The potential difference necessary to accelerate an He+ ion ?

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SUMMARY

The potential difference necessary to accelerate a He+ ion (charge +e, mass 4u) from rest to a speed of 1.2×10^6 m/s can be calculated using the kinetic energy formula. The kinetic energy acquired by the ion is given by ΔKE = q*ΔV, where q is the charge of the ion (1.6×10^-19 C). The discussion clarifies that the potential difference is measured in volts, which is equivalent to joules per coulomb. The correct approach involves equating the kinetic energy to the work done by the electric field on the ion.

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The potential difference necessary to accelerate an He+ ion...?

Homework Statement



"What potential difference is needed to accelerate a He+ ion (charge +e, mass 4u) from rest to a speed of 1.2×10^6 m/s?"

Homework Equations



K_f + q*V_f = K_i + q*V_i

The Attempt at a Solution



I figure that:

q = 1.6*10^-19 for the charge of an electron I believe...?

I started off thinking these were somehow true:

V_f = (1/2) m*vf^2

V_i = (1/2) m*vi^2

but now I realize that's just silly, and they both equal K...which is the kinetic energy, sure...but I still have *no* idea what the V's are going to be!

I'm honestly kind of banging my head against the book here...it's not really teaching me what I need to do this problem in any reasonably understandable way. Can anyone help out with this?
 
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What are the units of (electrical) potential difference? Volts, right? But the Volt is a "compound" unit comprising [Joules]/[Coulomb].

So a charge q accelerating though potential difference V will acquire kinetic energy qV:

ΔKE = q*ΔV
 


You're absolutely right :D Got it, ty!
 

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