Правка разделов «Statement», «Solution», «Answer»
en/14.5.5.md
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| ### Statement | |||
| − | $14.5.5.$ [Insert the problem statement] | ||
| + | $14.5.5.$ At what kinetic energy of electrons and positrons (in $MeV$) in experiments with colliding beams is the production of a proton‑antiproton pair observed: $e^- + e^+ \to p + \bar{p}$ ? And the production of a $\pi^0$‑meson: $e^- + e^+ \to \pi^0$ ? | ||
| ### Solution | |||
| + | First, the problem implies that the energies of the electron and positron are equal. Also note that the minimum kinetic energy of the electrons and positrons is achieved when the reaction products are at rest: no energy needs to be spent on accelerating them. | ||
| − | 1 | ||
| + | We use the particle masses in $MeV$ as commonly adopted in nuclear physics, i.e., having the dimension of energy (obtained by multiplying the usual mass by $c^2$). This approach is described in problem [14.5.10](https://savchenkosolutions.com/ru/14.5.10), from which the necessary numerical values can also be taken. | ||
| − | |||
| + | Then the energy conservation law can be written compactly: | ||
| + | $$ | ||
| + | 2m_e + 2E_{k1} = 2 m_p, | ||
| + | $$ | ||
| + | $$ | ||
| + | 2m_e + 2E_{k2} = m_{\pi^0}. | ||
| + | $$ | ||
| + | Hence the answers: | ||
| + | $$ | ||
| + | E_{k1} = m_p - m_e \approx 938 \ \text{MeV}, | ||
| + | $$ | ||
| + | $$ | ||
| + | E_{k2} = \frac{m_{\pi^0}}{2} - m_e \approx 67 \ \text{MeV}. | ||
| + | $$ | ||
| + | Or, in the more familiar form using masses in kilograms: | ||
| + | $$ | ||
| + | E_{k1} = (m_p - m_e)c^2, | ||
| + | $$ | ||
| + | $$ | ||
| + | E_{k2} = \left(\frac{m_{\pi^0}}{2} - m_e\right)c^2. | ||
| + | $$ | ||
| + | $Note:$ Savchenko's numerical answer is correct, but, as in many places at the end of the book, there is a misprint in the formula. | ||
| − | [Insert a concise answer or boxed result] | ||
| + | #### Answer | ||
| + | $$ | ||
| + | \boxed{E_{k1} \approx 938 \ \text{MeV}, \qquad E_{k2} \approx 67 \ \text{MeV}.} | ||
| + | $$ | ||
| @@ -1,11 +1,36 @@ | |||
| ### Statement | ### Statement | ||
| $14.5.5.$ [Insert the problem statement] | $14.5.5.$ At what kinetic energy of electrons and positrons (in $MeV$) in experiments with colliding beams is the production of a proton‑antiproton pair observed: $e^- + e^+ \to p + \bar{p}$ ? And the production of a $\pi^0$‑meson: $e^- + e^+ \to \pi^0$ ? | ||
| ### Solution | ### Solution | ||
| First, the problem implies that the energies of the electron and positron are equal. Also note that the minimum kinetic energy of the electrons and positrons is achieved when the reaction products are at rest: no energy needs to be spent on accelerating them. | |||
| 1 | We use the particle masses in $MeV$ as commonly adopted in nuclear physics, i.e., having the dimension of energy (obtained by multiplying the usual mass by $c^2$). This approach is described in problem [14.5.10](https://savchenkosolutions.com/ru/14.5.10), from which the necessary numerical values can also be taken. | ||
| Then the energy conservation law can be written compactly: | |||
| $$ | |||
| 2m_e + 2E_{k1} = 2 m_p, | |||
| $$ | |||
| $$ | |||
| 2m_e + 2E_{k2} = m_{\pi^0}. | |||
| $$ | |||
| Hence the answers: | |||
| $$ | |||
| E_{k1} = m_p - m_e \approx 938 \ \text{MeV}, | |||
| $$ | |||
| $$ | |||
| E_{k2} = \frac{m_{\pi^0}}{2} - m_e \approx 67 \ \text{MeV}. | |||
| $$ | |||
| Or, in the more familiar form using masses in kilograms: | |||
| $$ | |||
| E_{k1} = (m_p - m_e)c^2, | |||
| $$ | |||
| $$ | |||
| E_{k2} = \left(\frac{m_{\pi^0}}{2} - m_e\right)c^2. | |||
| $$ | |||
| $Note:$ Savchenko's numerical answer is correct, but, as in many places at the end of the book, there is a misprint in the formula. | |||
| [Insert a concise answer or boxed result] | #### Answer | ||
| $$ | |||
| \boxed{E_{k1} \approx 938 \ \text{MeV}, \qquad E_{k2} \approx 67 \ \text{MeV}.} | |||
| $$ | |||