Electric Potential Difference homework

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The discussion focuses on calculating the kinetic energy of an alpha particle moving from point A at +220 V to point B at -120 V. The relevant equation for kinetic energy is KE = q(VA - VB), where q is the charge of the alpha particle, which is 2e. The potential difference is calculated as VA - VB, resulting in a value of 340 V. Participants clarify that the charge should not be converted to Coulombs for the final answer in electron volts (eV). The correct approach leads to determining the kinetic energy in eV based on the potential difference.
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Point A is at a potential of +220 V, and point B is at a potential of -120 V. An alpha-particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An alpha-particle starts from rest at A and accelerates toward B. When the alpha-particle arrives at B, what kinetic energy (in electron volts) does it have? Plz help. thanks.

Here are my thoughts so far... i know the mass of the alpha particle is 4 amu. The charge of the alpha particle is 2e. 6.25*10^18=1e so the charge of the alpha particle is 3.2*10^-19 C. I think i should use the following equation to solve this problem: KE=q(VA-VB). this will give me the answer in volts, however i need eV so 1V=1eV/(1.6*10^-19 C) When i plug in these numbers i am not getting the right answer. please let me know what i am doing wrong. thanks.
 
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Since you want the answer in eV, no point in converting charge to Coulombs (and back). Your equation is correct: KE=q(VA-VB). What's q? What's Va-Vb?
 
would q be equal to 2 since their are 2 e's in an alpha particle? And VA=220 and VB=-120 so KE=2(220-120)? thanks.
 
Almost.

q = 2e. Va = 220V; Vb = -120V; so, Va-Vb = (220V) - (-120V).
 

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