- #1
tigerguy
- 32
- 0
Hi, I'm stuck on the following question. Maybe someone can point out where I'm going wrong.
Point A is at a potential of +300 V, and point B is at a potential of -140 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 B, what kinetic energy (in electron volts) does it have?
My steps for this solution was to use the conservation of energy approach. So I said that the EPEa = EPEb + KE. Upon doing that, I solved this to qVa-qVb = 1/2mv^2, where q = 2 times the charge of a proton, or 3.20 x 10^-19. When I solved for the Kinetic Energy, I get 1.41 x 10^-16, which is wrong.
Are my units wrong (the answer has to be in eV), or am I missing a step?
Thanks.
Point A is at a potential of +300 V, and point B is at a potential of -140 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 B, what kinetic energy (in electron volts) does it have?
My steps for this solution was to use the conservation of energy approach. So I said that the EPEa = EPEb + KE. Upon doing that, I solved this to qVa-qVb = 1/2mv^2, where q = 2 times the charge of a proton, or 3.20 x 10^-19. When I solved for the Kinetic Energy, I get 1.41 x 10^-16, which is wrong.
Are my units wrong (the answer has to be in eV), or am I missing a step?
Thanks.