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\begin{document}
G. Gamov Phys. Rev. {\bf 49,} 895 \hfill {\large \bf 1936}\\
\vspace{2cm}
\begin{center}
{\bf \Large Selection Rules for the $\beta$-Disintegration}
\end{center}
\begin{center}
G. GAMOW AND E. TELLER,\\
George Washington University, Washington D. C. \\
(Received March 28, 1936)
\end{center}
\begin{abstract}
\S~1. The selection rules for $\beta$-transformations are stated on the basis of
the neutrino theory outlined by Fermi. If it is assumed that the spins of the
heavy particles have a direct effect on the disintegration
these rules are modified. \S~2. It is shown that whereas the original selection
rules of Fermi lead to difficulties if one tries to assign spins to the members of
the thorium family the modified selection
rules are in agreement with the available experimental evidence.
\end{abstract}
\centerline{\large \S~1}
According to the theory of $\beta$-disintegration given
by Fermi\footnote{Fermi, Zeits. f. Physik {\bf 88,} 161 (1934).}
no change of the total nuclear spin should occur in
the most probable transformations, i.e., in transformations
located on the first Sargent
curve\footnote{Sargent, Proc. Roy. Soc.{\bf A139,} 659 (1933).}.
The transformations corresponding to the second Sargent curve approximately 100
times less probable should correspond to changes $\pm$ 1 or 0 of
the angular momentum of the nucleus. One may expect the
existence of still lower curves for higher changes in the
nuclear spin. This selection principle is based on the
assumption that the spin of the heavy particles does not enter
in the part of the Hamiltonian which is responsible for the
$\beta$-disintegration. The same assumption was made in the
modified theory of Konopinski and
Uhlenbeck\footnote{Konopinski and Uhlenbeck, Phys. Rev. {\bf 48,} 7 (1935).} who
introduced the derivative of the neutrino wave function in the
Hamiltonian in order to get a better fit with the experimental
curves of the energy distribution in $\beta$-spectra. We should like
to note here that this selection rule will be changed if the
spins of the heavy particles are
introduced into the Hamiltonian, a possibility proposed in
many discussions about this subject.
We shall first give the derivation of Fermi's selection rule
in a somewhat generalized form. The probability of
$\beta$-disintegration is proportional to the square of the matrix element.
\begin{equation}
M_1 = \sum \limits_l~ \int (\Omega_l^{N, P} \psi_i) \psi_f^{\ast} \delta
q_l \{ 0(\psi_{\nu}^{\ast} \psi_{\epsilon}^{\ast})\}.
\end{equation}
Here $\psi_i$ and $\psi_f$ are the proper functions of the heavy particles,
protons and neutrons, for the initial and final state,
respectively. These functions depend on the positions of the
heavy particles, on their spins, and on a third
variable\footnote{Introduced by Heisenberg, Zeits. f. Physik {\bf 77,} 1 (1932).}
which corresponds to the charge of the heavy particles and which is
capable of two values, in a manner similar to the spin
variable, the value 1 corresponding to a proton and the value
0 to a neutron. The operator $\Omega_l^{N,P}$ acts on this last variable
converting the $l$th particle in $\psi_i$ into a proton if it was a
neutron and giving $\Omega_l^{N,P} \psi_i = 0$ if the $l$th particles is already a
proton. The integration in (1) includes summation over the
spin and charge coordinates of the heavy particles. $\psi_{\nu}$
and $\psi_{\epsilon}$ are the proper functions of the
neutrino and the electron. $O$ is an operator acting on
these functions but not involving the heavy particles
and the delta function $\delta q_l$ substitutes the position
coordinate $q_l$ of the $l$th heavy particle for the
coordinates of the electron and neutrino.
In Fermi's paper the operator $O$ was simply a
summation over certain products of the four Dirac
components of the electron wave function and
components of the neutrino wave function. The
Konopinski and Uhlenbeck operator involved in
addition the first derivative of the neutrino wave
function. In both cases, however
$\delta q_l \{O(\psi_{\nu}^{\ast} \psi_{\epsilon}^{\ast})\}$ is a
scalar function of $q_l$. This is necessary since in
$(\Omega_l^{N,P} \psi_i)\psi_f^{\ast}$
the summation over the spins of the heavy
particles gives also a scalar and the integral in (1) must
be a scalar.
Supposing at first that $\psi_{\nu}$ and $\psi_{\epsilon}$ are plane waves, the
same will be true for $\delta q_l \{ O(\psi_{\nu}^{\ast} \psi_{\epsilon}^{\ast})
\}$.
If we expand this wave in spherical harmonics, and suppose that the
nuclear radius $r_0$ is small compared to the wave-length
$\lambda$ then the amplitudes of the zeroth, first, second $\ldots$
spherical harmonics within the nucleus will have the
ratio $1 : (r_0/\lambda) : (r_0/\lambda)^2 \ldots$. Neglecting all but the
zero-order spherical harmonic $M_1$ will be different
from zero only if the angular momentum $i$ of the
nucleus does not change during the $\beta$-transformation
and if the nuclear proper function is even with regard
to reflection on the mass center before and after the
disintegration or if it is odd before and after. These
transitions will correspond to the first Sargent curve.
Taking into account the first-order spherical
harmonic in the development of
$\delta q_l\{O(\psi_{\nu}^{\ast} \psi_{\epsilon}^{\ast})\}$
further transitions become possible. The selection rules
for these additional transitions are those valid for a
polar vector: The change in angular momentum $\Delta i$ is
$\pm 1$ or 0 (but not $i=0 \rightarrow i=0$) and one of the two
combining states is even, the other odd. For all these
cases, however, the matrix element $M_1$ will be smaller
by $r_0/\lambda$ than for the zero-order spherical harmonic and
consequently the transitions will be less probable by
$(r_0/\lambda)^2$. Now for most $\beta$-disintegrations $r_0/\lambda$ is about
$10^{-2}$ and the transformations arising from the first-order
harmonic are ten thousand times less probable
than those arising from the zero-order harmonic.
Actually the proper function of the electron is not a
plane wave because of the Coulomb interaction
between the nucleus and the electron. Fermi has shown
that for the heavy elements, where this interaction is
the greatest, the result will be to increase the
probability of emitting an electron with unit angular
momentum, this event being only about 100 times less
probable than the emission of the light particles with
zero angular momentum, thus giving the second
Sargent curve. The situation will be similar if we
accept the Hamilton term introduced by Konopinski
and Uhlenbeck or any other expression of the type
given in the matrix element $M_1$.
We have therefore from a generalized treatment of
Fermi's theory the following selection rules,
{\footnotesize First Sargent curve: (1) $\Delta_i=0$; (2) proper functions,
even-even, or odd-odd.}
{\footnotesize Second Sargent curve: (1) $\Delta_i=0$ or
$\pm 1;$ (2) proper functions, even-odd.}
If we now assume that the spin of the proton and
neutron enters into the Hamilton term which is
responsible for the transformation we may substitute
$M_1$ by the more complicated expression
\begin{equation}
M_2 = \sum \limits_{\xi} \sum \limits_l~ (\Omega_l^{N,P} \alpha_l^{\xi}
\psi_i) \psi_f^{\ast} \delta q_l \{ O^{\xi} (\psi_{\nu}^{\ast} \psi_{\epsilon}^{\ast})\}.
\end{equation}
Hereby $\alpha_l^{\xi}$ operates on the spin of the $l$th heavy
particle and signifies the three Pauli
matrices :\footnote{A similar expression was introduced by Fermi (in order to
insure relativistic invariance) as an additional term. In his
expression, however, the $\alpha$'s stood for the Dirac matrices
which give only a small contribution as long as the velocity
of the heavy particles are small compared to $c$. It should also
be noted that Dirac's $\alpha$'s are the components of a polar vector
whereas Pauli's $\alpha$'s form an axial vector.}
\begin{equation}
\alpha^x = \Biggl|
\begin{array}{cc}
0&1\\
1&0
\end{array} \Biggr|; \quad \alpha^y = \Biggl|
\begin{array}{cc}
0&i\\
-i&0
\end{array} \Biggr|; \quad \alpha^z = \Biggl|
\begin{array}{cc}
1&0\\
0&1
\end{array} \Biggr|.
\end{equation}
The summation over $\xi$ is meant to include the three
values $x, y, z.$ The three operators $O^{\xi}$ are the three
components of a vector. This means that by a
coordinate transformation the three operators are
transformed in the same way as the $x,y$ and $z$
components of a vector. It is then seen that the matrix
element $M_2$ will again be scalar quantity.
Expanding the expressions
$\delta q_i\{O^{\xi}(\psi_{\nu}^{\ast} \psi_{\epsilon}^{\ast})\}$ into
spherical harmonics and retaining only the zero-order
functions integration over the coordinates of the heavy
particles shows that transitions are possible if (1)
$\Delta i = 0$ or $\pm 1$ (but not $i = 0 \rightarrow i = 0$), and if (2) The
transition is of an odd-odd or even-even type. These
are the same selection rules as those valid for an axial
vector. The corresponding transitions would be
located on the first Sargent curve.
From the first-order spherical harmonics we obtain
transitions with (1) $\Delta i = 0;~ \pm 1$ or $\pm 2$, (2) the transitions
are of the odd-even type.
Either the matrix element $M_1$ or the matrix element
$M_2$ or finally a linear combination of $M_1$ and $M_2$ will
have to be used to calculate the probabilities of the
($\beta$-disintegrations. If the third possibility is the correct
one, and the two coefficients in the linear combination
have the same order of magnitude, then all transitions
which would lie on the first Sargent curve according
to any one of the two sets of selection rules mentioned
above would now lie on the first curve. This would
mean that the selection rules are the same as for an
axial vector with the addition that also the $i=0 \rightarrow i=0$
even $\rightarrow$ even or odd $\rightarrow$ odd transitions are allowed.
We shall show now that if exchange forces of the
Majorana type\footnote{Majorana, Zeits. f. Physik {\bf 82,} 137 (1933).}
are acting between protons and
neutrons and if these forces have to be explained by a
($\beta$-disintegration of the neutron and a following
capture of the electron and neutrino by the proton then
the actual matrix element to be used is the sum of the
matrix elements $M_1$ and $M_2$. Indeed if we should have
only $M_1$ then the charges would be exchanged with the
spins remaining unaffected, i.e., we should obtain
Heisenberg forces. If on the other hand $M_2$ were the
correct expression then considering a system of one
neutron and one proton represented by
$\psi_N(q_1 \uparrow)\psi_P(q_2\uparrow)$\footnote{The arrows in
$\psi_N(q_1\uparrow)$ and $\psi_P(q_2\uparrow)$ represent the spins of the neutron and
proton.}
and applying first to $\psi_N(q_1 \uparrow)$ the operator
corresponding to $M_2$ and then the inverse operator to
$\psi_P(q_2 \uparrow)$, the expression $\psi_P(q_1 \uparrow) \psi_N(q_2
\uparrow)$ would be obtained. By a similar procedure $\psi_N(q_1 \uparrow)\psi_P(q_2
\uparrow)$ is transformed into $2 \psi_P(q_1 \downarrow)\psi_N(q_2 \uparrow)
- \psi_P(q_1 \uparrow) \psi_N (q_2 \downarrow).$ Now
$\psi_N (q_1 \uparrow) \psi_P(q_2 \uparrow)$ is according to both Majorana
\begin{figure}
\centerline{\resizebox{10cm}{!}{\includegraphics{fig1e.gif}}}
\caption{Schematic representation of the radioactive $\alpha$-and
$\beta$-disintegrations from Tn B to Th D, indicating various
$\beta$-transformations leading to excited states of product-nuclei.}
\end{figure}
and Heisenberg in exchange interaction with
\mbox{$\psi_P(q_1 \uparrow) \psi_N(q_2 \uparrow)$} whereas $\psi_N(q_1 \uparrow)
\psi_P(q_2 \downarrow)$ exchanges with
$\psi_P(q_1 \uparrow) \psi_N(q_2 \downarrow)$ according to Heisenberg and with
\mbox{$\psi_P(q_1 \downarrow) \psi_N(q_2 \uparrow)$} according to Majorana. The matrix
element $M_2$ will correspond therefore to a
superposition of the Majorana and the Heisenberg
forces in the ratio 2 to $-1.$ If we want to obtain pure
Majorana forces we must add $M_1$ and $M_2$ with equal
coefficients.\\
\centerline{\S~2}
We can now show that the new selection rules help
us to remove the difficulties which appeared in the
discussion of nuclear spins of radioactive
elements\footnote{Gamow, Proc. Roy. Soc. {\bf A146,} 217 (1934); Physik.
Zeits. {\bf 35,} 533 (1934).} by
using the original selection rule of Fermi.
We shall discuss the sequence of transformations in
the thorium family leading from Th B to Th D
(thorium lead) which is represented schematically in
Fig. 1. First of all we can conclude with a rather high
degree of certainty that the normal states of Th B, Th
C$^{\prime}$ and Th D nuclei, possessing {\it even} atomic numbers
and even mass numbers, have the spin
$i=0.$\footnote{For four elements of this type ($_2$He$^4$; $_6$C$^{12}$;
$_8$O$^{16}$; $_{16}$S$^{32}$) the
absence of spin is directly shown by the band spectra; other
13 investigated elements of this type ($_{48}$Cd$^{110}$; $_{48}$Cd$^{112}$;
$_{48}$Cd$^{114}$; $_{56}$Ba$^{136}$; $_{56}$Ba$^{138}$; $_{80}$Hg$^{200}$;
$_{80}$Hg$^{202}$; $_{80}$Hg$^{204}$; $_{82}$Pb$^{204}$; $_{82}$Pb$^{206}$;
$_{82}$Pb$^{208}$ = Th D) do not show any hyperfine structure
which makes it very probable that their spin is also zero.} The transformation
Th B $\rightarrow$ Th C gives rise to a continuous
$\beta$-spectrum with the observed upper limit
$E_{\beta} = 0.362$ MV and is accompanied by a very strong
$\gamma$-line $h \nu = 0.238$ MV along with several much weaker
lines. The number of secondary electrons (due to
internal conversion of the $\gamma$-line 0.238 MV in the $K$
level) is according to Ellis and Mott\footnote{Ellis and Mott,
Proc. Roy. Soc. {\bf A139,} 369 (1933).} about $N_{\beta} = 0.25$
per disintegration from which these authors conclude
that it must correspond to a quadripole radiation. In
fact the internal conversion coefficients for this
frequency are, according to calculations of Mott and
Taylor,\footnote{Mott and Taylor, Proc. Roy. Soc. {\bf A138,} 665 (1932).}
$\alpha_d = 0.026$ and $\alpha-q=0.205$ for dipole and
quadripole radiation, respectively. Thus for the total
number of $\gamma$-quanta radiated by nuclei $N_{\gamma} = N_{\beta/\alpha}$ we
should have according to these two possibilities 9.6 or
1.2. Since this number should not be larger than unity
we must exclude the possibility of dipole radiation and
consider the $\gamma$-line in question as due to quadripole
transition with the intensity almost one quantum per
disintegration. The fact that the observed value is 20
percent larger than unity must be due to errors in the
measurements of $N_{\beta}$ or the calculation of $\alpha$.
Accordingly we admit with Ellis and Mott that in this
case we have 100 percent excitation of the quantum
level 0.238 MV of the Th C nucleus. The total energy
of the transformation is $0.362+0.238=0.600$ MV and
the observed upper limit of the $\beta$-energies corresponds
to the transformation Th B$_{\mbox{norm}} \rightarrow$ Th C$_{\mbox{exe}}$.
The $\beta$-transformation from Th C to Th C$'$ corresponds
to the upper limit of the $\beta$-spectrum $E_{\beta} = 2.25 $
MV and is accompanied with only very weak
$\gamma$-radiation. Thus we conclude that in this case the main
transformation, 80 percent, takes place between
normal states Th C$_{\mbox{norm}}$ $\rightarrow$ Th C$^{\prime}_{\mbox{norm.}}$
Finally in the $\beta$-transformation between Th C$^{\prime \prime}$ and
Th D the level 3.202 MV of Th D nucleus is
(according to Ellis and Mott) excited to almost 100
percent, the transition to the normal state occurring by
emission of two $\gamma$-lines 0.582 MV and 2.620 MV both
with the absolute intensities of the order unity. Thus
the observed upper limit of the $\beta$-spectrum $E_{\beta} = 1.79$
MV corresponds to the transformation Th C$^{\prime \prime}_{\mbox{norm}}
\rightarrow$ Th D$_{\mbox{exe}}$. The total energy of transformation being
$1.79 + 3.202 = 4.99$ MV.
In Fig. 2 the logarithms of the partial decay
constants of different subgroups of $\beta$-spectra are
plotted against the logarithms of the corresponding
upper energy limits.\footnote{Upper energy limits of different $\beta$-subgroups
are obtained by subtracting the excitation-energies from the total
energy of transformation; partial decay constants are
estimated from the total decay constant and relative excitation
of different nuclear levels.}
The curves I and II correspond to Sargent's
permitted and nonpermitted transformations as
estimated from different members of three radioactive
families. We see that the main transformation Th B$_{\mbox{norm}}\rightarrow$
Th C$_{\mbox{exe}}$ corresponds to the curve I whereas
the main transformation Th C$_{\mbox{norm}} \rightarrow$ Th C$^{\prime}_{\mbox{norm}}$
corresponds to the curve II. From the original Fermi
selection rule: Curve I:
$\Delta i = 0;$ Curve II: $\Delta i = 0$ or $\pm 1.$ We conclude that
$i ({\rm Th ~C}_{\mbox{exe}}) = i({\rm Th ~B}_{\mbox{norm}}) = 0.$.
Since the $\gamma$-line represents a quadripole transition we have further
$i({\rm Th~ C}_{\mbox{norm}}) = 2$. In this case the transformation
Th C$_{\mbox{norm}} \rightarrow$ Th C$^{\prime}_{\mbox{norm}}$
should correspond to $\Delta i = 2,$ i.e.,
must belong to the third Sargent curve, which is in
contradiction with experimental evidence, this
transformation belonging to the curve II.
The difficulty will be still not removed if we take
the possibility into account that $\gamma$-line 0.238 MV
corresponds to a magnetic dipole radiation. For this
case the coefficients for internal conversion have been
calculated by Fisk and Taylor\footnote{Fisk and Taylor, Proc. Roy. Soc. {\bf
A146.} 178 (1934).} and are considerably
larger than the corresponding coefficients for electric
radiation. Accepting this possibility we obtain for the
number of $\gamma$-quanta $hv = 0.238$ mv a value small
compared with unity (weak excitation) and
\begin{figure}
\centerline{\resizebox{10cm}{!}{\includegraphics{fig2e.gif}}}
\caption{Logarithmic plot of the relation between partial
decay constants and corresponding upper energy limits for
various components of complex $\beta$-ray spectra.}
\end{figure}
should be forced to accept that the observed upper
limit of $\beta$-spectrum corresponds to transformation Th
B$_{\mbox{norm}} \rightarrow$ Th C$_{\mbox{norm}}$. This will lead again to
contradiction with Fermi's original selection rule first
because the transformation
\mbox{Th B$_{\mbox{norm}} \rightarrow$ Th C$_{\mbox{norm}}$ and Th C$_{\mbox{norm}}
\rightarrow$ Th C$^{\prime}_{\mbox{norm}}$}
corresponding to the same spin-change (because \mbox{$i ({\rm Th~ B}_{\mbox{norm}})
= i ({\rm Th~ C}^{\prime}_{\mbox{norm}}) = 0$)} would belong to different
Sargent curves and secondly because in this case both
transformations Th C$_{\mbox{norm}} \rightarrow$ Th C$_{\mbox{norm}}$ and Th
B$_{\mbox{norm}} \rightarrow$ Th C$_{\mbox{exe}}$ being of the first Sargent's class
would lead to the conclusion \mbox{$i({\rm Th ~C}_{\mbox{norm}}) = i ({\rm Th
~C}_{\mbox{exe}} = i ({\rm Th~ B}_{\mbox{norm}}) = 0$}
which would exclude the possibility of any $\gamma$-transition.
Applying our modified selection rule, curve I $ \Delta i = 0$
or $\pm 1$ curve II $\Delta i = 0, ~ \pm 1$ or $\pm 2$ we have the following
possibilities
$$
\begin{array}{c}
{\rm nucleus:}\\
{\rm spin:}
\end{array}
\Biggr|
\begin{array}{c}
{\rm Th ~B}_{\mbox{norm}}\\
0
\end{array} \Biggl|
\begin{array}{c}
{\rm Th ~C}_{\mbox{exe}}\\
0 \quad \mbox{or} \quad \pm 1
\end{array} \Biggr|
\begin{array}{c}
{\rm Th ~C}_{\mbox{norm}}\\
0, \pm 1 \quad {\rm or} \quad \pm 2
\end{array} \Biggl|
\begin{array}{c}
{\rm Th ~C}^{\prime}_{\mbox{norm}}\\
0
\end{array} \Biggr|
\eqno(4)
$$
The possibility $i({\rm Th~ C}_{\mbox{norm}}) = 0$ or $\pm 1$ must, however,
be excluded as in this case the transformation Th
B$_{\mbox{norm}} \rightarrow$ Th C$_{\mbox{norm}}$ would correspond to the curve I
and consequently, because of larger energy, be more
probable than Th B$_{\mbox{norm}} \rightarrow$ Th C$_{\mbox{exe}}$. There remains
only the possibility $i({\rm Th ~C}_{\mbox{norm}}) = 2$ which is in good
agreement with the quadripole-character of the
$\gamma$-transition. The transformation Th B$_{\mbox{norm}} \rightarrow$
Th C$_{\mbox{norm}}$
corresponding to $\Delta i = 2$ cannot belong now to the first
Sargent curve, i.e., it must be at least 100 times weaker
than the main transition, as can be seen from Fig. 2.
This accounts for the fact that the corresponding
``long range'' component of the continuous $\beta$-spectra of Th B
has never been observed. {\it Thus vie see that the new
selection principle removes the difficulty originated in
the case of the older rule.}
It must be pointed out, however, that according to
the above considerations it is not possible to assign
even proper functions to all nuclei with
even atomic number and even mass number. Because
if the proper function of Th B$_{\mbox{norm}}$ is even, the same
is true for Th C$_{\mbox{exe}}$ (since the transition \mbox{Th B$_{\mbox{norm}}
\rightarrow$ Th C$_{\mbox{exe}}$} lies on the first curve) and also for Th
C$_{\mbox{norm}}$ (since Th C$_{\mbox{exe}} \rightarrow$ Th C$_{\mbox{norm}}$
is a quadripole transition). But \mbox{Th C$_{\mbox{norm}} \rightarrow$
Th C$^{\prime}_{\mbox{norm}}$} lies on the
second curve and therefore the proper function of Th
C$^{\prime}_{\mbox{norm}}$ is odd. This is unsatisfactory since it would be
nice to substitute the rule that nuclei with even atomic
number and even mass number have $i = 0$ by the rule
that the proper functions of these nuclei remain
unchanged during {\it any} symmetry operation.
Turning our attention to the $\beta$-transformation
leading from Th C$^{\prime \prime}$" to Th D we see that the main
transformation Th C$^{\prime \prime}_{\mbox{norm}} \rightarrow$ Th D$_{\mbox{exe}}$
corresponds to the first Sargent curve from which we conclude that
$i({\rm Th C}^{\prime \prime}_{\mbox{norm}}) - i ({\rm Th D}_{\mbox{exe}}=0$
or $\pm 1.$ It can also be seen
from Fig. 2 that the transformation Th C$^{\prime \prime}_{\mbox{norm}}
\rightarrow$ Th D$_{\mbox{norm}}$ belongs at least to third, or still higher order,
curve which excludes the possibilities of $i({\rm Th C}^{\prime \prime}_{\mbox{norm}})$
being 0, or $\pm 1.$ The excited level 3.202 mv of Th D
nucleus is connected with the normal level by two
$\gamma$-transitions, 0.582 mv and 2.620 mv, from which the
second is surely quadripole. This indicates that its spin
will not be larger than 4, because by each $\gamma$-transition
$\Delta i \leq 2$. This gives for $i({\rm Th C}^{\prime \prime}_{\mbox{norm}})$
the upper limit $\leq 5$. Thus for the spin of normal state of Th C$^{\prime
\prime}$ nucleus we have the choice between 2, 3, 4 and 5; it seems
however, to be necessary to accept the largest possible
value $i({\rm Th C}^{\prime \prime}_{\mbox{norm}}) = 5$ in order to have a
sufficiently large spin difference between Th C$_{\mbox{norm}}(i=2)$ and Th
C$^{\prime \prime}_{\mbox{norm}}$ to explain the presence of strong fine structure
of $\alpha$-rays in the Th C $\rightarrow$ Th C$^{\prime \prime}$
transformation\footnote{Gamow and Rosenblum, Comptes rendus {\bf 197,} 1620
(1933).}
\end{document}
%ENCODED JUNUARY 2003 BY NIS;
%EDITED 2003 By NIS;