Electric Potential & Kinetic Energy

pinky2468

Here is the problem:
Point A is at a potential of +250 V, and Point B is at a potential of -150 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 point B, what kinetic energy( in electron volts) does it have?

So here is what I know: alpha particle +2 charge and atomic mass of 4. KE=1/2mv^2. I know that E=KE + PE and E final = E initial.
I am going around in circles with this one any advice on where to begin?

vincentchan

the potential energy is [tex] q\Delta V [/tex], use this definition and conservation of energy.....
the mass and velocity of the particel is irrelevent, since you are asking to find KE only... ... beware the unit you use.... we are not doing SI unit here...

pinky2468

So is KEf= KEi+EPEi-EPEf right? The answer is 800eV and if I do this...
KE=0 +(250)(+2)-(-150)(+2) ...I get 800J. So do I need to multiply the +2 charge by 1.6X10-19? If so then that changes the answer, I am missing a step somewhere right?

christinono

Everything in the formula KEf= KEi+EPEi-EPEf must be expressed in the same units. You should express EPEi and EPEf in J. Then, when you determine KE, which will be in J, just divide that answer by 1.6X10-19 to get eV.

pinky2468

If I do that it changes answer. I got 800 J and the answer is 800eV. If I divide that by 1.6E-19 I get 5E21?

apchemstudent

you got the right idea... however though, the charge is not 2+. its +2e...

where e = 1.6 * 10 -19 Coulombs...

christinono

Point A is at a potential of +250 V, and Point B is at a potential of -150 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 point B, what kinetic energy( in electron volts) does it have?

Ok, lt me try it.
(delta) V=EPE/q
400V = EPE/q
400V*q = Ekfinal (in J)
400V*3.2x10^(-19) = Ekfinal (in J) = 1.28x10^(-16) J

Convert this to eV (divide by the charge on 1 electron), and you get 800J.
Isn't that the answer?

apchemstudent

well the units are wrong 800 eV is not the same as 800 J....

pinky2468

No, the answer(according to the book) is 800eV

christinono

Sorry...small mistake. Change my last 2 lines (previous post) to:

Convert this to eV (divide by the charge on 1 electron), and you get 800eV.
Isn't that the answer?