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Taking the values of atomic masses from the tables, find the maximum kinetic energy of beta-particles emitted by `Be^(10)` nuclei formed directly in the ground state. |
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Answer» The reaction is `Be^(10) rarr B^(10)+e^(-)+overset(-)v_(e )` For maximum `KE` of electrons we can put the energy of `vec(v)_(e )` to be zero. The atomic massed are `Be^(10)=10.016711 amu` `B^(10)= 10.016114 amu` So the `K.E` of electrons is (see previous problem) `597xx10^(-6) amuxxc^(2)= 0.56MeV` The momentum of electrons with this `K.E` is `0.941(MeV)/(c )` and the recoil energy of the daughter is `((0.941)^(2))/(2xxm_(d)c^(2))=((0.941)^(2))/(2xx10xx938)MeV= 47.2eV` |
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