Question

1．一靜止的${\;}_{92}^{238}$U核經(jīng)α衰變成為${\;}_{90}^{234}$Th核，釋放出的總動(dòng)能為4.27MeV．問(wèn)此衰變后${\;}_{90}^{234}$Th核的動(dòng)能為多少M(fèi)eV（保留1位有效數(shù)字）？

Answer 1

分析 選擇正確的研究對(duì)象．根據(jù)動(dòng)量守恒定律列出等式解決問(wèn)題，根據(jù)能量守恒列出等式求解問(wèn)題．

解答 解：據(jù)題意知，此α衰變的方程為${\;}_{92}^{238}U→{\;}_{90}^{234}Th+{\;}_2^4He$
根據(jù)動(dòng)量守恒定律得：m_αv_α=m_Thv_Th …①
式中，m_α和m_Th分別為α粒子和Th核的質(zhì)量，v_α和v_Th分別為α粒子和Th核的速度的大�。�由題設(shè)條件知：$\frac{1}{2}{m_a}v_a^2+\frac{1}{2}{m_{Th}}v_{Th}^2={E_k}$…②
$\frac{m_a}{{{m_{Th}}}}=\frac{4}{234}$…③
式中E_k=4.27MeV是α粒子與Th的總動(dòng)能．
由①②③式得：$\frac{1}{2}{m_{Th}}v_{Th}^2=\frac{m_a}{{{m_a}+{m_{Th}}}}{E_k}$…④
代入數(shù)據(jù)得，衰變后${\;}_{90}^{234}Th$核的動(dòng)能為：$\frac{1}{2}{m_{Th}}v_{Th}^2=0.07MeV$…⑤
答：衰變后${\;}_{90}^{234}Th$核的動(dòng)能為0.07MeV．

點(diǎn)評(píng) 解決問(wèn)題首先要清楚研究對(duì)象的運(yùn)動(dòng)過(guò)程．我們要清楚運(yùn)動(dòng)過(guò)程中能量的轉(zhuǎn)化，以便從能量守恒角度解決問(wèn)題．把動(dòng)量守恒和能量守恒結(jié)合起來(lái)列出等式求解是常見的問(wèn)題．