If two alpha particles (each with two protons and two neutrons) are 0.500 m apart, what is the electric potential energy of the system?

U = 1.84 10⁻²⁷ J

Explanation:

Electric potential energy is

        U = k q₁q₂/ r

in this case the electric charge is

         q = 2 p

         q = 2 1.6 10⁻¹⁹

         q = 3.2 10⁻¹⁹ C

indicate the distance between the charges is 0.500 m

we calculate

         U = 9 10⁹  3.2 10⁻¹⁹  3.2 10⁻¹⁹ /0.500

         U = 1.84 10⁻²⁷ J

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