Module 6 - The Electron and The Nucleus
I.
J.J. Thomson and the Electron
J.J. Thomson demonstrated convincingly that cosmic rays emitted from metallic
surfaces when an electric potential was applied were composed of particles.
Thus, he is credited with discovery of the electron.
A.
Experimental Apparatus
+V
d
B
Vacc
B.
L
Analysis
First, Thomson sent the electron beam through the device without a magnetic
field.
y
x
Free Body Diagram
The acceleration of the electron is given by
ay 
Since the acceleration is constant, we can apply the kinematic equations to find
the speed of the electron in the y-direction
vy 
The time that the electron is between the plates can also be determined using the
kinematic equations:
L=
t
Using our results, we have a relationship for the charge to mass ratio of an
electron:
q

m
Since the deflection of the electron in the y-direction is negligible within the
distance spanned by the plates, we have that
tan θ 
Thus, we have a relationship for finding the charge to mass ratio of the electron if
we can determine the initial speed of the electron:
q v x 2 d tan θ

m
VL
To determine the initial speed of the electron, Thomson used a Wein filter from
PHYS2424. With the magnetic field set to cancel the force of the electric field, we
have that
vx 
Thus, our final relationship for the charge to mass ratio of the electron is
q

m
 V 




 Bd 
2
d tan θ
VL

V tan θ
B2 d L
For the small angles of deflection, tan θ  θ so we have the equation of a straight
line with slope q/m as shown below:

q  B2 d L

V
m
θ  
(See pg. 12 of Rohlf)
II.
Thomson Model of the Atom
A.
Experimental Facts:
1)
_____________________________ exist in atoms and are
______________________________ charged.
2)
________________________ is _______________________ so it must
also have an ______________________ amount of
_________________________ charge.
B.
Model (Plumb Pudding)
III.
Alpha Scattering Experiments
A.
Setup
Collimator
Radioactive
Source
Gold Foil
B.
C.
Experimental Facts
1)
Most alphas only suffer small angle deflections as expected.
2)
A few alphas suffer large angle deflections (incredible result)!
Analysis
The force between the electric field of the gold atom of charge Z2 and the
incoming alpha particle of charge Z1 = 4 is found by

F
We now use Gauss's Law (PHYS2424) to determine the electric field due to the
positive uniformly charged sphere of Thomson's model:
R
Case I : r < R
Case II: r > R
The maximum force would occur when the particle is located at the distance
r = R as is given by

F
To determine the impulse, we need to determine the time that the force interacts.
We will assume that the force stays at its maximum value for the time required for
the alpha to travel across the diameter of the gold atom. This is an approximation
but it should over-estimate the alpha deflection and not under-estimate it!
2R
Δt
Δ p  FΔ t 
We can now determine the angle of deflection by looking at the change in linear
momentum diagram
p

p
tan θ 
Δp

p
tan θ
Using the radius of the atom (1x10-10 m) for the size of the positive charge, the
angle of deflection of a 6.0 MeV alpha particle incident upon the gold foil is
tan θ
1.44 eV  nm 279  3.79 x10 4  θ
0.1nm  6 x106 eV 


Our results so far consider only a single collision between the alpha particle and a
single gold atom in the foil. We can consider the effect of multiple collisions upon
the alpha particle. On average, the alpha particle's deflections will cancel out. This
is "drunken walk" problem and is described by a Gaussian distribution. The
standard deviation is given by
θ
std
 Nθ
where N is the number of scattering centers.
For a 1.0 m gold foil, the number of scattering centers can be approximately
determined by
6
1
x
10
m 1x104
N
1x1010 m
Thus, we have
θ
std
 1x104  3.79 x10 4   3.79 x102




Summary: Rutherford realized that if the positive charge was concentrated
into a smaller radius that the scattering angle is increased. By placing the positive
charge in a smaller radius of 1 fm and using classical mechanics, Rutherford
developed the following scattering cross section (probability) relationship
dσ   Z1 Z 2 
dΩ  E 
2
1
sin 4  θ 
 2


which was verified by careful experiments of his students Geiger and Marsden.
IV.
Rutherford's (Planetary) Model of the Atom
Small Positive
Charged Nucleus
Electron
A.
The model successfully explains alpha scattering as well as Thomson's results.
B.
However, the atom is STABLE while this model is UNSTABLE.
A force exists between the electron and the nucleus so the electron must be
_________________________________.
Thus, the electron must be ______________________________ according to
classical E&M.
Thus, the electron must lose _____________________ ____________________
and therefore it will ___________________ ______________________. The
electron will __________________ into the atom.
Classical E&M Picture
We would expect to see a continuous spectrum as the electron spiraled into the
nucleus in a few fractions of a second.