Question regarding particle charge and electric fields

OK, I understood most of this, but I don't get why in the equation v=Q/4piEor which is thereby substituted by W=VQ. I'm assuming that the charge that is multiplied by two is that of the alpha particle, but why is the Q multiplied by 79, I'm assuming it has something to do with gold, but I would like an explanation to understand which variable correspond to values of which elements/particles.
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In essence, MS is showing the use of conservation of energy to solve the problem.
Loss of KE of alpha particle = Gain electric potential energy
KE of alpha particle = Q_Au × q_α / 4π ϵ₀ r
Q_Au = 79 × 1.60 × 10^‒19 (79 protons in Au)
q_α = 2 × 1.60 × 10^‒19 (2 protons in α)

"Gain electric potential energy"

Could this be interpreted as electric potential needed to overcome repulsive force between two charged particles hence why we use E=kQq/r ?