If an alpha particle (two protons and two neutrons) is given an initial (nonrelativistic) velocity v at a very far distance and is aimed directly at a gold nucleus (Z=79), what is the closest distance d the alpha particle will come to the nucleus? In this problem you can estimate that the mass of the proton mp is equal to the mass of the neutron and only consider the effects of a single gold nucleus. Assume that the alpha particle comes close enough so that the nucleus is not substantially screened by inner-shell electrons.

From the law of conservation of energy, the total mechanical energy of the alpha particle at a very far distance, E equals the total mechanical energy at its closest point, E'

E = E'

U + K = U' + K' where

U = electric potential energy at very large distance = 0,
K = initial kinetic energy of alpha particle,
U = electric potential energy of alpha particle at closest distance,d and
K = final kinetic energy of alpha particle = 0 (since it stops at the closest distance, d)

So, U + K = U' + K'

0 + K = U' + 0

K = U' where

K = 1/2Mv² where
M = mass of alpha particle = 4mp where
mp = mass of proton and
v = speed of alpha particle and
U = kQq/d where
k = electric constant,
Q = charge on gold atom = Ze,
q = charge on alpha particle = 2e where
e = electron charge and
d = closest distance of alpha particle

1/2Mv² = kQq/d

1/2× 4mpv² = k × Ze × 2e/d

2mpv² = k × 79e × 2e/d

mpv² = 79ke²/d

Closest distance the alpha particle will come to the nucleus

Making d subject of the formula, we have

d = mpv²/79ke²

So, the closest distance the alpha particle will come to the nucleus is d = mpv²/79ke²

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