An alpha particle is fired at a uranium atom

[newageanubis]

Homework Statement

"If an alpha particle is fired straight at the nucleus of an uranium atom (q = 92e), at a velocity of 5 x 10^5 m/s, how close will it get to the uranium nucleus?

Homework Equations

Ek = (mv^2)/2
Ee = (kq1q2)/d

The Attempt at a Solution

Since the alpha particle and uranium atom are both positively charged, they will naturally repel each other. Firing the alpha particle at the uranium atom will therefore store electrical energy. At the closest point the alpha particle gets to the nucleus, all of its kinetic energy has been converted into electrical potential energy. Therefore, I equated Ee and Ek, and solved for d, taking the mass of an alpha particle to be quadruple that of a proton. However, the answer I got was 1.02 x 10^-10 m, but the correct answer is 5.1 x 10^-11 as per the provided solution.

 

[Xisune]

5.5 x 10^-11 is correct, do the math again.

 

[TSny]

Make sure you use the correct values for the charge and mass of the alpha particle.

 

[newageanubis]

I accidentally used q = 4e for the alpha particle (since m = 4 x mass of proton)... ^.^'''

Thanks, guys!

 

[Nickie]

I would first check my calculations and make sure I didn't make any errors. I would also consider the fact that the uranium atom is not a point charge and has a finite size, which could affect the distance at which the alpha particle gets closest to the nucleus. Additionally, I would take into account any other factors that could affect the motion of the alpha particle, such as the presence of other particles or external forces. I would also consider the limitations of the equations I used and whether there are more accurate or appropriate equations that could be used in this scenario. Overall, I would continue to analyze and refine my solution to better understand the behavior of the alpha particle and uranium atom in this situation.