A. \(14,{{85.10}^{6}}m/s\).
B. \(18,{{49.10}^{6}}\text{ }m/s\).
C. \(37,{{96.10}^{6}}m/s\).
D. \(16,{{93.10}^{6}}m/s\).
D
Đáp án D
\(Q=\left( {{m}_{P}}+{{m}_{Li}}-2{{m}_{\alpha }} \right){{c}^{2}}=\left( 1,0073+7,0144-2.4,0015 \right).931=17,41MeV\)
Ta có:
\(\left\{ \begin{align} & {{K}_{P}}+Q=2{{K}_{\alpha }} \\ & {{K}_{\alpha }}=\frac{1}{2}{{m}_{\alpha }}v_{\alpha }^{2} \\ \end{align} \right.\Leftrightarrow \frac{1}{2}{{m}_{P}}v_{P}^{2}+Q=2.\frac{1}{2}{{m}_{\alpha }}v_{\alpha }^{2}\)
\(\Leftrightarrow 0,5.1,0073.931{{\left( \frac{{{v}_{P}}}{{{3.10}^{8}}} \right)}^{2}}+17,41=4,0015.931{{\left( \frac{21,{{37.10}^{6}}}{{{3.10}^{8}}} \right)}^{2}}\Rightarrow {{v}_{P}}=16,{{93.10}^{6}}m/s\)
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