- #1

FelaKuti

- 19

- 0

## Homework Statement

An alpha-particle with velocity 3.5 x 10

^{6}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 m

_{p}= m

_{n}= 1.67 x 10

^{-27}

## Homework Equations

p = mv

KE = 1/2mv

^{2}

EPE = kq

_{1}q

_{2}/r

^{2}

## 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 10

^{6}) = 2.3 x 10

^{-20}.

The particle would repel the nucleus, at the closest approach they must have the same velocity. So p = m

_{au}v' + m

_{alpha}v'

v' = p/m

_{au}+ mu

^{alpha}

v' = 2.3 x 10

^{-20}/ 201 x 1.67 x 10

^{-27}

v' = 6.4 x 10

^{4}

So the change in KE in KE

_{v}minus KE

_{v'}.

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.