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Question

Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:

12N12C+e++v

12C12C+v (4.43MeV)

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Solution

12N 12C*+e++v

12C* 12C+v (4.43 MeV)

Net reactions:

12N 12C+e++v+v (4.43 MeV)

Energy of (e++v)=Energy of N12(C12+v)

Energy of (e++v)=12.018613 u(12)u4.43

=0.018613u4.43u

=17.3284.43=12.89 MeV

Maximum energy of e

(Assuming 0 energy for v)

=12.89MeV


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