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Electric Potential Difference involving Alpha Decay question

  • Thread starter Eloc Jcg
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  • #1
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Homework Statement


A plutonium-239 nucleus, initially at rest, undergoes alpha decay to produce a uranium-235 nucleus. The uranium-235 nucleus has a mass of 3.90x10^-25 kg, & moves away from the location of the decay with a speed of 2.62x10^5 m/s. Determine the minum electric potential difference that is required to bring the alpha particle to rest.


Homework Equations


V = ? = E/q
E = mc^2
m = mparent - mproducts
q = elementry charge x #protons

The Attempt at a Solution


m = 239.052156u - 235.0439299u - 4.002603u = -0.9943769u
E = -0.9943769u x 3x10^8m/s = 8.9493921x10^16 eV
q = 1.6x10^19C x 92protons = 1.472x10^-17 C

V = 8.9493921x10^16eV / 1.472x10^-17C = 6.08 x 10^33 V

Am I way off here in my approach to solving this problem? Or should this problem be solved using the Ek = 1/2mv^2 formula? Any help is appreciated as this is my last assignment I am needing to do, which assuming I pass my final, I will be done high school. I've been up all night working on this problem, & flipping through my high school physics text book & my college electronics text book, but I am not sure what to expect as my final answer.

Thanks.
 

Answers and Replies

  • #2
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I was thinking maybe use the F = E q = Vq/d = ma ?
But that means any non zero potential will be able to bring it to rest (lower potential will simply take long to do it and the nucleus will travel a larger distance while decelerating).
 

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