How to Calculate Potential and Speed of a Proton Near a Barium Nucleus?

[XxXDanTheManXxX]

A barium nucleus has a charge of q = +56e (where e is the proton charge).
a. find the potential V at a radius of 10-12 m from the Nucleus. If a Proton is released from this point, how fast will it be moving when it is 1 m from the nucleus? Note: Assume V (infinity=0)
Any help?

 

[Davorak]

I suggested in some of you other threads gaussian pill box. Learn to use it and love it. As an undergrad it looks like you have 1 or two more semsters of E&M and it won't go away. It might chagne a little though
$$\nabla \vec{E} = \frac{\rho}{\epsilon_0}$$

 

[thepatientmental]

Sure, I can help with this problem. To find the potential V at a radius of 10^-12 m from the nucleus, we can use the formula V = kq/r, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q is the charge of the nucleus (+56e), and r is the distance from the nucleus (10^-12 m). Plugging in these values, we get V = (9x10^9)(56e)/(10^-12) = 5.04x10^20 V.

To find the speed of a proton when it is 1 m from the nucleus, we can use the conservation of energy principle. At a distance of 10^-12 m, the proton has a potential energy of qV = (1.6x10^-19 C)(5.04x10^20 V) = 8.064x10^2 J. At a distance of 1 m, the potential energy will be converted into kinetic energy, so we can set the two equal to each other: qV = (1/2)mv^2, where m is the mass of the proton (1.67x10^-27 kg) and v is the speed we are looking for. Solving for v, we get v = √(2qV/m) = √(2(8.064x10^2 J)/(1.67x10^-27 kg)) = 1.6x10^9 m/s.

I hope this helps! Let me know if you have any other questions.