Nuclear Decay Forest NucPhys I

Alpha Decay

The spontaneous emission of an alpha particle[math]({4\atop 2 }He_{2})[/math] is the result of a natural decay process which can be described as the tunneling of energy ( in the form of the alpha particle) through the coulomb barrier. In other words, if a collection of nucleons within a nucleus finds itself sufficiently close to the nuclear force potential well limit, then a coulomb repulsion force can begin to dominant and facilitate the tunneling of this collection of nucleons ( an alpha particle) through the confining potential well.

The decay process can be represented by the following reaction notation

[math]{A \atop Z }X_{N} \rightarrow {A-4 \atop Z-2 }Y_{N-2} + \alpha[/math]

Q-value

The "Q-value" represents the net mass energy released in a nuclear reaction.

In the above example the Q value is calculated :

[math]E_i = E_f[/math]

[math]m_Xc^2 +T_X = m_Yc^2 + T_Y + m_{\alpha}c^2 + T_{\alpha}[/math]

[math]T_X = 0[/math] : assume nucleus is initially at rest

[math]Q \equiv m_Xc^2 -m_Yc^2 - m_{\alpha}c^2 = T_Y + T_{\alpha}[/math]

A positive Q value (Q>0) identifies a reaction as exothermic (exoergonic) which means that energy is given off and that the reaction is spontaneous

A negative Q value (Q<0) identifies the reaction as endothermic (endoergonic) which means that energy is required to for the reaction to take place.

Example

[math]{232 \atop 92 }U_{140} \rightarrow {228 \atop 90 }Th_{138} + \alpha[/math]

[math]Q = (232.0371463 - 228.0287313 - 4.002603 )uc^2 \frac{931.502 MeV}{uc^2} = 5.414 MeV[/math]

The positive Q value (Q>0) identifies the reaction as exothermic (exoergonic) which means that energy is given off and that the reaction is spontaneous

A negative Q value (Q<0) identifies the reaction as endothermic (endoergonic) which means that energy is required to for the reaction to take place.

Kinetic energy of alpha

Since the original nucleus was at rest, the final nuclei will have the same momentum in opposite directions in order to conserve momentum.

[math]T_{Y} = \frac{p^2_{Y}}{2m_Y}= \frac{p^2_{\alpha}}{2m_Y} = T_{\alpha} \frac{m_{\alpha}}{m_Y}[/math]

[math]Q = = T_Y + T_{\alpha} = T_{\alpha} \left ( \frac{m_{\alpha}}{m_Y} + 1\right )[/math]

[math]=T_{\alpha} \left ( \frac{4}{A-4} + 1\right ) = T_{\alpha} \left ( \frac{A}{A-4} \right )[/math]

[math] \Rightarrow T_{\alpha} = Q \left (1- \frac{4}{A} \right )[/math]

Example

[math]{232 \atop 92 }U_{140} \rightarrow {228 \atop 90 }Th_{138} + \alpha[/math]

[math] T_{\alpha} = Q \left (1- \frac{4}{A} \right )= 5.414 MeV \left (1- \frac{4}{228} \right ) = 5.32 MeV[/math]

Notice

The alpha particle caries away most of the kinetic energy.

The nuclear fragment (Y) does have a non-negligible amount of energy which can be sufficient to escape the material it is embedded in if it is on the order of a few microns from the materials surface. Heavy nuclei loose energy quickly when traveling through material.

Kinetic energy of alpha

Geiger-Nuttal Law

In 1911 Geiger and Nuttal noticed that the decay half life ([math]T_{1/2})[/math] of nuclei that emmitt alpha particles was related to the disentegration energy [math](Q)[/math].

[math]\log_{10}(T_{1/2}) = a + \frac{b}{\sqrt{Q}}[/math]

It works best for Nuclei with Even [math]Z[/math] and Even[math] N[/math]. The trend is still there for Even-Odd, Odd-Even, and Odd-odd nuclei but not as pronounced.

cluster decays

The Gieger-Nuttal Law has been extended to describe the decay of Large A (even-even and odd A) nuclei into clusters in which Silicon or Carbon are one of the clusters.

http://prola.aps.org/pdf/PRC/v70/i3/e034304

Beta Decay

Forest_NucPhys_I