An alpha particle is fired at a uranium atom

AI Thread Summary
An alpha particle fired at a uranium nucleus will experience repulsion due to both being positively charged, converting kinetic energy into electrical potential energy as it approaches. The user initially calculated the closest distance to the nucleus as 1.02 x 10^-10 m, but the correct answer is 5.1 x 10^-11 m. The error stemmed from using an incorrect charge value for the alpha particle, mistakenly applying q = 4e instead of the appropriate charge. It is crucial to use accurate values for both charge and mass in calculations. Properly addressing these parameters will yield the correct closest approach distance.
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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.
 
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5.5 x 10^-11 is correct, do the math again.
 
Make sure you use the correct values for the charge and mass of the alpha particle.
 
I accidentally used q = 4e for the alpha particle (since m = 4 x mass of proton)... ^.^'''

Thanks, guys!
 
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