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FelaKuti
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Homework Statement
An alpha-particle with velocity 3.5 x 106 strikes a block of gold, atomic number 79 and mass number 197. Find the distance of the closest possible approach between the alpha particle and a gold nucelus, assuming Coulomb's law holds over such distances. Assume mp = mn = 1.67 x 10-27
Homework Equations
p = mv
KE = 1/2mv2
EPE = kq1q2/r2
The Attempt at a Solution
Considering conversation of momentum and conservation of energy I worked out p = mv = (4 x 1.67 x 10-27)(3.5 x 106) = 2.3 x 10-20.
The particle would repel the nucleus, at the closest approach they must have the same velocity. So p = mauv' + malpha v'
v' = p/mau + mualpha
v' = 2.3 x 10-20 / 201 x 1.67 x 10-27
v' = 6.4 x 104
So the change in KE in KEv minus KEv'.
At this point I believe I've gone wrong as I've calculated the change in KE to be negligible yet the online question pp I'm using tells me to consider conversation of momentum (here). As I understand the remainder of the question is simply equate change KE to change in EPE, rearrange to find r and plug in values.