What is the formula for calculating electrical potential?

[jjson775]

Homework Statement

4.00 MeV alpha particles (2 protons, 2 neutrons) scatter off gold nuclei (79 protons, 118 neutrons). If an alpha particle scatters backward at 180 deg., determine a) the distance of the closest approach to the gold nucleus and b) the max. force exerted on the alpha particle assuming the gold nucleus remains fixed throughout.

Relevant Equations

V = Ke q/r
F = Ke q1xq2/r^2

V = Ke x q/r
2x 1.602x10^-19 = 8.99 x 10^9 x (79x1.602x10^-19)/r

 

[haruspex]

Homework Statement:: 4.00 MeV alpha particles (2 protons, 2 neutrons) scatter off gold nuclei (79 protons, 118 neutrons). If an alpha particle scatters backward at 180 deg., determine a) the distance of the closest approach to the gold nucleus and b) the max. force exerted on the alpha particle assuming the gold nucleus remains fixed throughout.
Relevant Equations:: V = Ke q/r
F = Ke q1xq2/r^2

V = Ke x q/r
2x 1.602x10^-19 = 8.99 x 10^9 x (79x1.602x10^-19)/r

Per forum rules, you must post an attempt.

 

[jjson775]

The attempt is to solve for r as shown. It is clearly wrong

 

[kuruman]

V = Ke x q/r
2x 1.602x10^-19 = 8.99 x 10^9 x (79x1.602x10^-19)/r

Can you explain what the above two equations mean? If the second equation follows from the first, then you are asserting that the electrostatic potential at r is equal to twice the electron charge. That's not even dimensionally correct. You need to rethink your strategy and consider using the fact that the alpha particle has energy 4 MeV.

 

[jjson775]

The second equation does not follow from the 1st. If I can find r, it will give me the answer to part (b. I have no idea how to find r, the closest distance to the nucleus. In the attempt, I tried to equate the electrical forces between the particles. No luck. I don't see the relevance of the alpha particle energy. Should I be looking at conservation of energy or momentum?

 

[kuruman]

You should be looking at mechanical energy conservation. Assume that the alpha particle has 4.00 MeV of kinetic energy when it is infinitely far from the nucleus. What happens to that kinetic energy when it is as close to the nucleus as it will ever get?

Momentum is not conserved because the gold nuclei are not allowed to move.

 

[jjson775]

Got it. For some reason it wasn't clear to me that 4.00 MeV is the kinetic energy of the alpha particles. I also found the correct formula for the electrical potential of the 2 charges. Thanks.