Người ta dùng một proton bắn phá hạt nhân X

* Hướng dẫn giải

Ta có:

\(\begin{array}{l}
\overrightarrow {{p_p}}  = \overrightarrow {{p_{\alpha 1}}}  + \overrightarrow {{p_{\alpha 2}}}  \Rightarrow {m_p}{v_p} = 2{m_\alpha }{v_\alpha }\cos \left( {\frac{\beta }{2}} \right)\\
 \Rightarrow \cos \left( {\frac{\beta }{2}} \right) = \frac{1}{2}\left( {\frac{{{m_p}}}{{{m_\alpha }}}} \right)\left( {\frac{{{v_p}}}{{{v_\alpha }}}} \right)(1)\\
\Delta E = 2{K_\alpha } - {K_p} > 0 \Rightarrow 2\left( {\frac{1}{2}{m_\alpha }v_\alpha ^2} \right) > \frac{1}{2}{m_p}v_p^2 \Rightarrow \frac{{{v_p}}}{{{v_\alpha }}} < \sqrt {\frac{{2{m_\alpha }}}{{{m_p}}}} (2)\\
(1),(2) \Rightarrow \cos \left( {\frac{\beta }{2}} \right) < \frac{1}{2}\left( {\frac{{{m_p}}}{{{m_\alpha }}}} \right)\sqrt {\frac{{2{m_\alpha }}}{{{m_p}}}}  = \frac{1}{2}\left( {\frac{1}{4}} \right)\sqrt {\frac{{2.\left( 4 \right)}}{1}}  \approx 0,35 \Rightarrow \beta  > {139^0}
\end{array}\)