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

Theory of alpha emission

Barrier problem

Decay half life

The disintegration constant for \alpha emission may be expressed as

[math]\lambda = f P = \frac{0.693}{t_{1/2}}[/math]

where

f = number of times the alpha particle tries to escape the well by interacting with the barrier ~ [math]\frac{\hbar}{\Delta E}[/math] P = probability that the alpha particle escapes when it hits the barrier

the half life [math](t_{1/2})[/math] is then proportional to [math]\lambda[/math].

Example

Curium

Beta Decay

Types of decay

negative beta decay

[math]n \rightarrow p + e^- + \bar{\nu_e}[/math]

positive beta decay

[math]p \rightarrow n + e^+ + \nu_e[/math]

electron capture

[math]p+e^- \rightarrow n + \nu_e[/math]

negative beta decay

[math]{A \atop Z }X_{N} \rightarrow {A \atop Z+1 }Y_{N-1} + \beta^- + \bar{\nu_e}[/math]

[math]Q_{\beta^-} = \left [ m_N \left ({A \atop Z }X \right)-m_N\left({A \atop Z+1 }Y\right) -m_e \right ]c^2[/math]

let

[math]m\left({A \atop Z }X\right)c^2 \equiv \mbox{Atomic mass} = m_N\left({A \atop Z }X\right)c^2 + Zm_ec^2 - \sum_i^Z B_i [/math]

where

[math]B_i[/math] = ith electron binding energy

then

[math]Q_{\beta^-} = m\left({A \atop Z }X\right)c^2 -Zm_ec^2 + \sum_i^Z B_i - m\left({A \atop Z+1 }Y\right)c^2 +(Z+1)m_ec^2 -\sum_i^{(Z+1)}B_i - m_ec^2 [/math]

[math]\;\;\;\;\;= \left [ m\left({A \atop Z }X\right)- m\left({A \atop Z+1 }Y\right) \right] c^2 - B_Z[/math]

[math]\;\;\;\;\;= T_{\beta^-} + T_{\bar{\nu}}[/math] = energy shared by electron and neutrino

[math]B_Z =[/math] binding energy of most outer electron in element "Y"

negative beta decay Example

[math]{3 \atop 1 }H_{2} \rightarrow {3 \atop 2 }He_{1} + \beta^- + \bar{\nu_e}[/math]

[math]Q_{\beta^-} = \left [ m\left({3 \atop 1 }H\right)- m\left({3 \atop 2 }He\right) \right] c^2 - B_Z[/math]

[math]= \left [ 3.0160493 - 3.0160293 \right] 931.502 \frac{\mbox{MeV}}{\mbox {u}} - 24.6[/math] eV

[math]= 18.63 - 0.0246 keV = 18.61 keV[/math]

[math]Q_{\beta^-} = 18.591 \pm 0.001[/math] keV

[math]\;\;\;\;\;= T_{\beta^-} + T_{\bar{\nu}}[/math] = energy shared by electron and neutrino

positive beta decay

[math]{A \atop Z }X_{N} \rightarrow {A \atop Z-1 }Y_{N+1} + \beta^+ + \nu_e[/math]

[math]Q_{\beta^+} = \left [ m_N \left ({A \atop Z }X \right)-m_N\left({A \atop Z-1 }Y\right) -m_e \right ]c^2[/math]

let

[math]m\left({A \atop Z }X\right)c^2 \equiv \mbox{Atomic mass} = m_N\left({A \atop Z }X\right)c^2 + Zm_ec^2 - \sum_i^Z B_i [/math]

where

[math]B_i[/math] = ith elctron binding energy

then

[math]Q_{\beta^+} = m\left({A \atop Z }X\right)c^2 -Zm_ec^2 + \sum_i^Z B_i - m\left({A \atop Z-1 }Y\right)c^2 +(Z-1)m_ec^2 -\sum_i^{(Z-1)}B_i - m_ec^2 [/math]

[math]\;\;\;\;\;= \left [ m\left({A \atop Z }X\right)- m\left({A \atop Z+1 }Y\right) \right] c^2 - 2m_e c^2 +B_Z[/math]

[math]\;\;\;\;\;= T_{\beta^+} + T_{\bar{\nu}}[/math] = energy shared by electron and neutrino

[math]B_Z =[/math] binding energy of most outer electron in element "X"

positive beta decay Example

[math]{11 \atop 6 }C_{5} \rightarrow {11 \atop 5 }B_{6} + \beta^+ + \nu_e[/math]

[math]Q_{\beta^+} = \left [ m\left({11 \atop 6 }C\right)- m\left({11 \atop 5 }B\right) \right] c^2 - 2m_ec^2+B_Z[/math]

[math]= \left [ 11.0114338- 11.0093055 \right] 931.502 \frac{\mbox{MeV}}{\mbox {u}} -2 \times 511 keV+ 8.3[/math] eV

[math]= 1.983 MeV - 1.022 MeV = 0.961MeV[/math]

[math]Q_{\beta^+} = 960.2 [/math] keV

[math]\;\;\;\;\;= T_{\beta^+} + T_{\bar{\nu}}[/math] = energy shared by electron and neutrino

electron capture

An electron, originally in the K (N=1), L(N=2), or M(N=3) shell, is captured by the nucleus. After the capture, the other electrons will move down the shill in order to fill the vacancy and emit characteristic X-rays in the process.

[math]{A \atop Z }X_{N} + \beta^- \rightarrow {A \atop Z-1 }Y_{N+1} + \nu_e[/math]

[math]Q_{\epsilon} = \left [ m \left ({A \atop Z }X \right)-m\left({A \atop Z-1 }Y\right) \right ]c^2 -B_N[/math]

where

[math]B_N[/math] = captured electron binding energy

Note

those are atomic masses above

Also

If the captured electrons leaves the nucleus in an excited state

Then : [math]Q_{excited} = Q_{ground} - E_{excitation}[/math]

electron capture Example

[math]{7 \atop 4 }Be_{3} + \beta^- \rightarrow {7 \atop 3 }Li_{4} + \nu_e[/math]

[math]Q_{\epsilon} = \left [ m\left({7 \atop 4 }Be\right)- m\left({7 \atop 3 }Li\right) \right] c^2 - 259 eV[/math]

[math]= \left [ 7.0169292- 7.0160040 \right] 931.502 \frac{\mbox{MeV}}{\mbox {u}} - 0.259 keV[/math]

[math]= 861.83 keV -0.259 keV= 861.57 keV[/math]

[math]Q_{\beta^+}(experiment) = 862 [/math] keV

Conservation rules

baryon number is conserved

[math]B = \frac{n_q - n_{\overline{q}}}{3} [/math]

where

[math]n_q \ [/math] is the number of constituent quarks, and

[math]n_{\overline{q}}[/math] is the number of constituent antiquarks.

beta decay just changes p to n or n to p so the number of quarks dont change just the flavor (isospin).

Up and down quarks each have isospin [math]\vec{I} = \vec{\frac{1}{2}}[/math], and isospin z-components

[math]I_z =\left \{ {\frac{1}{2} \;\;\;\; \mbox{up quark} \atop \frac{-1}{2} \;\;\;\; \mbox{down quark}} \right .[/math]

All other quarks have I = 0. In general

[math]I_z=\frac{1}{2}(n_u-n_d)[/math]

Lepton number is conserved

[math]L \equiv n_{\ell} - n_{\overline{\ell}}[/math]

so all leptons have assigned a value of +1, antileptons −1, and non-leptonic particles 0.

[math]\begin{matrix} & n & \rightarrow & p & + & e^{-} & + & {\overline{\nu}}_e \\ L: & 0 & = & 0 & + & 1 & - & 1 \end{matrix}[/math]

Angular momentum

Consider first that the net angular momentum is zero (only consider spins)

The \beta and neutrino are spin 1/2 objects, therefore their spins may be either parallel or anti-parallel.

Fermi decay

A [math]\beta[/math] decay in which the [math]\beta[/math] and neutrino spins are anti-parallel is known as Fermi decay.

This means

[math]\Delta I = 0 \Rightarrow[/math] no change in the spin of the nucleus

Examples

[math]{14 \atop 8 }O_{6} \rightarrow {14 \atop 7 }N^*_{7} + \beta^+ + \nu[/math]

[math]I_i = 0^+ \Rightarrow I_f = 0^+ \Rightarrow \Delta I = 0[/math]

also [math]\Delta \pi = 0 \Rightarrow[/math] parity is conserved: [math]\pi (Y_{\ell,m}) =(-1)^{\ell}[/math].

[math]{14 \atop 7 }N^*_{7}[/math] = excited state of N

Gamow-Teller decay

A [math]\beta[/math] decay in which the [math]\beta[/math] and neutrino spins are parallel is known as Gamow-Teller decay. Interactions which take place are known as axial-vector.

In terms of total angular momenum \vec{I} the transition is

[math]\vec{I_i} \rightarrow \vec{I_f} + \vec{1}[/math]

[math]\Delta I =\left \{ {0 \;\;\;\;\;\;\; I_i = I_f =0 \atop 1 \;\;\;\; I_{i} =0 \mbox{and} I_{f}=1} \right .[/math]

Examples

[math]{6 \atop 2 }He_{4} \rightarrow {6 \atop 3 }Li_{3} + \beta^- + \bar{\nu}[/math]

[math]I_i = 0^+ \Rightarrow I_f = 1^+ \Rightarrow \Delta I = 1[/math]

also [math]\Delta \pi = 0 \Rightarrow[/math] parity is conserved: [math]\pi (Y_{\ell,m}) =(-1)^{\ell} \Rightarrow[/math] the final Li-6 [math]1^+[/math] state has [math]\vec{L} =1[/math] and the [math]\beta + \bar{\nu}[/math] state has [math]\vec{S}=1[/math] states which couple to an even parity state.

Mixed Fermi and Gamow-Teller decay

It is possible that [math]\beta[/math] decay can be a mixture of the two decay types. Some of the time the remaining nucleus is in an exited state other times the decay is directly to the ground state.

Examples

[math]{21 \atop 11 }Na_{10} \rightarrow {21 \atop 10 }Ne_{11} + \beta^+ + \nu_e[/math]

[math]I_i = 3/2^+ \Rightarrow I_f = 3/2^+ \Rightarrow \Delta I = 0[/math]

or

[math]{21 \atop 11 }Na_{10} \rightarrow {21 \atop 10 }Ne^*_{11} + \beta^+ + \nu_e[/math]

[math]I_i = 3/2^+ \Rightarrow I_f = 5/2^+ \Rightarrow \Delta I = 1[/math]

The above reaction involves "mirror" nuclei, nuclei in which the number of protons and neutrons is just interchange.

One can measure the angular distributions of [math]\beta[/math] particles to determine what the mixture is between the two decay types (Fermi and Gamow-Teller).

The mixture can be expressed as a ratio of matrix elements (Fermi's golden rule relates transitions to Matrix elements)

[math]y \equiv \frac{g_F M_F}{g_{GT} M_{GT}}[/math]

The interesting observation is that "y" for mirror nuclei is on the order of the value of "y" for neutron decay while nonmirror nuclear decays tend to be an order of magnitude less.

What does this mean?

The CVC (Conservation of Vector Current) hypothesis was born. The Fermi decay is the result of a vector current and is dominant in the decay of the neutron to a proton while the Gammow-Terller decay is an an axial-current transition. CVC is the assumption that the weak vector current responsible for the decay is conserved. Another observation is that the mirror nuclei transition ( mostly Fermi) illustrates how the nucleons inside the nucleus interact as free particles despite being surrounded by mesons mediating the nuclear force.

Forbidden decays

The Fermi decays [math](\Delta I = 0 )[/math] are often times refered to as the superallowed decays which Gamow Teller [math](\Delta I =1 )[/math] are simple "allowed" decays.

Forbidden decays are those which are substantially more improbable, due to parity violation, and as a result have long decay times.

Now the ANGULAR momentum [math](L)[/math] of the[math] \beta + \nu[/math] systems can be non-zero (the [math]\beta[/math] particle has an orbit radius [math]R[/math] about the nucleus and momentum [math]p[/math]).

first-forbidden[math] \Rightarrow \vec{L} = \vec{1}\;\;\; \Delta I=0,1,2 \;\;\;\Delta \pi =1[/math]

second-forbidden[math] \Rightarrow \vec{L} = \vec{2}\;\;\; \Delta I=2,3 \;\;\;\Delta \pi =0[/math]

third-forbidden[math] \Rightarrow \vec{L} = \vec{3}\;\;\; \Delta I=3,4 \;\;\;\Delta \pi =1[/math]

fourth-forbidden[math] \Rightarrow \vec{L} = \vec{4}\;\;\; \Delta I=4,5 \;\;\;\Delta \pi =0[/math]

Each of the above have Fermi [math](\vec{S}=0)[/math] and Gamow-Teller [math](\vec{S}=1)[/math] decays.

So for the "first-forbidden" transitions you have

[math]\vec{I} = \vec{L} + \vec{S} = \vec{1} + \vec{0} \Rightarrow \Delta I = 0,1[/math] Fermi

and

[math]\vec{I} = \vec{L} + \vec{S} = \vec{1} + \vec{1} \Rightarrow \Delta I = 0,1,2[/math] Gamow-Teller

systems.

Notice [math]\Delta \pi = 1 \Rightarrow[/math] parity Violating

The half life of the decay increases with each order

[math]{22 \atop 11 }Na_{11} (3^+)\rightarrow {22 \atop 10 }Ne_{12}(2^+) + \beta^+ + \nu_e \;\;\; t_{1/2} = 2.6 years[/math]

[math]{115 \atop 49 }In_{76} (9/2^+)\rightarrow {115 \atop 50 }Sn_{75}(1/2^+) + \beta^- + \bar{\nu_e} \;\;\; t_{1/2} = 10^{14} years[/math]

Decay rate

A calculation of the [math]\beta[/math] emmission decay rate is quite different from a calculation of [math]\alpha[/math] decay. In \alpha decay the nucleons of the original nucleus are used to form the fnial state [math]\alpha[/math] particle (He-4). In [math]\beta[/math] decay the [math]\beta[/math] and neutrino particles are the result of a nucleon transformation into its isospin complement [math](n \rightarrow p \mbox{or} p \rightarrow n)[/math]. Below is a list of the differences

1. the [math]\beta[/math] and neutrino did not exist before the decay
2. The [math]\beta[/math] and neutrino are relativistic (nuclear decay energy usually no enough to make heavy \alpha nucleus relativistic)
3. The light decay products can have continuous energy distributions. (before assuming the [math]\alpha[/math] carried away most of the eergy was usually

a good approximation)

The [math]\beta[/math] decay rate calculation was developed by Fermi in 1934 and was based on Pauli's neutrino hypothesis. Fermi's Golden rule(see Forest_FermiGoldenRule_Notes) says that the transition rate [math]W[/math] is given by a transition matrix element (or "Amplitude") [math]M_{i,f}[/math] weighted by the phase space and Plank's constant [math]\hbar[/math] such that

[math]W = \frac{2 \pi}{\hbar} |M_{i,f}|^2 \times [/math](Phase Space) [math]= \frac{\ln(2)}{t_{1/2}}[/math]

The underlying assumption is that the transition is a weak purturbation of the system. This assumption appears to be true based on the very short time scale ([math]10^{-20}[/math] sec) it takes for the formation of quasi-stationary nuclear states compared with the time it takes for a [math]\beta[/math] decay ( half lives ranging from seconds to days)

[math]|M_{i,f}|^2 = \lt \psi_{Daughter} \phi_{\beta} \psi_{\nu} \left | \hat{H_{int}}\right|\psi_{\mbox{Parent}}\gt [/math]

[math]\hat{H_{int}} =\left \{ {\hat{1} \hat{\tau} \;\;\;\; \mbox{Fermi decay} \atop \hat{\sigma} \hat{\tau} \;\;\;\; \mbox{Gamow-Teller Decay}} \right .[/math]

[math]\hat{\tau}[/math] = isospin transition matrix which turn protons to neutrons and vis-versa

Forest_NucPhys_I