Electric Potential Difference homework

[spoonthrower]

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.

 

[Doc Al]

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?

 

[spoonthrower]

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.

 

[Doc Al]

Almost.

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