What is the Minimum Electric Potential Needed to Stop an Alpha Particle?

[coreluccio]

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 3.90 * 10^-25 kg, and moves away from the location of the decay with a speed of 2.62 * 10⁵ m/s.

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

Homework Equations

The Attempt at a Solution

m_u-235 = 3.90 * 10^-25 kg
v_u-235 = 2.62 * 10⁵ 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-235v_u-235 = m_apv_ap
v_ap = m_u-235v_u-235/m_ap
v_ap = (3.90 * 10^-25 kg)(2.62 * 10⁵ m/s)/(6.65 * 10^-27 kg)
v_ap = 1.536541353 * 10⁷ m/s
E_k = 1/2m_apv_ap²
E_k = 1/2(6.65 * 10^-27 kg)(1.536541353 * 10⁷ m/s)²
E_k = 7.85018977 * 10^-13 J
E_k = q_apV_stop
V_stop = E_k/q_ap
V_stop = (7.85018977 * 10^-13 J)/(3.20 * 10^-19 C)
V_stop = 2.45 * 10⁶ V

The minimum electric potential difference that is required to bring the alpha particle to rest is 2.45 * 10⁶ V.

 

[coreluccio]

I posted my work/answer. I just want to confirm that it is in fact right.

 

[diazona]

You're not being ignored, sometimes it just takes a while ;-) Anyway, the procedure you used looks reasonable, and I don't see any obvious errors in what you've done.