## Friday, December 9, 2011

### Beta decay

Beta decay can involves three processes, in all the three processes atomic number of nucleus becomes one unecay dit greater or smaller but the mass number remains the same.

Problems in explaining beta decay

(1) non conservation of energy

(2) non conservation of angular momentum

To explain this Pauli suggested that a second particle is emitted with beta particles simultaneously but the sum of kinetic energies of two particles must be always equal to the energy difference between the parent and daughter nuclei. Thus the principle of conservation of energy is not voilated even the beta particle do not carry same energy.

The maximum beta particle energy is equal to the energy difference between the parent and daughter nuclei.

Beta decay Theory (Fermi's Theory):-

Electron, positron and ν does not as such exists inside the nucleus but are formed at the time of the decay. Fermi assumed that β-decay results from some form of nutrino, electrons and nucleus. This type of interaction is known as weak interaction. Constant required to express its strength is g=10-47erg cm3 . It is because of the weakness of this interaction that the β-decay does not take place instantly in the case when it is energitically possible.

This theory must include some relationship between the particles of initial and final nuclei. The relationship is expressed by means of matrix element. It involves the wave function of initial and final nuclear states and hence their spin, parities and the arrangement of the nucleons. When two states are very different from each other |Mif|2 becomes smaller. The total available β-decay energy E0 , which is the energy difference between initial and final states, can be divided between electrons and ν in large number of ways which affects the shape of beta ray spectrum.

The number of ways of distributing total available energy between electron and ν per unit total energy E0 is dN/dE0.

Now , number of ways in which electron may be given volume V and having momentum between pe and pe+dpe is given by

${\mathrm{dn}}_{e}=4\pi {p}_{e}^{2}V\frac{d{p}_{e}}{{h}^{3}}$

Similarly , number of ways in which neutrino can be arranged between volume V having momentum between pν and pν+dpν is

${\mathrm{dn}}_{\nu }=4\pi {p}_{\nu }^{2}V\frac{d{p}_{\nu }}{{h}^{3}}$

Number of ways in which β-decay can lead to an electron having momentum between pe and pe+dpe and ν having momentum between pν and pν+dpν is

dN=dne.dnν

$\mathrm{dN}=\frac{16{\pi }^{2}{V}^{2}}{{h}^{6}}{p}_{e}^{2}{p}_{\nu }^{2}d{p}_{e}d{p}_{\upsilon }$

Relativistic momentum of particle of rest mass m is given by

$p=\frac{{\left[E\left(E+2m{c}^{2}\right]}^{1/2}}{c}$

since mass of neutrino is almost zero

p=$\frac{E}{c}$

Therefore momentum of neutrino is

${p}_{\nu }=\frac{{E}_{\nu }}{c}=\frac{{E}_{0}-{E}_{e}}{c}$

and hence,

${\mathrm{dp}}_{\nu }=\frac{d{E}_{0}}{c}$ for given E

This shows that ,$\frac{\mathrm{dN}}{d{E}_{0}}=\frac{16{\pi }^{2}{V}^{2}}{{h}^{6}{c}^{3}}{\left({E}_{0}-{E}_{e}\right)}^{2}{p}_{e}^{2}d{p}_{e}$

We shall now also consider the role of coulomb's barrier in letting out electron and positron against beta decay. The coulomb barrier aids the escape of positron but hinders the escape of electrons. The effect of Coulomb barrier depends on the atomic number Z and the energy of electron or positron. The fermi factor represented by F(Z,Ee) is a complex function.

Now Fermi's theory finally gives the probability of decay with the emission of an electron having a given momentum pe by the expression which involves the nucleon-beta-neutrino force constant g , the matrix element |Mif|2 and the function of Fermi factor.