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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

, charge-number

, and mass-number

. Such a nucleus thus decays to produce a daughter nucleus of charge-number

and mass-number

, and an $\alpha$ -particle of charge-number

and mass-number

. Let the characteristic energy of the $\alpha$ -particle be

. 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 , is trapped in a potential well of radius by the potential barrier

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

(354)

for

Making use of the WKB approximation (and neglecting the fact that 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

and $r_2 = Z_1 Z_2 e^2/(4\pi \epsilon_0 E)$ . Here,

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

then there are only

at time

, 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

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, , where is the charge number of the nucleus, and 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, , and the $\alpha$ -particle energy, , 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: Square Potential Well Up: One-Dimensional Potentials Previous: Cold Emission

Richard Fitzpatrick 2010-07-20