If an α particle were scattered by many electrons and N electrons results in .
The number of atoms across the thin gold layer of 6 × 10−7 m:
Assume the distance between atoms is
and there are .
That gives .
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even if the α particle scattered from all 79 electrons in each atom of gold.
The experimental results were not consistent with Thomson’s atomic model.
Rutherford proposed that an atom has a positively charged core (nucleus) surrounded by the negative electrons.
The positive charge is concentrated in
a nucleus with a radius much smaller than 10-10 m
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Scattering experiments help us study matter too small to be observed directly.
There is a relationship between the impact parameter b and the scattering angle θ.
When b is small,
r gets small.
Coulomb force gets large.
θ can be large and the particle can be repelled backward.
4.2:
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Any particle inside the circle of area πb02 will be similarly scattered.
The cross section σ = πb2 is related to the probability for a particle being scattered by 1 nucleus.
The number of scattering nuclei per unit area .
The fraction of incident particles scattered is
f = ntAs/A
Rutherford Scattering
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In actual experiment a detector is positioned from θ to θ + dθ that corresponds to incident particles between b and b + db.
The number of particles scattered per unit area is
Rutherford Scattering Equation
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4.3:
Let’s consider atoms as a planetary model.
The force of attraction on the electron by the nucleus and Newton’s 2nd law give
where v is the tangential velocity of the electron.
The total energy is
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The Classical Atomic Model
Consider an atom as a planetary system.
The Newton’s 2nd Law force of attraction on the electron by the nucleus is:
where v is the tangential velocity of the electron:
The total energy is then:
This is negative, so the system is bound, which is good.
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The Planetary Model is Doomed
From classical E&M theory, an accelerated electric charge radiates energy (electromagnetic radiation) which means total energy must decrease. Radius r must decrease!!