Question

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?

Answer 1

U = 1.84 10⁻²⁷ J

Step-by-step 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