What is the speed of a proton accelerated by a potential difference of 250 MV?

Question

perl · Answer

A proton (q = e , m = 1.67*10^-27 kg )  moves "downhill" through a potential difference of 250 MegaVolts  (250 * 10^6 Volts)

perl · Answer

ok so far?

calleyleeann · Answer

I'm really confused about the "accelerated by a potential difference of 250 MV" part... :/ I don't understand what it means

calleyleeann · Answer

and I'm unsure how I can utilize that information

perl · Answer

The proton moves "downhill" through a 250*10^6 V potential difference.
The loss in PE must equal to the gain in KE.

PE= q*V = (1.6 * 10^-19 C) (250 *10^6 V) = 4.0*10^-11 J 
KE = 1/2 * m * v^2

solve PE = KE

calleyleeann · Answer

What is q?

perl · Answer

q is the charge of an electron in coulombs

calleyleeann · Answer

Oh okay. I understand now. 
Thank you!!

perl · Answer

I got 2.18 * 10^8  m/s

calleyleeann · Answer

Alrighty.  I'm working on it... slowly 
but surely