Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 32 - Elementary Particles - Problems - Page 945: 26

Answer

$$52.3\ \mathrm{MeV}$$

Work Step by Step

The two neutrinos must move together, in the direction which is opposite of the electron, in order for the electron to have the maximum kinetic energy, and thus $\text{( total momentum of neutrinos = magnitude to momentum of electron).}$ Since a neutrino is $\textbf{mass-less}$, we have $$E_{v}=p_{v} c .$$ The muon is at rest when it decays. Use momentum and conservation of energy , along with their relativistic relationship. $\therefore$ $$p_{\mathrm{e}^{-}}=p_{\overline{v}_{\mathrm{e}}}+p_{v_{\mu}}$$ $$m_{\mu^{-}} c^{2}=E_{\mathrm{e}^{-}}+E_{\overline{v}_{\mathrm{e}}}+E_{v_{\mu}}=E_{\mathrm{e}^{-}}+p_{\overline{v}_{\mathrm{e}}} c+p_{v_{\mu}} c=$$$$E_{\mathrm{e}^{-}}+\left(p_{\overline{v}_{\mathrm{e}}}+p_{v_{\mu}}\right) c=E_{\mathrm{e}^{-}}+p_{\mathrm{e}^{-}} c \quad \rightarrow$$ $$m_{\mu^{-}} c^{2}-E_{\mathrm{e}^{-}}=p_{\mathrm{e}^{-}} c \rightarrow\left(m_{\mu^{-}} c^{2}-E_{\mathrm{e}^{-}}\right)^{2}=\left(p_{\mathrm{e}^{-}} c\right)^{2}=E_{\mathrm{e}^{-}}^{2}-m_{\mathrm{e}^{-}}^{2} c^{4} \quad \rightarrow$$ $$m_{\mu^{-}} c^{4}-2 m_{\mu^{-}} c^{2} E_{\mathrm{e}^{-}}+E_{\mathrm{e}^{-}}^{2}=E_{\mathrm{e}^{-}}^{2}-m_{\mathrm{e}^{-}}^{2} c^{4} \quad \rightarrow $$$$\quad E_{\mathrm{e}^{-}}=\frac{m_{\mu^{-}}^{2} c^{4}+m_{\mathrm{e}^{2}}^{2} c^{4}}{2 m_{\mu^{-}} c^{2}}=\mathrm{KE}_{\mathrm{e}^{-}}+m_{\mathrm{e}^{-}} c^{2} \rightarrow$$ $$\mathrm{KE}_{\mathrm{e}^{-}}=\frac{m_{\mu}^{2} c^{4}+m_{\mathrm{e}^{2}}^{2} c^{4}}{2 m_{\mu^{-}} c^{2}}-m_{\mathrm{e}}^{-} c^{2}$$$$=\frac{(105.7 \mathrm{MeV})^{2}+(0.511 \mathrm{MeV})^{2}}{2(105.7 \mathrm{MeV})}-(0.511 \mathrm{MeV})=$$ $$=52.3\ \mathrm{MeV}$$
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