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: 24

Answer

$$69.3\ \mathrm{MeV}$$

Work Step by Step

$\text{Since the pion decays from rest, the momentum before the decay is zero. }$$\text{Thus the momentum after the decay is also zero. Then}$ $$m_{\pi^{+}} c^{2}=E_{\mathrm{e}^{+}}+E_{v} ; \quad p_{\mathrm{e}^{+}}=p_{v} \quad \rightarrow \quad\left(p_{\mathrm{e}^{+}}^{2} c^{2}\right)=\left(p_{v}^{2} c^{2}\right) \quad \rightarrow$$$$ \quad E_{\mathrm{e}^{+}}^{2}-m_{\mathrm{e}^{+}}^{2} c^{4}=E_{v}^{2}$$ $$E_{\mathrm{e}^{+}}^{2}-m_{\mathrm{e}^{+}}^{2} c^{4}=\left(m_{\pi^{+}} c^{2}-E_{\mathrm{e}^{+}}\right)^{2}=m_{\pi^{+}}^{2} c^{4}-2 E_{\mathrm{e}^{+}} m_{\pi^{+}} c^{2}+E_{\mathrm{e}^{+}}^{2} \quad \rightarrow $$$$\quad 2 E_{\mathrm{e}^{+}} m_{\pi^{+}} c^{2}=m_{\pi^{+}}^{2} c^{4}+m_{\mathrm{e}^{+}}^{2} c^{4}$$ we continue solution $$E_{\mathrm{e}^{+}}=\frac{1}{2} m_{\pi^{+}} c^{2}+\frac{m_{\mathrm{e}^{+}}^{2} c^{2}}{2 m_{\pi^{+}}} \rightarrow $$$$\mathrm{KE}_{\mathrm{e}^{+}}+m_{\mathrm{e}^{+}} c^{2}=\frac{1}{2} m_{\pi^{+}} c^{2}+\frac{m_{\mathrm{e}^{+}}^{2} c^{2}}{2 m_{\pi^{+}}} \rightarrow$$ $$\mathrm{KE}_{\mathrm{e}^{+}}=\frac{1}{2} m_{\pi^{+}} c^{2}-m_{\mathrm{e}^{+}} c^{2}+\frac{m_{\mathrm{e}}^{2} c^{2}}{2 m_{\pi^{+}}}=$$$$= \frac{1}{2}(139.6 \mathrm{MeV})-0.511 \mathrm{MeV}+\frac{\left(0.511 \mathrm{MeV} / c^{2}\right)(0.511 \mathrm{MeV})}{2\left(139.6 \mathrm{MeV} / c^{2}\right)}$$ $$=69.3\ \mathrm{MeV}$$ $\text{alternate solution will be, using the momentum.}$ $$Q=\left[m_{\pi^{+}}-\left(m_{\mathrm{e}^{+}}+m_{v}\right)\right] c^{2}=139.6 \mathrm{MeV}-0.511 \mathrm{MeV}=139.1 \mathrm{MeV}$$ $$\mathrm{KE}_{\mathrm{e}^{+}}=E_{\mathrm{e}^{+}}-m_{\mathrm{e}^{+}} c^{2}=\sqrt{\left(p_{\mathrm{e}^{+}} c\right)^{2}+\left(m_{\mathrm{e}^{+}} c^{2}\right)^{2}}-m_{\mathrm{e}^{+}} c^{2} ;$$$$ \quad \mathrm{KE}_{v}=E_{v}=p_{v} c ; \quad p_{\mathrm{e}^{+}}=p_{v}=p$$ $$Q=\mathrm{KE}_{\mathrm{e}^{+}}+\mathrm{KE}_{v} \rightarrow 139.1 \mathrm{MeV}=$$$$\sqrt{(p c)^{2}+(0.511 \mathrm{MeV})^{2}}-(0.511 \mathrm{MeV})+p c$$ $$139.6-p c=\sqrt{(p c)^{2}+(0.511 \mathrm{MeV})^{2}} \rightarrow$$$$(139.6)^{2}-2(139.6) p c+p^{2} c^{2}=(p c)^{2}+(0.511 \mathrm{MeV})^{2}$$ $$\frac{(139.6 \mathrm{MeV})^{2}-(0.511 \mathrm{MeV})^{2}}{2(139.6 \mathrm{MeV})}=p c=69.8 \mathrm{MeV}$$ $$E_{v}=\mathrm{KE}_{v}=69.8 \mathrm{MeV} ; $$$$\quad Q-\mathrm{KE}_{v}=\mathrm{KE}_{\mathrm{e}^{+}}=139.1 \mathrm{MeV}-69.8 \mathrm{MeV}$$ $$=69.3\ \mathrm{MeV}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.