Saved Bookmarks
| 1. |
Considerthe 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 advancedbyW . Pauli to perdict the existence of a thirdparticle in the decay products of beta -decay . This particleis 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) |
|
Answer» Solution :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 positionis 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 . |
|