Download Some properties of Thomson`s atom

Some properties of Thomson’s atom
In Thomson’s model for the Hydrogen atom, the positive charge e is uniformly distributed within a
sfere of radius a0 . The electron, having charge −e, is considered to be a point particle and is located
inside the sphere.
a) Find the electric field and the potential generated by the positive charge and the equilibrium
position for the electron (assumed to have zero angular momentum).
b) Find the ionization energy UI (i.e. the energy needed to remove the electron from the atom) and
the related value of a0 consistent with the experimental value UI = 13.6 eV = 2.18 × 10−18 J.
c) Find the oscillation frequency of the electron around its equilibrium position.
d) Find the polarizzability α of the atom and the dielectric constant ² in the “solid state” (in which
all atoms are adiacent to each other forming a lattice), using the mean field approximation.
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Solution
a) The field is radial and can be found by Gauss’s theorem applied to an uniformly charged sphere
of density ρ+ = e/(4πa30 /3):
½
e
r/a30 (r < a0 )
.
×
E(r) =
1/r2 (r > a0 )
4π²0
The electric potential V = V (r) satisfies E = −∂r V and thus we obtain by integrating
½
e
(3/2a0 − r2 /2a30 ) (r < a0 )
×
V (r) =
,
1/r
(r > a0 )
4π²0
where we assumed V (r = ∞) = 0. The position of stable equilibrium is at r = 0.
b) Assuming that the electron is at r = 0, its total energy is −eV (0). The energy needed to take
the electron to infinity is
UI = eV (0) =
3k0 e2
,
2a0
where k0 = 1/4π²0 ' 9 × 109 SI units. We thus obtain
a0 =
3k0 e2
3 × 9 × 109 × (1.6 × 10−19 )2
m = 1.6 × 10−10 m .
=
2UI
2 × 10−18
The actual atomic dimensions are about one third of this value.
c) The equation of motion of the electron is
e
k0 e2
dvr
= − E(r) = −
r ≡ −ω 2 r ,
dt
me
me a30
thus the electron motion is harmonic, with a frequency
s
k0 e2
' 7.9 × 10−15 s .
ω=
3
m e a0
This value approaches the experimental one just as an order of magnitude estimate.
d) Let us find the electric dipole p induced by an external field E0 . Due to this latter, the equilibrium
position for the electron shifts to the value req given by E(req ) = E0 , from which we obtain req =
4π²0 a30 E0 /e. The induced dipole is thus
α = 4π²0 a30 .
p = −ereq ≡ αE0 ,
The number density of atoms in the “solid” state is n = 1/(2a0 )3 ; thus
²r = 1 +
nα
π
=1+ .
²0
2
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