Consider the decay of a free neutron at rest :
` n to p +e^(-)`
Show that the two body decay of this type must necessary give an electron of fixed energy and therefore cannot account distribution in the `beta - `decay of a neutron or a nucleous as shown in figure .

[ Note : The simple result of this exercise was one among the several arguments advanced by W . Pauli to perdict the existence of a third particle in the decay products of `beta - ` decay . This particle is known as neutrino spin `1/2 ` ( like `e^(-)` p or n ) but is neutral ad either massless or having an extremely small mass (compared to the mass of electron) and which interacts very weakly with matter . The correct decay process of neutron is : `n to p+e(-)+v)`

Explanation

Suppose `Deltam` is used for a decay of neutron to a proton and electron energy released is E .
but `E = Deltamc^(2)`
where `Deltam + "mass of neutron "-("mass of proton "+ " mass of electron")`
`= 1.6747 xx10^(-24) - (1.6724 xx10^(-24)+9.11 xx10^(-28))`
` = 1.6747 xx10^(-24) -1.6733 xx10^(-24)`
` = 0.0014 xx10^(-24)`fram
Now , `E = Deltamc^(2)`
`= 0.0014 xx10^(-24) xx(3xx10^(-10))^(2)`
` = 0.0126 xx10^(-4)` erg
` =(0.0126 xx10^(-4))/(1.6 xx10^(-12))eV`
`= 0.007875 xx 10^(8)` eV
` = 0.7875 xx10^(6) ` eV
` = 0.79` MeV
Mass of a electron and a position is same and charges are also same but they are opposite .
When the electron and a positron comes to close they destroyed each other and their mass converted into energy as according to Einstein relation and the energy in the form of gamma rays produced .
`E= 2mc^(2) = 2xx9.1 xx10^(-31) xx(3xx10^(8))^(2)`
` = 1.638 xx 1^(-13)J `
` =(1.638 xx10^(-13))/(1.6xx10^(-19))`MeV
Photon must have minimum energy for pair production .