# Homework Help: Kinetic Energy (in eV) of He with given change in potential

1. Jun 30, 2009

### xcmntgeek

1. The problem statement, all variables and given/known data
Point A is at a potential of +230 V, and point B is at a potential of -140 V. An -particle is a helium nucleus that contains two protons and two neutrons; the neutrons are electrically neutral. An -particle starts from rest at A and accelerates toward B. When the -particle arrives at B, what kinetic energy (in electron volts) does it have?

hence:
Change in potential=370 Volts

2. Relevant equations
Change in potential= -work/q

and

1e- moving through one volt of potential= 1.6e-19 Joules

3. The attempt at a solution

He= 2 protons = 2 (1.6e-19 coulombs)= 3.2e-19c

(370 Volts)(3.2e-19c)= 1.184e-16

Of course, this isn't right or I wouldn't be posting this. My ideas for where I've gone astray:

-the charge (in coulombs) of He is different than 3.2e-19
-The answer asks for the kinetic energy, but I thought eV was a measure of Potential Electric Energy?

2. Jun 30, 2009

### LowlyPion

Welcome to PF.

If they want the answer in eV ...

... isn't it simply 2*370 eV ?

3. Jun 30, 2009

### xcmntgeek

Thanks for the welcome.

Isn't 370 volts just the difference in potential from A-->B ?

4. Jul 1, 2009

### LowlyPion

Yes.

And if W = KE = q*ΔV

So ... q = 2 electrons worth and ΔV = 370 V ...

5. Jul 1, 2009

### xcmntgeek

I had tried that originally, but it didn't work. Here's how:
2(1.6e-19)*(370)=1.184e-16

Thanks for the help

6. Jul 1, 2009

### LowlyPion

That answer is in Joules, not eV.

7. Jul 1, 2009

### xcmntgeek

Ohhhh, got it. Much more simple than I made it out to be.

Thanks again