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Beta decay Problem

  • Thread starter neelakash
  • Start date
511
1
1. Homework Statement

Consider a beta decay:

[tex]\ {X_Z^A} \rightarrow\ {Y_{Z+1}^A} \ + {\beta_{-1}^{0}}\ + {\bar\nu_e}[/tex]


To show that the KE of the recoil nucleus is

[tex]\ E = \frac{\ Q + \ 2 \ m \ c^2}{\ 2 \ M_Y \ c^2}\ {T_{max}}[/tex]

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

2. Homework 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:

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

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 [tex]\ {T_y} + \ {T_\beta}[/tex]

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:

Answers and Replies

511
1
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.
 

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