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Alpha Decay

Many types of heavy atomic nucleus spontaneously decay to produce daughter nucleii via the emission of $\alpha $-particles (i.e., helium nucleii) of some characteristic energy. This process is know as $\alpha $-decay. Let us investigate the $\alpha $-decay of a particular type of atomic nucleus of radius $R$, charge-number $Z$, and mass-number $A$. Such a nucleus thus decays to produce a daughter nucleus of charge-number $Z_1=Z-2$ and mass-number $A_1=A-4$, and an $\alpha $-particle of charge-number $Z_2=2$ and mass-number $A_2=4$. Let the characteristic energy of the $\alpha $-particle be $E$. Incidentally, nuclear radii are found to satisfy the empirical formula
\begin{displaymath}
R = 1.5\times 10^{-15} A^{1/3} {\rm m}=2.0\times 10^{-15} Z_1^{1/3} {\rm m}
\end{displaymath} (353)

for $Z\gg 1$.

In 1928, George Gamov proposed a very successful theory of $\alpha $-decay, according to which the $\alpha $-particle moves freely inside the nucleus, and is emitted after tunneling through the potential barrier between itself and the daughter nucleus. In other words, the $\alpha $-particle, whose energy is $E$, is trapped in a potential well of radius $R$ by the potential barrier

\begin{displaymath}
V(r) = \frac{Z_1 Z_2 e^2}{4\pi \epsilon_0 r}
\end{displaymath} (354)

for $r>R$.

Making use of the WKB approximation (and neglecting the fact that $r$ is a radial, rather than a Cartesian, coordinate), the probability of the $\alpha $-particle tunneling through the barrier is

\begin{displaymath}
\vert T\vert^{ 2} = \exp\left(-\frac{2\sqrt{2 m}}{\hbar}\int_{r_1}^{r_2}
\sqrt{V(r)-E} dr\right),
\end{displaymath} (355)

where $r_1=R$ and $r_2 = Z_1 Z_2 e^2/(4\pi \epsilon_0 E)$. Here, $m=4 m_p$ is the $\alpha $-particle mass. The above expression reduces to
\begin{displaymath}
\vert T\vert^{ 2} = \exp\left(-2 \sqrt{2} \beta \int_{1}^{E_c/E}\left[\frac{1}{y}-\frac{E}{E_c}\right]^{1/2} dy\right),
\end{displaymath} (356)

where
\begin{displaymath}
\beta = \left(\frac{Z_1 Z_2 e^2 m R}{4\pi \epsilon_0 \hbar^2}\right)^{1/2} = 0.74 Z_1^{2/3}
\end{displaymath} (357)

is a dimensionless constant, and
\begin{displaymath}
E_c = \frac{Z_1 Z_2 e^2}{4\pi \epsilon_0 R} = 1.44 Z_1^{2/3}  {\rm MeV}
\end{displaymath} (358)

is the characteristic energy the $\alpha $-particle would need in order to escape from the nucleus without tunneling. Of course, $E\ll E_c$. It is easily demonstrated that
\begin{displaymath}
\int_1^{1/\epsilon}\left[\frac{1}{y} - \epsilon\right]^{1/2} dy \simeq
\frac{\pi}{2 \sqrt{\epsilon}}-2
\end{displaymath} (359)

when $\epsilon\ll 1$. Hence.
\begin{displaymath}
\vert T\vert^{ 2} \simeq \exp\left(-2 \sqrt{2} \beta\left[\frac{\pi}{2}\sqrt{\frac{E_c}{E}}-2\right]\right).
\end{displaymath} (360)

Now, the $\alpha $-particle moves inside the nucleus with the characteristic velocity $v= \sqrt{2 E/m}$. It follows that the particle bounces backward and forward within the nucleus at the frequency $\nu\simeq v/R$, giving

\begin{displaymath}
\nu\simeq 2\times 10^{28}  {\rm yr}^{-1}
\end{displaymath} (361)

for a 1 MeV $\alpha $-particle trapped inside a typical heavy nucleus of radius $10^{-14}$m. Thus, the $\alpha $-particle effectively attempts to tunnel through the potential barrier $\nu $ times a second. If each of these attempts has a probability $\vert T\vert^{ 2}$ of succeeding, then the probability of decay per unit time is $\nu \vert T\vert^2$. Hence, if there are $N(t)\gg 1$ undecayed nuclii at time $t$ then there are only $N+dN$ at time $t+dt$, where
\begin{displaymath}
dN = - N \nu \vert T\vert^2 dt.
\end{displaymath} (362)

This expression can be integrated to give
\begin{displaymath}
N(t) = N(0) \exp(-\nu \vert T\vert^2 t).
\end{displaymath} (363)

Now, the half-life, $\tau$, is defined as the time which must elapse in order for half of the nuclii originally present to decay. It follows from the above formula that
\begin{displaymath}
\tau = \frac{\ln 2}{\nu \vert T\vert^2}.
\end{displaymath} (364)

Note that the half-life is independent of $N(0)$.

Finally, making use of the above results, we obtain

\begin{displaymath}
\log_{10}[\tau ({\rm yr})] = -C_1 - C_2 Z_1^{ 2/3} + C_3 \frac{Z_1}{\sqrt{E({\rm MeV})}},
\end{displaymath} (365)

where
$\displaystyle C_1$ $\textstyle =$ $\displaystyle 28.5,$ (366)
$\displaystyle C_2$ $\textstyle =$ $\displaystyle 1.83,$ (367)
$\displaystyle C_3$ $\textstyle =$ $\displaystyle 1.73.$ (368)

Figure 15: The experimentally determined half-life, $\tau _{ex}$, of various atomic nucleii which decay via $\alpha $ emission versus the best-fit theoretical half-life $\log_{10}(\tau_{th}) = -28.9 - 1.60 Z_1^{ 2/3} + 1.61 Z_1/\sqrt{E}$. Both half-lives are measured in years. Here, $Z_1=Z-2$, where $Z$ is the charge number of the nucleus, and $E$ the characteristic energy of the emitted $\alpha $-particle in MeV. In order of increasing half-life, the points correspond to the following nucleii: Rn 215, Po 214, Po 216, Po 197, Fm 250, Ac 225, U 230, U 232, U 234, Gd 150, U 236, U 238, Pt 190, Gd 152, Nd 144. Data obtained from IAEA Nuclear Data Centre.
\begin{figure}
\epsfysize =4in
\centerline{\epsffile{Chapter05/fig06.eps}}
\end{figure}

The half-life, $\tau$, the daughter charge-number, $Z_1=Z-2$, and the $\alpha $-particle energy, $E$, for atomic nucleii which undergo $\alpha $-decay are indeed found to satisfy a relationship of the form (365). The best fit to the data (see Fig. 15) is obtained using

$\displaystyle C_1$ $\textstyle =$ $\displaystyle 28.9,$ (369)
$\displaystyle C_2$ $\textstyle =$ $\displaystyle 1.60,$ (370)
$\displaystyle C_3$ $\textstyle =$ $\displaystyle 1.61.$ (371)

Note that these values are remarkably similar to those calculated above.


next up previous
Next: Square Potential Well Up: One-Dimensional Potentials Previous: Cold Emission
Richard Fitzpatrick 2010-07-20