * (1.602 * 10 ^ -19) ^ 2 / (9.06e-15) ^ 2 = 2.81391606 Newton!