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\begin{document}
G.~Gamow, ZP, {\bf 51,} 204 \hfill {\large \bf 1928}\\
\vspace{2cm}
\begin{center}
{\Large \bf Quantum Theory of the Atomic Nucleus}\\
\end{center}
\vspace{0.5cm}
\begin{center}
G.~Gamov\\
~~~\\
(Received 1928)
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\vspace{2cm}
It has often been suggested that non--Coulomb attractive forces play a very
important role inside atomic nuclei. We can make many hypotheses concerning
the nature of these forces. They can be the attractions between the magnetic
moments of the individual constituents of the nucleus or the forces engendered
by electric and magnetic polarization.
In any case, these forces diminish very rapidly with increasing distance from
the nucleus, and only in the immediate vicinity of the nucleus do they outweigh
the Coulomb force.
From the scattering of \mbox{$\alpha$--particles} we may conclude that for heavy
elements the forces of attraction are still not measurable down to a distance of
$\sim 10^{-12}$ cm. We may therefore take the potential energy as being
correctly represented by the curve in [Fig. 1].
Here $r^{\prime \prime}$ gives distance down to which it has been shown
experimentally that the Coulomb repulsion alone exists. From $r^{\prime}$ down
the deviation ($r^{\prime}$ is unknown and perhaps much smaller than
$r^{\prime \prime}$) from the
Coulomb force is pronounced and the $U$--curve has a maximum at $r_0$.
For $r < r_0$ the attractive forces dominate; this region, the particle circles
around the rest of the nucleus like a satellite.
This motion, however, is not stable since the particle's energy is positive,
and, after some time, the \mbox{$\alpha$-particle} will fly out
(\mbox{$\alpha$ -emission}). Here, however, we meet a fundamental difficulty.\\
\begin{figure}[h]
\centerline{\epsfig{file=fig1e.eps}}
\caption{}
\end{figure}
To fly off, the
\mbox{$\alpha$--particle} must overcome a potential barrier of height
$U_0$ [Fig. 1]; its energy may not be less than u$_0$. But the energy of the
emitted
\mbox{$\alpha$--particle}, as verified experimentally, is much less. For
example, we find on analyzing the scattering of
$Ra - c^{\prime}$ -- \mbox{$\alpha$--particles} by uranium that for the uranium nucleus the
Coulomb law is
valid down to distances of $3.2 \times 10^{-12}$ cm. on the other hand, the
\mbox{$\alpha$-particles} emitted by uranium itself have an energy which
represents a
distance of $3.2 \times 10^{-12}$ cm ( r$_2$ in [Fig. 1] ) on the repulsive
curve. If an
\mbox{$\alpha$-particle}, coming from the interior of the nucleus, is
to fly away, it must pass through the region $r_1$ and $r_2$ where its kinetic
energy would be negative, which, naturally, is impossible classically.
To overcome this difficulty, Rutherford assumed that the
\mbox{$\alpha$-particles}
in the nucleus are neutral, since they are assumed to have two electrons there.
Only at a certain distance from the nucleus, on the other side of the potential
barrier's maximum, do they, according to Rutherford. lose their two electrons.
which fall back into the nucleus while the
\mbox{$\alpha$-particles} fly on impelled
by the Coulomb repulsion. But this assumption seems very unnatural, and can
hardly be a true picture.
If we consider the problem from the wave mechanical point of view, the above
difficulties disappear by themselves. In wave mechanics a particle always has a
finite probability, different from zero, of going from one region to another
region of the same energy, even through the two regions are separated by an
arbitrarily large but finite potential barrier.
As we shall see further, the probability for such a transition, all things
considered, is very small and, in fact, is smaller the higher the potential
barrier is. To clarify this point, we shall analyze a simple case
[see Fig. 2].
\begin{figure}[h]
\centerline{\epsfig{file=fig2e.eps}}
\caption{}
\end{figure}
We have a rectangular potential barrier and we wish to find the solution of
Schroedinger's equation which represents the penetration of the particle from
right to left. For the energy E we write the wave function $\psi$ in the
following form:
$$
\psi = \Psi(q) e^{(2 \pi i/h)Et},
$$
where $\Psi(q)$ satisfies the amplitude equation
\begin{equation}
\frac{\partial^2 \Psi}{\partial q^2} + \frac{8 \pi^2 m}{h^2}~ (E - U)~
\Psi = 0
\end{equation}
For the region I we have the solution
$$
\Psi_1 = A \cos \cdot (kq + \alpha)
$$
where $A$ and $\alpha$ are two arbitrary constants and
$$
k = \frac{2 \pi \sqrt{2m}}{h} ~ \cdot \sqrt{E} \eqno (2a)
$$
In the region II the solution reads
$$
\Psi_{II} = B_1 e^{- k' q} + B_2 e^{k'q}
$$
where
$$
k^{\prime} = \frac{2\pi \sqrt{2m}}{h} \cdot \sqrt{U_0 - E} \eqno (2b)
$$
At the boundary $q = 0$ the following conditions apply:
$$
\Psi_I(0) = \Psi_{II}(0) \qquad \mbox{and} \qquad
\left[ \frac{\partial \Psi_I}{\partial q}\right]_{q = 0} =
\left[ \frac{\partial \Psi_{II}}{\partial q}\right]_{q = 0},
$$
from which we easily obtain
$$
B_1 = \frac{A}{2 \sin \theta} \cdot \sin(\alpha + \theta); ~~~ B_2 = -
\frac{A}{2 \sin \theta} \cdot \sin(\alpha - \theta),
$$
where
$$
\sin \theta = \frac{1}{\sqrt{1 + (\frac{k}{k^{\prime}})^2}}
$$
The solution in region II therefore reads
$$
\Psi_{II} = \frac{A}{2\sin \theta} \cdot [\sin(\alpha + \theta)
\cdot e^{k^{\prime}q} -
\sin(\alpha - \theta) e^{k^{\prime}q}].
$$
In III we again have
$$
\Psi_{III} = C \cos(kq + \beta)
$$
At the boundary $q = l$ we have from the boundary conditions
$$
\frac{A}{2 \sin \theta} \cdot [\sin(\alpha + \theta)e^{- lk^{\prime}}]
- \sin(\alpha - \theta)e^{+lk^{\prime}}] = C \cos(kl + \beta)
$$
and
$$
\frac{A}{2 \sin \theta} k^{\prime} \cdot
[\sin(\alpha + \theta)e^{- lk^{\prime}}]
- \sin(\alpha - \theta)e^{+lk^{\prime}}] = - kC \sin(kl + \beta).
$$
Hence
$$
C^2 = \frac{A^2}{4 \sin^2 \theta} \cdot \left\{ \left[1 +
\left( \frac{k^{\prime}}{k} \right)^2 \right] \cdot
\sin^2(\alpha - \theta) \cdot e^{2lk^{\prime}} \right.
$$
$$
- \left[1 - \left(\frac{k^{\prime}}{k}\right)^2\right] \cdot
2\sin(\alpha - \theta) \cdot \sin(\alpha + \theta)
$$
$$
\left.+ \left[1 + \left(\frac{k^{\prime}}{k}\right)^2\right] \cdot
\sin^2(\alpha + \theta) \cdot e^{-2lk^{\prime}}
\right\}. \eqno (3)
$$
The calculation of $\beta$ is of no interest to us. We are interested only in
the case in which $lk^{\prime}$ is very large so that we need consider only the
firs term in (3).
We thus have the following solution:
~~~~~~~~~~~left:~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~right:
$$
A \cos(kq + \alpha) \ldots A~ \frac{\sin(\alpha - \theta)}
{2 \sin \theta} \cdot
\left[1 + \left(\frac{k^{\prime}}{k}\right)^2\right]^{1/2} \cdot
e^{+ lk^{\prime}} \cos(kq + \beta).
$$
If we now write $\alpha - \frac{\pi}{2}$ instead of $\alpha$ multiply the
obtained solution by $i$, and then add the two solutions, we obtain on the left
$$
\Psi = A e^{i(kq + \alpha)}, \eqno (4a)
$$
on the right, however,
$$
\Psi = \frac{A}{2 \sin \theta} \cdot
\left[1 + \left(\frac{k^{\prime}}{k}\right)^2\right]^{1/2} \cdot
e^{+lk^{\prime}} \left \{\sin(\alpha - \theta) \cos(kq + \beta) -
\right.
$$
$$
\left. i \cos(\alpha + \theta) \cos(kq + \beta^{\prime}) \right\}, \eqno (4b)
$$
where $\beta^{\prime}$ is the new phase.
If we multiply this solution by $e^{2 \pi i\frac{E}{h}t}$, we obtain for $\Psi$
on the left the (from right to left) advancing wave; on the right, however, the
complex oscillatory phenomenon with a very large amplitude $(e^{lk^{\prime}})$
that departs only slightly from a standing wave. This means nothing other than
that the wave coming from the right is partly reflected and partly transmitted.
We thus see that the amplitude of the transmitted wave is smaller the smaller is
the total energy $E$, and in fact the factor
$$
e^{- lk^{\prime}} = e^{\frac{\displaystyle 2 \pi \cdot \sqrt{2m}}{h}
\sqrt{U_0 - E }~ .l}
$$
plays an important role in this connection.
We can now solve the problem for two symmetrical potential barriers
[Fig. 3]. We shall seek two solution.
\begin{figure}[h]
\centerline{\epsfig{file=fig3e.eps}}
\caption{}
\end{figure}
One solution is to be valid for positive $q$, and for $q > q_0 + l$
is to give the wave:
$$
Ae^{i(\frac{\displaystyle 2 \pi E}{\displaystyle h} t - kq + \alpha)}
$$
The other solution is valid for negative $q$, and for $q < - (q_0 + l)$
gives the wave
$$
Ae^{i \left(\frac{\displaystyle 2 \pi El}{\displaystyle h} +
q k' - \alpha \right)}.
$$
We cannot attach the two solutions each other continuously at
$q = 0$ since we have
here two boundary conditions to fulfill and only one arbitrary constant $\alpha$
to adjust. The physical reason for this impossibility is that the $\Psi$ --
function constructed from these two solutions does not satisfy the conservation
law
$$
\frac{\partial}{\partial t}~ \int\limits_{- (q_0 + l)}^{+ (q_0 + l)}
\psi \bar{\psi} dq = 2 \cdot \frac{- h}{4 \pi \cdot i \cdot m}
\left[\psi~ \mbox{grad} ~\bar{\psi}
- \bar{\psi}~ \mbox{grad} ~\psi\right]_I
$$
To overcome these difficulties we must assume that the vibrations
are damped and
make $E$ complex
$$
E = E_0 + i \frac{h \lambda}{4 \pi}
$$
where $E_0$ is the usual energy and $\lambda$ is the damping decrement (decay
constant). We then see from the relations (2a) and (2b) that
$k$ and k$^{\prime}$
are complex, that is, that the amplitude of our wave also depends exponentially
on the coordinate $q$. For example, for the running wave the amplitude in the
direction of the diverging wave will increase. This means nothing more than that
if the vibrations are damped at the source of the wave, the amplitude of the
wave segment that left earlier must be larger. We can now determine $\alpha$ so
that the boundary conditions are fulfilled. But the exact solution does not
interest us. If $\lambda$ is small compared to $\frac{E}{h}$
$(\mbox{for} Ra - c'~{\frac{E}{h} \cong \frac{10^{-5}}{10^{-27}}
\quad \mbox{sec}^{-1} =
10^{22} \mbox{sec}^{-1} \quad \mbox{and} \quad \lambda
= 10^{5} \mbox{sec}^{-1})$ the change in $\Psi(q)$ is
very small and we can simply multiply the old solution with
$e^{- \frac{\displaystyle \alpha}{\displaystyle 2}t}$.
The conservation principle then reads
$$
\frac{\partial}{\partial t} ~ e^{- \lambda t}
\int\limits_{-(q_0 + l)}^{+ (q_0 + l)} ~\Psi^{(q)}_{II, III} \cdot
\Psi^{(q)}_{II,~ III}~ d q = - 2 ~\frac{A^2 h}
{4 \pi \cdot i \cdot m} \cdot 2ik
\cdot e^{-\lambda t},
$$
from which we obtain
$$
\lambda = \frac{4 h k \sin^2 \theta}
{\pi m \left[
1 + \left( \frac{k'}{k^0} \right)^2 \right]
2(l + q_0)_k} \cdot e^{-\frac{\displaystyle 4\pi l \sqrt{2m}}
{\displaystyle h} \sqrt{U_0 - E}},\eqno (5)
$$
where $k$ is a number of order of magnitude one.
This formula gives the dependence of the decay constant on the decay energy for
our simple nuclear model.
Now we can go over to the case of the actual nucleus.
We cannot solve the corresponding wave equation because we do not know the exact
formula for the potential in the neighborhood of the nucleus. But even without
an exact knowledge of the potential we can carry over to the actual nucleus
results obtained from our simple model.
As usual, in the case of a central force, we seek the solution in polar
coordinates and, in fact, in the form
$$
\Psi = u(\theta , \phi) \chi(r).
$$
For $u$ we obtain the spherical harmonics, and $\chi$ must satisfy the
differential equation
$$
\frac{d^2 \chi}{d r^2} + \frac{2}{r} \frac{d \chi}{dr} + \frac{8 \pi^2 m}{h^2}
\cdot \left[
E - U - \frac{h^2}{8 \pi^2 m} \cdot \frac{n(n + 1)}{r^2}
\right] \cdot \chi = 0,
$$
where $n$ is the order of the spherical harmonic. We can place $n = 0$,
since if
$n> 0$, this would really be just as through the potential energy were
enlarged, and because of this damping for these oscillations is much smaller.
The particle must first pass over to the state
$n = 0$ and can only then fly away.
It is quite possible that such transitions are the cause of the $\gamma$ rays
which always accompany $\alpha$ -- emission. The probable shape of
$U$ is shown in
[Fig. 4].
For large values of $r$, we shall take for $\chi$ the solution
$$
\chi I = \frac{A}{r} \cdot e^{i (\frac{\displaystyle e\pi E}
{\displaystyle h} t - kr)}
$$
Even through we cannot obtain the exact solution of the problem in this case, we
can still say that on the average in the regions I and I$^{\prime}$, $\chi$ does
not decrease very rapidly
(in the three dimensional case like $l/r$).
\begin{figure}[h]
\centerline{\epsfig{file=fig4e.eps}}
\caption{}
\end{figure}
In the region III, however, $\chi$ decreases exponentially, and in analogy with
our simple case we may state that the relation between the amplitude decrease
and $E$ is given by the factor
$$
e^{- \frac{\displaystyle 2 \pi \sqrt{2m}}{\displaystyle h}}
\int\limits^{r_2}_{r_1} \sqrt{U - E}~ dr
$$
If we use the conservation principle we can again write down the formula
$$
\lambda = D ~ . ~ e^{- \frac{\displaystyle 2 \pi \sqrt{2m}}{\displaystyle h}}
\int\limits^{r_2}_{r_1} \sqrt{U - E} dr \eqno (6)
$$
where $D$ depends on the particular properties of the nuclear model. We can
neglect the dependence of $D$ on $E$ compared to its exponential dependence.
We may also replace the integral
$$
\int\limits_{r_1}^{r_2} \sqrt{U - E}~ dr
$$
by the approximate integral
$$
\int\limits^{\frac{\displaystyle 2Z e^2}{\displaystyle E}}_{0}
\sqrt{\frac{2 Z e^2}{r} - E} \cdot d r
$$
The relative error we introduce in this way is of the order of
$\sqrt{\frac{r_1}{r_2}}$. Since $r_1/r_2$ is small, this error is not very
large. Since E does not differ much for different radioactive elements, we write
as an approximation
$$
\log \lambda = \mbox{const}_E + B_E ~ . ~ \Delta E,
$$
$\ldots$ where
$$
B_E = \frac{\pi^2 \sqrt{2m} \cdot 2 Ze^2}{h E^{3/2}}. \eqno (7)
$$
We wish to compare this formula with the experimental results. It is known that
if we plot the logarithm of the decay constant against the energy of the emitted
\mbox{$\alpha$-particle}, all the points for a definite radioactive
family fall on a
straight line. For different families we obtain different parallel lines. The
empirical formula reads
$$
\log \lambda = \mbox{const} + b E
$$
where $b$ is a constant that is common to all radioactive families.
The experimental value of $b$ is $b_{exp} = 1.02 \times 10^7$ (calculated from
Ra--$A$ and Ra).
If we put the energy value for Ra--$A$ into our formula, we get
$$
b_{theoretical} = 0.7 \times 10^7
$$
This order of magnitude agreement shows that the basic assumptions of our
theory must be correct $\ldots$
%SB. = ENCODED 17 JULE 2000 BY NIS;
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