Electric Potential alpha particles

[tigerguy]

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.

 

[tigerguy]

Disregard my question, I've managed to solve this. I needed to divide by 1.60E-19, to get a final answer of 880 eV.

 

[kyle8921]

Your approach using the conservation of energy is correct. However, there are a few things that need to be taken into consideration in your calculation.

Firstly, the potential difference between points A and B should be calculated as Vb - Va, not the other way around. This will give a negative value for the potential difference, which is the correct sign for the alpha particle to accelerate towards the negative potential.

Secondly, when using the equation qΔV = 1/2mv^2, the charge q should be the charge of the alpha particle, which is twice the charge of a proton (as you correctly stated). However, the mass m should be the mass of the alpha particle, which is 4 times the mass of a proton (since an alpha particle contains 2 protons and 2 neutrons). This will give a higher value for the kinetic energy.

Lastly, when solving for kinetic energy, the units for the potential difference should be in joules (J) and not in electron volts (eV). To convert from joules to eV, you can use the conversion factor 1 eV = 1.602 x 10^-19 J.

Taking all of these into consideration, the correct calculation should be:

qΔV = 1/2mv^2
(2 x 3.20 x 10^-19 C)(-440 V) = 1/2(4 x 1.67 x 10^-27 kg)v^2
-7.04 x 10^-16 J = 3.34 x 10^-27 kg v^2
v^2 = (2.11 x 10^11 m/s)^2
v = 4.59 x 10^5 m/s

To convert this velocity to eV, we can use the equation:

1/2mv^2 = KE
1/2(4 x 1.67 x 10^-27 kg)(4.59 x 10^5 m/s)^2 = 3.34 x 10^-16 J
KE = 2.10 x 10^-16 J
KE = 1.31 x 10^3 eV

Therefore, the kinetic energy of the alpha particle when it arrives at point B is 1.31 x 10^3 eV. I hope this helps in understanding and correcting your calculation.