(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A plutonium-239 nucleus, initially at rest, undergoes alpha decay to produce a uranium-235 nucleus. The uranium-235 nucleus has a mass of 3.90 * 10^{-25}kg, and moves away from the location of the decay with a speed of 2.62 * 10^{5}m/s.

Determine the minimum electric potential difference that is required to bring the alpha particle to rest.

2. Relevant equations

3. The attempt at a solution

m_{u-235}= 3.90 * 10^{-25}kg

v_{u-235}= 2.62 * 10^{5}m/s

m_{ap}= 6.65 * 10^{-27}kg

q_{ap}= +2e = +2(1.60 * 10^{-19}C) = +3.20 * 10^{-19}C

|p_{u-235}| = |p_{ap}|

m_{u-235}v_{u-235}= m_{ap}v_{ap}

v_{ap}= m_{u-235}v_{u-235}/m_{ap}

v_{ap}= (3.90 * 10^{-25}kg)(2.62 * 10^{5}m/s)/(6.65 * 10^{-27}kg)

v_{ap}= 1.536541353 * 10^{7}m/s

E_{k}= 1/2m_{ap}v_{ap}^{2}

E_{k}= 1/2(6.65 * 10^{-27}kg)(1.536541353 * 10^{7}m/s)^{2}

E_{k}= 7.85018977 * 10^{-13}J

E_{k}= q_{ap}V_{stop}

V_{stop}= E_{k}/q_{ap}

V_{stop}= (7.85018977 * 10^{-13}J)/(3.20 * 10^{-19}C)

V_{stop}= 2.45 * 10^{6}V

The minimum electric potential difference that is required to bring the alpha particle to rest is 2.45 * 10^{6}V.

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# Homework Help: Minium Electric Potential

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