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Radioactivity

Problem 6.242

Find the kinetic energy of the recoil nucleus in the positronic decay of a \(\mathrm{N}^{13}\) nucleus for the case when the energy of positrons is maximum.

Reveal Answer
 6.242. T≈1/2Q(Q+2mc2)/MNc2=0.11keV, where Q==(MN−MC−2m)c2,m is the mass of an electron. \begin{aligned} &\text { 6.242. } T \approx 1 / 2 Q\left(Q+2 m c^{2}\right) / M_{\mathrm{N}} c^{2}=0.11 \mathrm{keV}, \text { where } Q=\\ &=\left(M_{\mathrm{N}}-M_{\mathrm{C}}-2 m\right) c^{2}, m \text { is the mass of an electron. } \end{aligned}
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