Question

20．某星球上存在著一種叫做“氦3（$\left.\begin{array}{l}{3}\\{2}\end{array}\right.$He）”的化學(xué)元素，如果可以開發(fā)的話，將為人類帶來取之不盡的能源．“氦3（$\left.\begin{array}{l}{3}\\{2}\end{array}\right.$He）”與氘核（$\left.\begin{array}{l}{2}\\{1}\end{array}\right.$H）通過核反應(yīng)生出氦4（$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He）和質(zhì)子，該核反應(yīng)方程為$\left.\begin{array}{l}{3}\\{2}\end{array}\right.$He+$\left.\begin{array}{l}{2}\\{1}\end{array}\right.$H→$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He+${\;}_{1}^{1}$H，該核反應(yīng)是聚變（填“聚變”或“裂變”）反應(yīng)．

Answer 1

分析 氦3（$\left.\begin{array}{l}{3}\\{2}\end{array}\right.$He）”與氘核（$\left.\begin{array}{l}{2}\\{1}\end{array}\right.$H）通過核反應(yīng)生出氦4（$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He）和質(zhì)子，這是聚變反應(yīng)；由質(zhì)量數(shù)守恒和電荷數(shù)守恒書寫反應(yīng)方程式．

解答 解：氦3（$\left.\begin{array}{l}{3}\\{2}\end{array}\right.$He）”與氘核（$\left.\begin{array}{l}{2}\\{1}\end{array}\right.$H）通過核反應(yīng)生出氦4（$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He）和質(zhì)子，這是聚變反應(yīng)．
由質(zhì)量數(shù)守恒和電荷數(shù)守恒，那么反應(yīng)方程為：$\left.\begin{array}{l}{3}\\{2}\end{array}\right.$He+$\left.\begin{array}{l}{2}\\{1}\end{array}\right.$H→$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He+${\;}_{1}^{1}$H；
故答案為：$\left.\begin{array}{l}{3}\\{2}\end{array}\right.$He+$\left.\begin{array}{l}{2}\\{1}\end{array}\right.$H→$\left.\begin{array}{l}{4}\\{2}\end{array}\right.$He+${\;}_{1}^{1}$H，聚變．

點(diǎn)評 本題考查了核反應(yīng)方程的書寫規(guī)律，以及聚變反應(yīng)，注意與裂變的區(qū)別，屬于簡單基礎(chǔ)題目，平時練習(xí)中對這類問題注意多加訓(xùn)練，不可忽視．