Change in Electric Potential Question

[r_swayze]

In Ernest Rutherford's scattering experiment, described in detail in an earlier homework problem in this chapter, alpha particles (with a mass of 6.68×10–27 kg and a charge of 3.20×10–19 C) come from a few milligrams of radium (to be very precise, its decay product radon). They leave with an initial speed of 2.25×107 m/s before being directed to strike a thin sheet of gold foil. What is the change in electric potential (watch the sign of your answer) that would be needed to accelerate an alpha particle, starting at rest, up to this speed? Ignore relativistic effects.

my work so far,

change in V = W/q = F*(change in x)/q = m*a*(change in x) / q

since vf^2 = vi^2 + 2a*(change in x)

then (vf^2 - vi^2) / 2 = a(chng in x)

So, chng in V = m((vf^2)/2) / q

Am I using the formulas correctly? For some reason my answer was wrong.

 

[jbsteele]

Yes, you are using the formulas correctly. The change in electric potential for an alpha particle with the given parameters is V = (6.68*10^-27 kg) * (2.25*10^7 m/s)^2 / (3.20*10^-19 C) = 8.39*10^5 J/C.