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| Experimental design of α-particle scattering on gold foil |
Lord
Rutherford realized that the information regarding the internal structure of
the matter could be revealed by bombarding matter with high speed α-particles.
The α-particles are doubly ionized helium atoms, i.e. He2+.
Geiger and Marsden carried out the experiment under the suggestion of
Rutherford by bombarding a very thin foil (thickness of the order of 10-7
m) of gold (Z = 79) with a narrow beam of α-particles of high energy
incoming from radium, a radioactive material. Behind the target foil, fluorescent
screen was placed to detect the fate of α-particles after bombardment. It
was found that close to 99% of the incident particles passed the target gold foil
being undeflected. Only a very small fraction got deflected by quite large
angles and a very few of them were deflected by 90o or more.
Deflection
by these large angles could not be explained from Thomson's atomic model in
which positive charges were assumed to remain uniformly distributed throughout
the volume of atomic dimension. The massive positive charge matter is
responsible to deflect He2+. Hence, if the model were correct we
could have the observations as:
(i)
The fate of all the incident α-particles would be the same as
the interacting positive charge matter is uniformly distributed throughout the
space.
(ii)
The α-particles should experience only a very weak
electrostatic repulsive barrier from the positive charge matter, as it is
diluted due to its uniform distribution throughout the volume. Such with
interactions can never produce deflection by large angles such as 90° or more.
The
Thomson's atomic model could not survive in the experimental verification and
Rutherford proposed his nuclear model in 1911 to explain the experimental
observations. According to this model:
(i)
All the positive charges of an atom along with most of the atomic mass are
concentrated within a very small central region of atom. The central region is
called the nucleus.
(ii)
The electrons which balance the positive charge a revolving around the nucleus
in different circular orbits just like the planets around the sun. Due to the
revolving nature, they acquire a centrifugal force (= mu2/r, where m
and u are the mass and velocity of electron respectively, and r is the radius
of the orbit) which balances is the electrostatic attractive force towards the
nucleus. Thus it saves them from falling in the nucleus and ensures the
stability of the atom.
The
proposed model explains the observations of α-particles scattering. When the α-particles collide with the nucleus or pass through the vicinity of the nucleus which
occupies only a very small region in the whole atomic dimension are deflected
by large angles. In this model, most of the space is vacant and transparent to
the α-particles. The negatively charged very light
particles, i.e. the electrons, have nothing to do with the high energy massive α-particles.
They fail to attract the high energy massive α-particles.
Here
it is worth mentioning that the foil must be sufficiently thin so that,
(i)
cross sections of adjacent nuclei do not overlap,
(ii)
an α-particle suffers its entire deflection from an
encounter with a single nucleus,
(iii)
it does not reduce the velocity of the α-particle which traverses it.
In the treatment, it is assumed that the scattering
nucleus remains at rest and this approximation is only reasonably valid for the
heavier nucleus like that of gold.
Also,
there is a reason why gold foil is chosen for this particular experiment. The
reasons are:
Condition of
malleability: In
the α-particle scattering experiment, the foil was
required to be very thin (10-7 m). The reasons behind the
requirement of such a thin foil have been explained already. To make such a
thin foil, the material must be highly malleable. The metals which possess the fcc structure are highly malleable
because of the existence of large number of slip systems. Gold having the fcc structure satisfies the condition of
malleability.
Condition of
massive scattering nucleus:
During the interaction between the approaching α-particle and scattering nucleus,
the scattering nucleus must be sufficiently massive to satisfy the condition.
Condition of high
positive charge of the scattering nucleus: To detect the α-particles, by a large angle the
scattering nucleus should have high positive charge and this condition is
satisfied for gold having Z = 79.
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| Trajectory of α-particle Scattering |
An
α-particle
of mass ma and charge q approaching with a velocity u along AO, when
a nucleus of charge Ze is placed at D. Let the position of the particle α-particle
at a particular instant at A be denoted by (r, Ф) where r gives the distance
between the particle and the nucleus. The repulsive force is given by
F
= Zeq/4πε0r2 (SI unit)
F
= Zeq/ r2, (CGS unit)
It
can be mathematically proved that under the repulsive force, the α-particle
will move on a hyperbolic path along ABC with nucleus at the focus D. If the
perpendicular distance (i.e. shortest distance) between the nucleus (D) and the
original line of approach (AO) of the α-particle is given by b, called impact
parameter, then we have:
Cot(θ/2)
= 4πε0mau2b/Zeq (SI unit)
= mau2b/Zeq
(CGS unit)
The
above equations indicate that the value of impact parameter (b) increases as θ
decreases. In fact, when b→0, then θ→180o, i.e., the particle moving
in the direction of the nucleus will be deflected back in the same direction.
So
it appears that, all the α-particles approaching a target nucleus with an
impact parameter in the range zero to b, will be deflected by an angle ≥ θ. The
circular area (Ï€b2) of radius b is called the cross- section for
interaction. Any particle approaching towards the nucleus within this area will
suffer deflection by θ or more. If, the thickness of the foil target is t, and
the number of nuclei is n per unit volume then the number of nuclei present in
the cylinder of length t and cross-section πb2 is πb2nt.
As we are dealing with a very large number of α-particles, it is reasonably
expected that the fraction (f) of the α-particles deflected by θ or more is
given by:
f
= πb2nt
Nuclear Dimension
from the α-particle Scattering Experiment
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| Impact parameter and α-particle Scattering |
It
is known that the α-particle follows the trajectory ABC of a hyperbola due to
the electrostatic repulsive force between the target nucleus and the positively
charged α-particle. The nucleus D is placed at the outer focus. Finally, the
scattering α-particle follows the asymptote OC, after suffering a deflection by
θ, the angle between the initial path AO which is the path of asymptote of the
hyperbolic trajectory and the final path OC.
The
shortest distance between the nucleus D and the original path AO of the
particle is called impact parameter (b) which has been already defined. The
physical significance of b is that all the α-particles lying in the cross
section of area πb2 will suffer a deflection by θ or more. When b
becomes zero, θ becomes 180o. Under this condition the kinetic
energy of the α-particle is equal to the electrostatic repulsive energy at the
closest distance r0 between the α-particle and the target nucleus.
Hence,
1/2
mau2 = Zeq/4πε0r0
r0 = Zeq/[4πε0(1/2)mau2]
If
the maximum kinetic energy (=1/2 mau2) is 7.7 MeV (7.7
MeV = 7.7 × 106 eV × 1.6 × 10-19 joules/eV = 1.23 ×10-12
joules; ε0 = 8.854 × 10-12; q = 2e for α-particles), then
by substituting these values we get, r0 = 3.8 × 10-16 Zm.
For the gold foil, Z = 79, and r0 = 3.8 × 10-16 ×79 m, as
r0 gives the minimum possible distance between the α-particle (of
7.7 MeV) and then the target nucleus. It indicates that the nucleus is about
1/104 th of the dimension of the atom.
Drawbacks of
Rutherford's Model:
(i) Instability of the Rutherford atom: Though it was attempted to
explain the stability of the atoms by considering to strong foundation stones
of the classical physics, i.e. Newton's law of motion and Coulomb's law of
electrostatic force, it could not be supported by the electromagnetic theory, a
revolving electron under an acceleration will radiate its energy in the form of
electromagnetic radiation. Thus by losing its energy, it will gradually spiral
down into the nucleus and the atom will get destroyed. As a matter of fact, if
it occurs so, it can be shown mathematically that the revolving electron will
require about one hundred-millionth of a second to get plunged into the
nucleus.
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| Fate of the revolving electron according to electromagnetic theory in Rutherford model |
(ii) No explanation
of atomic line spectra: From Rutherford's model, according to the electromagnetic theory, the energy
radiated in the form of electromagnetic radiation from the revolving electron
should lead to a continuous spectrum covering all frequencies in a certain range.
But experimentally, some well-defined spectral lines corresponding to certain
frequencies are noted instead of continuous spectrum.
1) Concise inorganic chemistry by J. D. Lee.
2) Inorganic Chemistry by James E. Huheey, Ellen A Keither, Richard L. Keither, Okhil K. Medhi.
3) Shriver and Atkins Inorganic Chemistry.
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