(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 x 10^{-25}kg, and moves away from the location of the decay with a speed of 2.62 x 10^{5}m/s.

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

2. Relevant equations

From the question, is "the minimum electric potential difference...required to bring the alpha particle to rest" the stopping voltage of the alpha particle?

If this is the case, the formula would be

V_{stopping}= Ekmax/q

V_{stopping}= (1/2mv^{2})/2

V_{stopping}= mv^{2}/2q

3. The attempt at a solution

(239/94)Pu --> (235/92)U + (4/2)He

According to conservation of momentum, the alpha particle's momentum must exactly counteract that of the uranium nucleus, as the initial momentum of the system was zero.

Since we know the mass and velocity of the uranium nucleus, its' momentum (p=mv) is:

1.0218 x 10^{-19}kgm/s

Therefore, the momentum of the alpha particle must be negative this amount.

Knowing the mass of the alpha particle to be 6.65 x 10^{-27}kg (from data sheet), velocity of the alpha particle (v = p/m) is -1.54 x 10^{7}m/s. Also from the data sheet is the charge (q) of the alpha particle, +2e, or 3.2 x 10^{-19}C.

Using the stopping potential formula given above, the Vstopping of the alpha particle is:

mv^{2}/2q

(6.65 x 10^{-27}kg)(-1.54 x 10^{7}m/s)2 / 2(3.2 x 10^{-19}C)

= 2.45 x 10^{-8}V

I am very uncertain of this result, and I don't have the luxury of knowing the correct answer. I think my approach may be flawed - should I be using E = mc^{2}rather than stopping potential?

Any help will be greatly appreciated! Thanks!

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# Homework Help: Minimum Electrical Potential Difference required to stop Alpha particle

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