# Beta decay Problem

1. Apr 8, 2008

### neelakash

1. The problem statement, all variables and given/known data

Consider a beta decay:

$$\ {X_Z^A} \rightarrow\ {Y_{Z+1}^A} \ + {\beta_{-1}^{0}}\ + {\bar\nu_e}$$

To show that the KE of the recoil nucleus is

$$\ E = \frac{\ Q + \ 2 \ m \ c^2}{\ 2 \ M_Y \ c^2}\ {T_{max}}$$

m and T(max) is the mass and maximum KE of beta particle

2. Relevant equations

3. The attempt at a solution

writing the expression for Energy conservation,we see that the electron rest energy terms cancels and the resulting equation is:

$$\ {T_y} + \ {T_\beta} + \ {T_{\bar\nu}}=\ Q = [ \ {M_x} - \ {M_y} - \ {M_{\beta}} - \ {M_{\bar\nu}} ] \ {c^2}$$

For beta particle kinetic energy to be maximum, the kinetic energy of the neutrinos must be zero.(The kinetic energy of the recoiling nucleus assumed non-zero).

This gives an equation with known $$\ {T_y} + \ {T_\beta}$$

But we need another equation to solve for the kinetic energy of Y.

I also used conservation of momentum---disregarding the momentum of the neutrinos.But that did not help.Can anyone please tell how to do it?

Last edited: Apr 8, 2008
2. Apr 10, 2008

### neelakash

P²=p² [P---momentum of Y,p---momentum of beta particle]

P²c²=p²c²

(E+E')(E-E')=(e+e')(e-e') [E---energy of Y,E'---rest energy of Y, e---energy of

beta particle and e'---rest energy of beta particle]

K(K+2Mc²)=T(T+2mc²) [K-KE of Y and T---KE of beta particle]

T²-K²+2mc²T=2KMc²=(T+K)(T-K)+2Tmc² ~QT+2Tmc² where (T-K)~T

K= [(Q+2mc²)T]/(2Mc²)

Hence proved.