Electric Potential & Kinetic Energy

AI Thread Summary
The discussion revolves around calculating the kinetic energy of an alpha particle moving from a potential of +250 V to -150 V. Participants clarify that the potential difference is 400 V, and using the formula for potential energy (EPE = qΔV), the kinetic energy can be derived. The charge of the alpha particle, being +2e, is emphasized, and it is noted that energy must be converted from joules to electron volts by dividing by the charge of an electron. Ultimately, the correct kinetic energy of the alpha particle when it reaches point B is confirmed to be 800 eV. The conversation highlights the importance of unit consistency in calculations.
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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?
 
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the potential energy is q\Delta V, 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...
 
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?
 
pinky2468 said:
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?
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.
 
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?
 
pinky2468 said:
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?

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

where e = 1.6 * 10 -19 Coulombs...
 
Last edited:
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?
 
christinono said:
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?

well the units are wrong 800 eV is not the same as 800 J...
 
No, the answer(according to the book) is 800eV
 
  • #10
christinono said:
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?
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?
 
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