Consider the decay of a free neutron at rest: n`top+e^(-)` Show that the tow-body dacay of this type must necessarily give an electron of fixed energy and, therefore, cannot for the observed continous energy distribution in the `beta`-decay of a neutron or a nucleus.

Explanation

Let the masses of the electron and proton be m and M respectively. Let v and V be the velocities of electroni and proton respectively. Using law of conservation of momentum. Momentum of electron + momentum of proton= momentum of neutron
`therefore mv+MV=0 rArr V=-m/Mv`
Clearly, the electronand the proton move in opposite directions. If mass `Deltam` has been converted into energy in th reaction, then
`1/2mv^(2)+1/2MV^(2)= Deltam xx c^(2)`
or `1/2mv^(2)+1/2M[-m/M]^(2)v^(2)=Deltamc^(2)`
or `1/2mv^(2)[1+m/M]=Deltamc^(2)`
or `v^(2)=(2MDeltamc^(2))/(m(M+m))`
Thus, it is proved that the value of `v^(2)` is fixed since all the quantities in right hand side are constant it establishes that the emitted electron must have a fixed energy and thus we cannot account for the continous energy distribution in the `beta`-decay of a neutron.