Question

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

${}^{12}\text{N}{\to }^{12}{\text{C}}^{\ast }+e^{+}+v$

${}^{12}{\text{C}}^{\ast }{\to }^{12}\text{C}+v\left(4.43\text{MeV}\right)$

Answer 1

${}^{12}\text{N}\to {}^{12}\text{C*}+{\text{e}}^{+}+v$

${}^{12}\text{C*}\to {}^{12}\text{C}+v\left(4.43\text{MeV}\right)$

$\text{Net reactions}:$

${}^{12}\text{N}\to {}^{12}\text{C}+{\text{e}}^{+}+v+v\left(4.43\text{MeV}\right)$

$\text{Energy of}(e^{+}+v)=\text{Energy of}{\text{N}}^{12}-({\text{C}}^{12}+v)$

$\text{Energy of}(e^{+}+v)=12.018613u-\left(12\right)u-4.43$

=$0.018613u-4.43u$

=$17.328-4.43=12.89\text{MeV}$

${\text{Maximum energy of e}}^{-}$

$\text{(Assuming 0 energy for v)}$

$=12.89\text{MeV}$