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E.~ Rutherford, Philos. Mag, {\bf 6,} 21 \hfill {\large \bf 1911}\\
\vspace{2cm}
\begin{center}
{\Large The Scattering of $\alpha$ and $\beta$ Particles by Matter and\\
the Structure of the Atom}\\
\end{center}
\begin{center}
E. Rutherford\\
University of Manchester\footnote{Communicated by the author. A brief account of
this paper was communicated to the Manchester Literary and Philosophical Society
in February, 1911.}\\
(Received April 1911)\\
\end{center}
\vspace{1cm}
\section*{\S~1}
It is well known that the $\alpha$ and $\beta$ particles suffer deflexions from
their rectilinear paths by encounters with atoms of matter. This scattering is
far more marked for the $\beta$ than for the $\alpha$ particle on account of the
much smaller momentum and energy of the former particle. There seems to be no
doubt that such swiftly moving particles pass through the atoms in their path,
and that the deflexions observed are due to the strong electric field traversed
within the atomic system. It has generally been supposed that the scattering of
a pencil of $\alpha$ or $\beta$ rays in passing through a thin plate of matter
is the result of a multitude of small scattering by the atoms of matter
traversed. The observations, however, of Geiger and Marsden\footnote{Proc. Roy.
Soc., LXXXII, p. 495 (1909)}
%\cite{Geiger1909}
on the scattering of $\alpha$
rays indicate that some of the $\alpha$ particles must suffer a deflexions of
more than a right angle at a single encounter. They found, for example, that a
small fraction of the incident $\alpha$ particles, about 1 in 20,000 turned
through an average angle of 90 degrees in passing through a layer of gold--foil
about .00004 cm. thick, which was equivalent in stopping power of the $\alpha$
particle to 1.6 millimetres of air. Geiger\footnote{Proc. Roy. Soc. LXXXIII. p.
492. (1910).}
%\cite{Geiger1910}
showed later that the most probable angle of
deflexions for a pencil of $\alpha$ particles traversing a gold--foil of this
thickness was about $0.87^{\circ}$. A simple calculation based on the theory of
probability shows that the chance of an $\alpha$ particle being deflected
through 90 degrees is vanishingly small. In addition, it will be seen later that
the distribution of the $\alpha$ particles for various angles of large deflexion
does not follow the probability law to be expected if such deflexions are made
up of a large number of small deviations. It seems reasonable to suppose that
the deflexion through a large angle is due to a single atomic encounter, for the
chance of a second encounter of a kind to produce a large deflexion must in most
cases be exceedingly small. A simple calculation shows that the atom must be a
seat of an intense electric field in order to produce such a large deflexion at
a single encounter.
Recently Sir J.J. Thomson\footnote{Camb. Lit. \& Phil. Soc. XV. pt. 5 (1910).}
%\cite{Thomson1910}
has put forward a theory to explain the scattering of
electrified in passing through small thickness of matter. The atom is supposed
to consist of a number $N$ of negatively charged corpuscles, accompanied by an
equal quantity of positive electricity uniformly distributed throughout a
sphere. The deflexion of a negatively electrified particle in passing through
the atom is ascribed to two causes -- (1) the repulsion of the corpuscles
distribution through the atom, and (2) the attraction of the positive
electricity in the atom. The deflexion of the particle in passing through the
atom is supposed to be small, while the average deflexion after a large number m
of encounters was taken as $\sqrt{m \cdot \theta},$ where $\theta$ is the
average deflexion due to a single atom. It was shown that the number $N$ of the
electrons within the atom could be deduced from observations of the scattering
of electrified particles. The accuracy of this theory of compound scattering was
examined experimentally by Crowther\footnote{Crowther, Proc. Roy. Soc. LXXXIV.
p.226 (1910).}
%\cite{Crowther1910}
in a later paper. His result apparently
confirmed the main conclusions of the theory, and he deduced, on the assumption
that the positive electricity was continuous, that the number of electrons in an
atom was about three times its atomic weight.
The theory of Sir J.J. Thomson is based on the assumption that the scattering
due to a single encounter is small, and the particular structure assumed for the
atom does not admit of a very large deflexion of an $\alpha$ particle in
traversing a single unless it be supposed that the diameter of the sphere of
positive electricity is minute compared with the diameter of the sphere of
influence of the atom.
Since the $\alpha$ and $\beta$ particles traverse the atom, it should be
possible from a close study of the nature of the deflexion to form some idea
of the constitution of the atom to produce the effects observed. In fact, the
scattering of high--speed charged particles by the atoms of matter is one of the
most promising methods of attack of this problem. The development of the
scintillation method of counting single $\alpha$ particles affords unusual
advantages of investigation, and the researches of H. Geiger by this method have
already added much to our knowledge of the scattering of $\alpha$ rays by
matter.\\
%\vspace{0.5cm}
\section*{
$\S~2$~}
We shall first examine theoretically the single
encounters \footnote{The deviation of a particle throughout a
considerable angle from an
encounter with a single atom will in this paper be called ``single'' scattering.
The deviation of a particle resulting from a multitude of small deviations will
be termed ``compound'' scattering} with an atom of simple structure, which is
able to produce large deflexions of an $\alpha$ particle, and then compare the
deductions from the theory with the experimental data available.
Consider an atom which contains a charge $\pm$ $Ne$ at its centre
surrounded by a
sphere of electrification containing a charge $\mp$ $Ne$ supposed uniformly
distributed throughout a sphere of radius is the fundamental unit of charge,
which in this paper is taken as $4.65 \times 10^{-10}$ E.S. unit. We shall
suppose that for distances less than $10^{-12}$ cm. the central charge and also
the charge on the $\alpha$ particle may be supposed to be concentrated at a
point. It will be shown that the main deductions from the theory are independent
of whether the central charge is supposed to be positive or negative. For
convenience, the sign will be assumed to be positive. The question of the
stability of the atom proposed need not be considered at this stage, for this
will obviously depend upon the minute structure of the atom, and on the motion
of the constituent charged parts.
In order to from some idea of the forces required to deflect an $\alpha$
particle through a large angle, consider an atom containing a positive charge
$Ne$ at its centre, and surrounded by a distribution of negative electricity
$Ne$
uniformly distributed within a sphere of radius $R$. The electric force $X$
and the
potential $V$ at a distance $r$ from the centre of an atom for a
point inside the atom, are given by
$$
X = Ne \cdot \left(\frac{1}{r^2} - \frac{r}{R^3}\right)
$$
$$
V = Ne \cdot \left(\frac{1}{r} - \frac{3}{2R} + \frac{r^2}{2R^3}\right)
$$
Suppose an $\alpha$ particle of mass $m$ velocity u and charge $E$ shot directly
towards the centre of the atom. It will be brought to rest at a distance $b$
from the centre given by
$$
1/2 mu^2 = NeE \cdot \left( \frac{1}{b} - \frac{3}{2R} + \frac{b^2}{2R^3}\right).
$$
It will be seen that $b$ is an important quantity in later
calculations. Assuming
that the central charge is $100 e$. it can be calculated that the value of b for
an $\alpha$ particle of velocity $2.09 \times 10^9$ cms. per second is
about $3.4 \times 10^{-12}$ cm. In this calculation $b$ is supposed to be
very small compared with $R$. Since $R$ is supposed to be
of the order of the radius
of the atom, viz. $10^{-8}$ cm., it is obvious that the $\alpha$ particle before
being turned back penetrates so close to the central charge, that the field due
to the uniform distribution of negative electricity may be neglected. In
general, a simple calculation shows that for all deflexions greater than a
degree, we may without sensible error suppose the deflexion due to the field of
the central charge alone. Possible single deviations due to the negative
electricity, if distributed in the form of corpuscles, are not taken into
account at this stage of the theory. It will be shown later that its effect is
in general small compared with that due to the central field.
Consider the passage of a positive electrified particle close to the centre of
an atom. Supposing that the velocity of the particle is not appreciably charged
by its passage through the atom, the path of the particle under the influence of
a repulsive force varying inversely as the square of the distance will be an
hyperbola with the centre of the atom $S$ as the external focus. Suppose the
particle to enter the atom in the direction $PO$ [Fig. 44--1], and that the
direction on motion on escaping the atom is $OP^{\prime}$.
$OP$ and $OP^{\prime}$
make equal angles with the line $SA$, where $A$ is the apse of the hyperbola.
$p = SN$ = perpendicular distance from centre on direction of initial motion of
particle.
Let angle $POA = \theta$.
Let $V$ = velocity of particle on entering the atom,
$\nu$ its velocity at $A$, then
from consideration of angular momentum.
$$
pV = SA \cdot v
$$
From conservation of energy
$$
1/2 m V^2 = 1/2 m v^2 + \frac{NeE}{SA},
$$
$$
v^2 = V^2(1 - \frac{b}{SA})
$$
%\vspace{1.5cm}
%\includegraphics{rutherford_11_1.gif}
%\\Fig. 1\\
%\vspace{1.5cm}
\begin{figure}[h]
\centerline{\epsfig{file=fig1.eps}}
\caption{}
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Since the eccentricity is see $\theta$
$$
SA = SO + OA = p ~~\mbox{cosec} \theta(1 + \cos \theta) = p \cot
\theta/2
$$
$$
p^2 = SA(SA - b) = p \cot \theta/2(p \cot \theta/2 - b),
$$
$$
b = 2p \cot \theta.
$$
The angle of deviation $\phi$ of the particle is $\pi - 2 \theta$ and
\begin{eqnarray}
\cot \phi/2 = \frac{2p^{\ast}}{b}\footnote{A simple consideration shows that the
deflexion is unaltered if the forces are attractive instead of repulsive.}
\end{eqnarray}
This gives the angle of deviation of the particle in terms of $b$, and the
perpendicular distance of the direction of projection from the centre of the
atom.
For illustration, the angle of deviation $\phi$ for different values of $p/b$
are shown in the following table:--
\vspace{0.5cm}
\begin{center}
\begin{tabular}{c c c c c c c c}\\
p/b &10&5&2&1&.5&.25&.125\\
$\phi$&5$^{\circ}$.7&11$^{\circ}.4$&28$^{\circ}$&53$^{\circ}$&
90$^{\circ}$&127$^{\circ}$&152$^{\circ}$\\
\hline
\end{tabular}
\end{center}
\vspace{0.5cm}
\section*{$\S~3$
{\it Probability of Single Deflexion Through any Angle}}
\vspace{0.5cm}
Suppose a pencil of electrified particles to fall normally on a thin screen of
matter of thickness $t$. With the exception of the few particles which are
scattered through a large angle, the particles are supposed to pass nearly
normally through the plate with only a small change of velocity. Let
$n$ = number
of atom in unit volume of material. Then the number of collisions of the
particle with the atom of radius $R$ is $\pi R^2 nt$ in the thickness $t$.
The probability $m$ of entering an atom within a distance $p$ of its
centre is given by
$$
m = \pi p^2 nt.
$$
Chance $dm$ of striking within radii $p$ and $p + dp$ is given by
\begin{eqnarray}
dm = 2 \pi p \cdot n \cdot t~dp = \frac{\pi}{4} ntb^2 \cot \phi/\mbox{cosec}^2
\phi/2 d \phi
\end{eqnarray}
since
$$
\cot \phi/2 = 2p/b.
$$
The value of $dm$ gives the {\it fraction} of the total number of particles which
are deviated between the angles $\phi$ and $\phi + d \phi.$
The fraction $\rho$ of the total number of particles which are deflected through
an angle greater than $\phi$ is given by
\begin{eqnarray}
\rho = \frac{\pi}{4} ntb^2 \cot^2 \phi/2
\end{eqnarray}
The fraction $\rho$ which is deflected between the angles $\phi_1$ and $\phi_2$
is given by
\begin{eqnarray}
\rho = \frac{\pi}{4} ntb^2(\cot^2 \frac{\phi_1}{2} - \cot^2 \frac{\phi_2}{2})
\end{eqnarray}
It is convenient to express the equation (2) in another from for comparison with
experiment. In the case of the $\alpha$ rays, the number of scintillation
appearing on a {\it constant} area of a zinc sulphide screen are counted for
different angles with the direction of incidence of the particles. Let $r$ =
distance from point of incidence of $\alpha$ rays on scattering material, then
if $Q$ be the total number of particles falling on the scattering material, the
number y of $\alpha$ particles falling on until area which are deflected through
an angle $\phi$ is given by
\begin{eqnarray}
y = \frac{Qdm}{2 \pi r^2 \sin \phi \cdot d \phi} = \frac{ntb^2 \cdot Q \cdot
\mbox{cosec}^4 \phi/2}{16 r^2}
\end{eqnarray}
Since
$$
b = \frac{2NeE}{mu^2},
$$
we see from this equation that a number of $\alpha$ particles (scintillations)
per unit area of zinc sulphide screen at a given distance $r$ from the point of
incidence of the rays is proportional to \\
\begin{itemize}
\item[(1)]
$\mbox{cosec}^4 \phi/2$ or $1/\phi^4$ if $\phi$ be small;
\item[(2)]
thickness of scattering material t provided this is small;
\item[(3)]
magnitude of central charge $Ne$;
\item[(4)]
and is inversely proportional to $(mu^2)^2$, or to the fourth power of velocity
if $m$ be constant.\\
\end{itemize}
In these calculations, it is assumed that the $\alpha$ particles scattered
through a large angle suffer only large deflexion. For this to hold, it is
essential that the thickness of the scattering material should be so small that
the chance of a second encounter involving another large deflexion is very
small. If, for example, the probability of a single deflexion $\phi$ in passing
through a thickness $t$ is 1/1000, the probability, of two successive deflexions
each of value $\phi$ is $1/10^6$, and is negligibly small.
The angular distribution of the $\alpha$ particles scattered from a thin
metal sheet affords one the simplest methods of testing the general correctness
of this theory of single scattering. This has been done recently for $\alpha$
rays by Dr. Geiger,\footnote{Manch. Lit. \& Phil. Soc. 1910.}
%\cite{Geiger1910a}
who found that the distribution for particles deflected
between 30 degrees and 150 degrees from a thin gold--foil was substantial
agreement with the theory. A more detailed account of these and other
experiments to test the validity of the theory will be published later.
%\vspace{0.5cm}
\section*{
$\S~4$ {\it Alteration of Velocity in an Atomic Encounter}}
%\vspace{0.5cm}
It has so far been assumed that an $\alpha$ or $\beta$ particles does not suffer
an appreciable change of velocity as the result of a single atomic encounter
resulting in large deflexion of the particle. The effect of such an encounter in
altering the velocity of the particle can be calculated on certain assumptions.
It is supposed that only two systems are involved, viz., the swiftly moving
particle and the atom which it traverses supposed initially at rest. It is
supposed that the principle of conservation of momentum and of energy applies
and that there is no appreciable loss of energy or momentum by radiation.
Let $m$ be mass of the particle
\begin{itemize}
\item[]
$\nu_1$ = velocity of approach,
\item[]
$\nu_2$ = velocity of recession,
\item[]
$M$ = mass of atom,
\item[]
$V$ = velocity communicated to atom as result of encounter.
\end{itemize}
Let $OA$ [Fig. 44--2] represent in magnitude and direction the momentum $m v_1$
of the entering particle, and $OB$ the momentum of the receding
particle which has
been turned through an angle $AOB$ = $\phi$. Then $BA$ represents in
magnitude and direction the momentum $MV$ of the recoiling atom.
$$
(MV)^2 = (m v_1)^2 + (m v_2)^2 - 2m2 v_1 v_2 \cos \phi. \eqno (1)
$$
By the conservation of energy
$$
MV^2 = m v_1^2 - m v_2^2 \eqno (2)
$$
%\vspace{1.5cm}
%\includegraphics{rutherford_11_2.gif}
%\\Fig. 1\\
%\vspace{1.5cm}
\begin{figure}[h]
\centerline{\epsfig{file=fig2.eps}}
\caption{}
\end{figure}
Suppose $M/m = K$ and $v_2 = \rho v_1,$ where $\rho$ is $<1$.
From (1) and (2),
$$
(K + 1) \rho^2 - 2 \rho \cos \phi = K - 1,
$$
or
$$
\rho = \frac{\cos \phi}{K + 1} + \frac{1}{K +1} \cdot \sqrt{K^2 - \sin^2 \phi}.
$$
Consider the case of an $\alpha$ particle of atomic weight 4, deflected through
an angle of 90 degrees by an encounter with an atom of gold of atomic weight
197.
Since $K = 49$ nearly,
$$
\rho = \sqrt{\frac{K - 1}{K +1}} = .979,
$$
or the velocity of the particle is reduced only about 2 per cent. by the
encounter.
In the case of aluminium $K$ = 27/4 and for $\phi = 90^{\circ},$ $\rho =
.86$,
It is seen that the reduction of velocity of the $\alpha$ particle becomes
marked on this theory for encounters with the lighter atoms. since the range of
an $\alpha$ particle in air or other matter is approximately proportional to the
cube of the velocity, it follows that an $\alpha$ particle of range 7 cms. has
its range reduced to 4.5 cms. after incurring a single deviation of 90
degrees in traversing an aluminium atom. This is of magnitude to be easily
detected experimentally. Since the value of $k$ is very large for
an encounter of
a $\beta$ particle with an atom, the reduction of velocity on this formula is
very small.
Some very interesting cases of the theory arise in considering the changes of
velocity and the distribution of scattered particles when the $\alpha$ particle
encounters a light atom, for example a hydrogen or helium atom. A discussion of
these and similar cases is reserved until the question has been examined
experimentally.\\
%\vspace{0.5cm}
\section*{
\S~5 {\it Comparison of single and compound scattering}}
%\vspace{0.5cm}
Before comparing the results of theory with experiment, it is desirable to
consider the relative importance of single and compound scattering in
determining the distribution of the scattering particles. Since the atom is
supposed to consist of a central charge surrounded by a uniform distribution of
the opposite sign through a sphere of radius $R$, the chance of encounters with
the atom involving small deflexions is very great compared with the chance of a
single large deflexion.
This question of compound scattering has been examined by Sir J.J. Thomson in
the paper previously discussed ($\S~1$). In the notation of this paper, the
average deflexion $\phi_1$ due to the field of the sphere of positive
electricity of radius $R$ and quantity $Ne$ was found by him to be
$$
\phi_1 = \frac{\pi}{4} \cdot \frac{NeE}{mu^2} \cdot \frac{1}{R}.
$$
The average deflexion $\phi_2$ due to the $N$ negative corpuscles supposed
distributed uniformly throughout the sphere was found to be
$$
\phi_2 = \frac{16eE}{5mu^2} \cdot \frac{1}{R} \cdot \sqrt{\frac{3N}{2}}.
$$
The mean deflexion due to both positive and negative electricity
was taken as
$$
(\phi^2_1 + \phi^2_2)^{1/2}.
$$
In a similar way, it is not difficult to calculate the average deflexion due to
the atom with a central charge discussed in this paper.
Since the radial electric field $X$ at any distance $r$ from the centre is given
by
$$
X = Ne \cdot \left( \frac{1}{r^2} - \frac{r}{R^3} \right),
$$
it is not difficult to show that the deflexion (supposed small) of an
electrified particle due to this field is given by
$$
\theta = \frac{b}{p} \cdot \left(1 - \frac{p^2}{R^2} \right)^{3/2},
$$
where $p$ is the perpendicular from the centre on the path of the particle and $
b$ has the same value as before. It is seen that the value of $\theta$
increases with diminution of $p$ and becomes great for small values of $\phi$.
Since we have already seen that the deflexions become very large for a particle
passing near the centre of the atom, it is obviously not correct to find the
average values by assuming $\theta$ is small.
Taking $R$ of the order $10^{-8}$ cm, the value of $p$ for a large deflexion is
for $\alpha$ and $\beta$ particles of the order $10^{-11}$ cm. Since the chance
of an encounter involving a large deflexion is small compared with the chance of
small deflexions, a simple consideration shows that the average small deflexion
is practically unaltered if the large deflexions are omitted. This is equivalent
to integrating over that part of the cross section of the atom where the
deflexions are small and neglecting the small central area. It can in this way
be simply shown that the average small deflexion is given by
$$
\phi_1 = \frac{3\pi}{8} \cdot \frac{b}{R}.
$$
This value of $\phi_1$ for the atom a concentrated central charge is three times
the magnitude of the average deflexion for the same value of $Ne$ in the type of
atom examined by Sir J.J. Thomson. Combining the deflexions due to the electric
field and to the corpuscles, the average deflexion is
$$
(\phi^2_1 + \phi^2_2)^{1/2} ~~~ \mbox{or} ~~~
\frac{b}{2R} \cdot \left( 5.54 + \frac{15.4}{N} \right)^{1/2}.
$$
It will be seen later that the value of $N$ is nearly proportional to the atomic
weight, and is about 100 for gold. The effect due to scattering of the
individual corpuscles expressed by the second term of the equation is
consequently small for
heavy atoms compared with that due to the distributed electric field.
Neglecting the second term, the average deflexion per atom is $3 \pi b/8R$. We
are now in a position to consider the relative effects on the distribution of
particles due to single and to compound scattering. Following J.J. Thomson's
argument, the average deflecion $\theta_t$ after passing through a thickness
$t$ of matter is proportional to the square of the number of encounters and is
given by
$$
\theta_t = \frac{3 \pi b}{8R} \cdot \sqrt{\pi R^2 nt} = \frac{3 \pi b}{8}
\cdot \sqrt{\pi nt},
$$
where $n$ as before is equal to the number of atoms per unit volume.
The probability $p_1$ for compound scattering that the deflexion of the particle
is greater than $\phi$ is equal $e^{- \phi/\theta^2_t}$. Consequently
$$
\phi^2 = - \frac{9 \pi^3}{64} \cdot b^2 nt~ \mbox{log}~ p_1.
$$
Next suppose that single scattering alone is operative. We have seen ($\S~3$)
that the probability $p_2$ of deflexion greater than $\phi$ is given by
$$
p_2 = \frac{\pi}{4} \cdot b^2 \cdot n \cdot t \mbox{cot}^2 \phi/2.
$$
By comparing these two equations
$$
p_2~ \mbox{log}~ p_1 = - 0.181 \phi^2 \mbox{ctg}^2 \phi/2,
$$
$\phi$ is sufficiently small that
$$
\mbox{tan}~ \phi/2 = \phi/2,
$$
$$
p_2~ \mbox{log}~ p_1 = - 0.72.
$$
If we suppose $p = 0.5$, than $p_1 = 0.24$. If $p_2 = 0.1$, $p_1 = 0.0004$.
It is evident from this comparison, that the probability for any given deflexion
is always greater for single than for compound scattering. The difference is
especially marked when only a small fraction of the particles are scattered
through any given angle. It follows from this result that the distribution of
particles due to encounters with the atoms is for small thickness mainly
governed by single scattering. No doubt compound scattering produces some effect
in equalizing the distribution of the scattered particles; but its effect
becomes relatively smaller, the smaller the fraction of the particles scattered
through a given angle.\\
\section*{
$\S~6$ {\it Comparison of Theory with Experiments}}
On the present theory, the value of the central charge $Ne$ is an important
constant, and it is desirable to determine its value for different atoms. This
can be most simply done by determining the small fraction of $\alpha$ or $\beta$
particles of known velocity falling on a thin metal screen, which are scattered
between $\phi$ and $\phi + d \phi$ where is the angle of deflexion. The
influence of compound scattering should be small when this fraction is small.
Experiments in these directions are in progress, but it is desirable at this
stage to discuss in the light of the present theory the data already published
on scattering of $\alpha$ and $\beta$ particles.
The following points will be discussed: --
\begin{itemize}
\item[(a)]
The ``diffuse deflexion'' of $\alpha$ particles, i.e. the scattering of $\alpha$
particles through large angles (Geiger and Marsden).
\item[(b)]
The variation of diffuse reflexion with atomic weight of the radiator (Geiger
and Marsden).
\item[(c)]
The average scattering of a pencil of $\alpha$ rays transmitted through a thin
metal plate (Geiger).
\item[(d)]
The experiments of Crowther on the scattering of $\beta$ rays different
velocities by various metals.\\
\end{itemize}
(a)In the paper of Geiger and Marsden on the diffuse reflexion of $\alpha$
particles falling on various substances it was shown that about 1/8000
of the $\alpha$ particles from radium $C$ falling on a thick
plate of platinum are
scattered back in the direction of the incidence. This fraction is deduced on
the assumption that the $\alpha$ particles are uniformly scattered in all
directions, the observations being made for a deflexion of about 90 degrees. The
form of experiment is not very suited for accurate calculation, but from the
data available it can be shown that the scattering observed is about that to be
expected on the theory if the atom of platinum has a central charge of
about $100e$.
(b) In their experiments on this subject, Geiger and Marsden gave the relative
number of $\alpha$ particles diffusely reflected from thick layers of different
metals, under similar conditions. The number obtained by them are given in [
Table 1] below, where $z$ represents the relative number of scattered
particles, measured by the number of scintillations per minute on a zinc
sulphide screen.\\
Table 1.\\
\begin{center}
\begin{tabular}{l c c c}\\
\hline
&&&\\
&Atomic weight,&&\\
Metal.&$A$.&$z$.&$z/A^{3/2}$\\
\hline
&&&\\
Lead&207&62&208\\
Gold&197&67&242\\
Platinum&195&63&232\\
Tin&119&34&226\\
Silver&108&27&241\\
Copper&64&14.5&225\\
Iron&56&10.2&250\\
Aluminium&27&3.4&243\\
\hline
&&Average&233\\
&&&\\
\end{tabular}
\end{center}
\vspace{0.3cm}
On the theory of single scattering, the fraction of the total number of $\alpha$
particles scattered through any given angle in passing through a thickness
$t$ is
proportional to $n$. $A^2$ $t$, assuming that the central charge is
proportional to
the atomic weight $A$. In the present case, the thickness of matter
from which the
scattered $\alpha$ particles are able to emerge and affect the zinc sulphide
screen depends on the metal. Since Bragg has shown that the stopping power of an
atom for an $\alpha$ particle is proportional to the square root of its atomic
weight, the value of $nt$ for different elements is proportional to
$1/\sqrt{A}$.
In this case $t$ represents the greatest depth from which the scattered $\alpha$
particles emerge. The number $z$ of $\alpha$ particles scattered back
from a thick
layer is consequently proportional to $A^{3/2}$ or $z/A^{3/2}$ should be a
constant.
To compare this deduction with experiment, the relative values of the latter
quotient are given in the last column. Considering the difficulty of the
experiments, the agreement between theory and experiment is reasonably good.
\footnote{The effect of charge of velocity in an atomic encounter is neglected
in this calculation.}
The single large scattering of $\alpha$ particles will obviously affect to some
extent the shape of the Bragg ionization curve for a pencil of $\alpha$ rays.
This effect of large scattering should be marked when the $\alpha$ rays have
traversed screens of metals of high atomic weight, but should be small for atoms
of light atomic weight.
(c) Geiger made a careful determination of the scattering of $\alpha$ particles
passing through thin metal foils, by the scintillation method, and deduced the
most probable angle through which the $\alpha$ particles are deflected in
passing through known thicknesses kind of matter.
A narrow pencil of homogeneous $\alpha$ rays was used as a source. After passing
through the scattering foil, the total number of $\alpha$ particles deflected
through different angles was directly measured. The angle for which the number
of scattered particles was a maximum was taken as the most probable angle. The
variation of the most probable angle with thickness of matter was determined,
but calculation from these data is some what complicated by the variation of
velocity of the $\alpha$ particles in their passage through the scattering
material. A consideration of the curve of distribution of the $\alpha$ particles
given in the paper . . . shows that an angle through which half the particles
are scattered is about 20 per cent greater than the most probable angle.
We have already seen that compound scattering may become important when about
half the particles are scattered through a given angle, and it is difficult to
disentangle in such cases the relative effect due to the two kinds of
scattering. An a approximate estimate can be made in the following way:
From ($\S~5$)
the relation between the probabilities $p_1$ and $p_2$ for compound and
single scattering respectively is given by
$$
p_2 ~\mbox{ln}~ p_1 = - 0.721.
$$
The probability $q$ of the combined effects may as a first
approximation be taken as
$$
q = (p^2_1 + p^2_2)^{1/2}.
$$
If q = 0.5, it follows that
$$
p_1 = 0.2 ~~~~\mbox{and}~~~~ p_2 = 0.46.
$$
We have seen that the probability $p_2$ of a single deflexion greater than
$\phi$ is given by
$$
p_2 = \frac{\pi}{4} \cdot n \cdot t \cdot b^2 ~\cot^2 ~\phi/2.
$$
Since in the experiments considered $\phi$ comparatively small
$$
\frac{\phi \sqrt{p_2}}{\sqrt{\pi n t}} = b = \frac{2NeE}{m u^2}.
$$
Geiger found that the most probable angle of scattering of the $\alpha$ rays in
passing through a thickness of gold equivalent in stopping power to about
.76 cm. of air was 1 degree 40$^{\prime}$. The angle $\phi$ through which half
the $\alpha$ particles are turned thus corresponds to 2 degrees nearly.
$$
t = 0.00017 cm.; ~~n = 6 \cdot \times 10^{22};
$$
$$
\mbox{u (average value)} = 1.8 \times 10^9.
$$
$$
e/m = 1.5 \times 10^{14}.~ \mbox{E.S. units};~~ e = 4.65 \times
10^{-10}
$$
Taking the probability of single scattering 0.46 and substituting the
above values in the formula, the value of $N$ for gold comes out to be 97.
For a thickness of gold equivalent in stopping power to 2.12 cms. of
air, Geiger found the most probable angle to be $3^{\circ} 40 ^{\prime}$. In
this
case $t = .00047, \phi = 4^{\circ} \cdot 4,$ and average $u = 1.7
\times 10^0$, and $N$ comes out to be 114.
Geiger showed that the most probable angle of deflexion for an atom was nearly
proportional to its atomic weigh. It consequently follows that the value of $N$
for different atoms should be nearly proportional to their atomic weights, at
any rate for atomic weight between gold and aluminium.
Since the atomic weight of platinum is nearly equal to that of gold, it follows
from these considerations that the magnitude of the diffuse reflexion of
$\alpha$ particles through more than 90 degrees from gold and the magnitude of
the average small angle scattering of a pencil of rays in passing through gold--
foil are both explained on the hypothesis of single scattering by supposing the
atom of gold has a central charge of about $100 e$.\\
\vspace{0.3cm}
(d) {\it Experiments of Crowther on scattering of $\beta$ rays.} ---
We shall now consider how far the experimental results of Crowther on scattering
of $\beta$ particles of different velocities by various materials can be
explained on the general theory of single scattering. On this theory, the
fraction of $\beta$ particles $p$ turned through an angle greater than $\phi$ is
given by
$$
p = \frac{\pi}{4} \cdot n \cdot t \cdot b^2 \mbox{cot}^2 \phi/2.
$$
In most of Crowther's experiments $\phi$ is sufficiently
small that $\mbox{tan} \phi/2$ may be put equal to $\phi/2$ without
much error. Consequently
$$
\phi^2 = 2 \pi n \cdot t \cdot b^2~~~~~~~~~~~~~~~~~\mbox{if}~~~~p = 1/2.
$$
On the theory of compound scattering, we have already seen that the chance $p_1$
that the deflexion of the particles is greater than $\phi$ is given by
$$
\phi^2/\mbox{log} p_1 = - \frac{9 \pi^3}{64} n \cdot t \cdot b^2.
$$
Since in the experiments of Crowther the thickness $t$ of matter was determined
for which $p_1 = 1/2$,
$$
\phi^2 = 0.96 \cdot \pi n \cdot t \cdot b^2.
$$
For a probability of 1/2, the theories of single and compound scattering are
thus identical in general from, but differ by a numerical constant. It is thus
clear that the main relations on the theory of compound scattering of Sir J.J.
Thomson, which were verified experimentally by Crowther, hold equally well on
the theory of single scattering.
For example, if $t_m$ be the thickness for which half the particles are
scattered through an angle $\phi$, Crowther showed that $\phi/\sqrt{t_m}$ and
also $m u^2/E \cdot \sqrt{t_m}$ were constants for a given material when $\phi$
was fixed. These relations hold also on the theory of single scattering.
Notwithstanding this apparent similarity in form, the two theories are
fundamentally different. In one case, the effects observed are due to cumulative
effects of small deflexions, while in the other the large deflexions are
supposed to result from a single encounter. The distribution of scattered
particles is entirely different on the two theories when the probability of
deflexion greater than $\phi$ is small.
We have already seen that the distribution of scattered $\alpha$ particles at
various angles has been found by Geiger to be in substantial agreement with the
theory of single scattering, but cannot be explained on the theory of compound
scattering alone. Since there is every reason to believe that the laws of
scattering of $\alpha$ and $\beta$ particles are very similar, the law of
distribution of scattered $\beta$ particles should be the same as for $\alpha$
particles for small thickness of matter. Since the value of $mu^2/E$ for the
$\beta$ particles is in most cases much smaller than the corresponding value for
the $\alpha$ particles, the chance of large single deflexions for $\beta$
particles in passing through a given thickness of matter is much greater than
for $\alpha$ particles. Since on the theory of single scattering the fraction of
the number of particles which are deflected through a given angle is
proportional to $kt$, where $t$ is the thickness supposed small and $k$ a
constant, the number of particles which are undeflected through this angle is
proportional to $1 - kt$. From considerations based on the theory of compound
scattering, Sir J.J. Thomson deduced that the probability of deflexion less than
$\phi$ is proportional to $1 - e^{-\mu/t}$ where $\mu$ is constant for any given
value of $\phi$.
The correctness of this latter formula was tested by Crowther by measuring
electrically the fraction $I/I_0$ of the scattered $\beta$ particles which
passed through a circular opening subtending an angle of 36$^{\circ}$ with the
scattering material. If
$$
I/I_0 = 1 - e^{- \mu/t},
$$
the value of $I$ should decrease very slowly at first with increase of $t$.
Crowther, using aluminium as scattering material, states that the variation of $
I/I_0$ was in good accord with this theory for small values of $t$. On the other
hand, if single scattering be present, as it undoubtedly is for $\alpha$ rays,
the curve showing the relation between $I/I_0$ and $t$ should be nearly linear
in the initial stages. The experiments
of Madsen\footnote{Phil. Mag. XVIII. p. 909 (1909).}
%\cite{Marsden1909}
on scattering of $\beta$ rays, although not made with quite so small a
thickness of aluminium as that used by Crowther, certainly support such a
conclusion. Considering the importance of the point at issue, further
experiments on this question are desirable.
From the table given by Crowther of the value $\phi/ \sqrt{t_m}$. for different
elements for $\beta$ rays of velocity $2.68 \times 10^{10}$ cms. per second, the
values of the central charge $Ne$ can be calculated on the theory of single
scattering. It is supposed, as in the case of the $\alpha$ rays, that for the
given value of $\phi/\sqrt{t_m}$ the fraction of the $\beta$ particles deflected
by single scattering through an angle greater than $\phi$ is
0.46 instead of 0.5.
The values of $N$ calculated from Crowther's data are given below.
\vspace{0.3cm}
Table 2.\\
\begin{center}
\begin{tabular}{l | c | c | r}
\hline
&&&\\
Element&Atomic weight&$\varphi/\sqrt{t_m}$&$N$\\
&&&\\
\hline
&&&\\
Aluminium&27&4.25&22\\
Copper&63.2&10.0&42\\
Silver&108&15.4&78\\
Platinum&194&29.0&138\\
\hline
\end{tabular}
\end{center}
\vspace{0.3cm}
It will be remembered that the values of $N$ for gold deduced from scattering of
the $\alpha$ rays were in two calculations 97 and 114. These numbers are
somewhat smaller than the values given above for platinum (viz. 138), whose
atomic weight is not very different from gold. Taking into account the
uncertainties involved in the calculation from the experimental data, the
agreement is sufficiently close to indicate that the same general laws of
scattering hold for the $\alpha$ and $\beta$ particles, notwithstanding the wide
differences in the relative velocity and mass of these particles.
As in the case of the $\alpha$ rays, the value of $N$ should be most simply
determined for any given element by measuring the small fraction of the incident
$\beta$ particles scattering through a large angle. In this way, possible errors
due to small scattering will be avoided.
The scattering data for the $\beta$ rays, as well as for the $\alpha$ rays.
indicate that the central charge in an atom is approximately proportional to its
atomic weight. This falls in with the experimental deductions of
Schmidt\footnote{Annal. d. Phys., IV. 23, p. 671 (1907)}.
%\cite{Schmidt1907}
In his
theory of absorption of $\beta$ rays, he supposed that in traversing a thin
sheet of matter, a small fraction $\alpha$ of the particles are stopped, and a
small fraction $\beta$ are reflected or scattered back in the direction of
incidence. From comparison of the absorption curves of different elements, he
deduced that the value of the constant $\beta$ for different elements is
proportional to $n A^2$ where n is the number of atoms per unit
volume and $A$ the
atomic weight of the element. This is exactly the relation to be expected on the
theory of single scattering if the central charge on an atom is proportional to
its atomic weight.\\
\section*{
$\S~7$ {\it General Considerations}}
In comparing the theory outlines in this paper with the experimental result, it
has been supposed that the atom consists of a central charge supposed
concentrated at a point, and that the large single deflexions of the $\alpha$
and $\beta$ particles are mainly due to their passage through the strong central
field. The effect of the equal and opposite compensating charge supposed
distributed uniformly throughout a sphere has been neglected. Some of the
evidence in support of these assumptions will now be briefly considered. For
concreteness, consider the passage of a high speed $\alpha$ particle through an
atom having a positive central charge $Ne$, and surrounded by a compensating
charge of $N$ electrons. Remembering that the mass, momentum, and kinetic energy
of the $\alpha$ particle are very large compared with the corresponding values
for an electron in rapid motion, it does not seem possible from dynamic
considerations that an $\alpha$ particles can be deflected through a large angle
by a close approach to an electron, even if the latter be in rapid motion and
constrained by strong electrical forces. It seem reasonable to suppose that the
chance of single deflexions through a large due to this cause, if not zero, must
be exceedingly small compared with that due to the central charge.
It is of interest to examine how far the experimental evidence throws light on
the question of the extent of the distribution of the central charge. Suppose,
for example, the central charge to be composed of $N$ unit charges distributed
over such a volume that the large single deflexions are mainly due to the
constituent and not to the external field produced by the distribution. It has
shown that the fraction of the $\alpha$ particles scattered through a large
angle is proportional to $(NeE)^2$, where $Ne$ is the central charge
concentrated at a point and $E$ the charge on the deflected particle.
If, however, this charge
is distributed in single units, the fraction of the $\alpha$ particles scattered
through a given angle is proportional to $Ne^2$ instead of $N^2 e^2$. In this
calculation, the influence of mass of the constituent particles has been
neglected, and account has only been taken of its electric field. Since it has
been shown that the value of the central point charge for gold must be about
100, the value of the distributed charge requires to produce the same proportion
of single deflexions through a large angle should be at least 10,000. Under
these conditions the mass of the constituent particle would be small compared
with that of the $\alpha$ particle, and the difficulty arises of the production
of large single deflexions at all. In addition, with such a large distributed
charge, the effect of compound scattering is relatively more important than that
of single scattering. For example, the probable small angle of deflexion of a
pencil of $\alpha$ particles passing through a thin gold--foil would be much
greater than that experimentally observed by geiger. The large and small angle
scattering could not then be explained by the assumption of a central charge of
the same value. Considering the evidence as a whole, it seems simplest to
suppose that the atom contains a central charge distributed through a very small
volume, and that the large single deflexions are due to the central charge as a
whole, and not to its constituents. At the same time, the experimental evidence
is not precise enough to negative the possibility that a small fraction of the
positive charge may be carried by satellites extending some distance from the
centre. Evidence on this point could be obtained by examining whether the same
central charge is required to explain the single
deflexions of $\alpha$ and $\beta$ particles; for the $\alpha$
particle must approach much closer to the
centre of the atom than the $\beta$ particle of average speed to suffer the same
large deflexion.
The general data available indicate that the value of this central charge for
different atoms is approximately proportional to their atomic weights, at any
rate atoms heavier than aluminium. It will be of great interest to examine
experimentally whether such a simple relation holds also for the lighter atoms.
In cases where the mass of the deflecting atom (for example, hydrogen, helium,
lithium) is not very different from that of the $\alpha$ particle, the general
theory of single scattering will require modification, for it is necessary to
take into account the movements of the atom itself (see $\S~4$).
It is interest to note that Nagaoka\footnote{Nagaoka, Phil.
Mag. VII. p. 445 (1904).}
%\cite{Nagaoka1904}
has mathematically considered the properties
of a ``Saturnian'' atom which he supposed to consist of a central attracting
mass surrounded by rings of rotating electrons. He showed that such a system was
stable if the attractive force was large. From the point of view considered in
this paper, the chance of large deflexion would practically be unaltered,
whether the atom is considered to be a disk or a sphere. It may be remarked that
the approximate value found for the central charge of the atom of gold ($100 e$)
is about that to be expected if the atom of gold consisted of 49 atoms of
helium, each carrying a charge $2 e$. This may be only a coincidence, but it is
certainly suggestive in view of the expulsion of helium atoms carrying two unit
charge from radioactive matter.
The deductions from the theory so far considered are independent of the sigh of
the central charge, and it has not so far been found possible to obtain definite
evidence to determine whether it be positive or negative. It may be possible to
settle the question of sign by consideration of the difference of the laws of
absorption of the $\beta$ particle to be expected on the two hypotheses, for the
effect of radiation in reducing the velocity of the $\beta$ particle should be
far more marked with a possible than with a negative centre. If the central
charge be positive, it is easily seen that a positively charge mass, if released
from the centre of a heavy atom, would acquire a great velocity in moving
through the electric field. It may be possible in this way to account for the
high velocity of expulsion of $\alpha$ particles without supposing that they are
initially in rapid motion within the atom.
Further consideration of the application of this theory to these and other
questions will be reserved for a later paper, when the main deductions of the
theory have been tested experimentally. Experiments in this directions are
already in progress by Geiger and Marsden.
\vspace{2cm}
{\it University of Manchester\\
April 1911.}
The preceding paper by Rutherford sets forth his theory of the scattering of
$\alpha$ particles by atoms composed of a small, centrally located, positively
charged nucleus surrounded by a sphere of equal but uniformly distributed
negative charge whose effect on the scattering of the particles is negligible.
The orders of magnitude envisioned were, roughly, for the nuclear radius about $
10^{-12}$ cm, and for the whole atom about $10^{-8}$ cm. If one imagines that
discrete negative electrons were present instead of a distribution of negative
charge, then the nuclear atom would be mostly empty space.
Rutherford's scattering formula (see the previous paper, equation [5]) predicted
that the fraction of the $\alpha$ particles scattered by a thin foil should be
proportional to: (1) the inverse $\sin^4 \phi/2,$ where $\phi$ is the angle
through which the $\alpha$ particle is deflected by its encounter with an atom
through which the $\alpha$ particle is deflected by its encounter with an atom
of the foil; (2) the thickness of the scattering foil, provided this is small ;
(3) the square of the nuclear charge $Ne$; (4) the inverse fourth power of the
velocity $\nu$ of the bombarding $\alpha$ particles.
Fig. 44--3 in Geiger and Marsden's paper, which follows, shows the very simple
apparatus that they set up to examine the theoretical predictions enumerated
above. Essentially the device consists of a scattering foil $F$, upon which
$\alpha$ particles from the source $R$, drawn into a stream of parallel trajectories
by a diaphragm at $D$, strike at right angles. The box $B$, which carries the
viewing microscope M, rotates around the axis of the foil F by means of the
ground--glass joint C. A cap over the end of vibrated they radiated energy until
they came to rest again. This picture is in complete accord with the classical
electromagnetic theory of the electron as developed by Maxwell and Lorentz, and
for this reason the Thomson model was favorably received by Thompson's
contemporaries.
Thomson's purpose in developing this model was to explain the ``scattering of
electrified particles in passing through small thickness of matter''. In
scattering experiments, the crucial criterion for the atom model is the angle
through which a charged particle is deflected from its original direction of
motion as it passes through a metal foil used as the scatterer. Thomson assumed
that the angle of deviation suffered by the charged particle was always caused
by a large number of collision with many atoms. Any single collision played only
a minimal role in the total deviation, which was a cumulative affect. It can be
shown that on the basis of the Thomson model the total deviation is not the
average deviation produced in a single collision multiplied by the number of
collisions; rather, the multiplicand is the {\it square root} of this sum of
collisions. Thus, if each collision resulted on the average in a deviation of 1
degree, . 100 collisions would give rise to a net deviation of only 10 degrees.
Rutherford pointed out the importance of this fact by calling attention to the
observations of Geiger and Marsden. They had found, in their experiments with
$\alpha$ particles passing through a layer of gold foil about 0.00006 cm thick,
that they could be scattered through an angle of 90 degrees or more. If only
small deviations occurred in each encounter, the $\alpha$ particle would have
had to undergo 10,000 of the lesser collisions to produce such a large total
deviation. This was highly improbable, as Rutherford pointed out, because of the
extreme thinness of the gold foil. Rutherford contended that such large
deviations must have been caused, therefore, by single direct collisions. He
then produced to analyze the theory of single collisions on the basis of a model
of the atom that is radically different from the Thomson model.
In this Rutherford model the positive electricity is not distributed over a
large volume but instead is concentrated in a very small nucleus at the center
of the atom. As Rutherford points out in the paper that follows, the actual
analysis is the same whether one assumes that the positive charge is
concentrated at the nucleus and the electrons are on the outside, or vice versa.
A model of this sort cannot be in static equilibrium, since the electrons would
all be dragged into the nucleus if they were not moving in stable orbits around
this nucleus. yet this kind of dynamical equilibrium is in serious contradiction
with classical electrodynamics. Rutherford was aware of this, but chose to
ignore the difficulty for the time being . He stated that the ``question of the
stability of the atom proposed need not be considered at this stage . . .''
By very simple but elegant arguments and with the most elementary mathematics,
Rutherford showed that his model of the atom gives rise to the kind of
deviations during single collisions that Geiger and Marsden had observed. The
paper itself is exemplary in its simplicity, yet so profound that none could
doubt that Rutherford's ideas must serve as the basis of a new and correct
picture of the structure of matter.
We should note, however, that the picture of the atom that Rutherford drew was
still very tentative and vague. He speculates that not all of the positive
charge is in the nucleus; ``a small fraction of the positive charge may be
carried by satellites extending some distance from the center.'' Although the
values he obtained for the charge on the nuclei of different metals are all too
large, for example, 100 for gold, he correctly concludes that the nuclear
charge should be `` approximately proportional'' to the atomic weight of the
atom. But Rutherford was not sure that this would hold for the light elements
and indicated that for such elements his simple theory of atomic collisions is
no longer applicable.
Although throughout most of the paper he does not specifically mention the
planetary theory of the atom, it is clear from his reference to the work of
Nagaoka that Rutherford had this planetary model in mind, and it is here that we
have the starting point of modern atomic theory.
It has often happened in the past that authors in referring to this paper of
Rutherford's speak of his experiments on the scattering of $\alpha$ particles;
however, the experiments used by Rutherford were not own but those of Geiger and
Marsden. Rutherford's great contribution lay in showing that the Thomson model
of the atom cannot possible explain the large number of large--angle
scatterings, whereas the nuclear model can.
%SB. = ENCODED 3 MAY 1998 BY NIS;
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