题目内容
用加速后动能为EK0的质子![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_ST/1.png)
【答案】分析:根据电荷数守恒质量数守恒写出核反应方程.根据能量守恒求出释放的核能,通过爱因斯坦质能方程求出质量亏损.
解答:解:根据电荷数守恒、质量数守恒有:
→
.
释放的核能△E=2Ek+hv-Ek0.
根据爱因斯坦质能方程得,△E=△mc2,
解得
.
答:核反应方程为
→
,核反应中的质量亏损为
.
点评:解决本题的关键知道在核反应过程中电荷数守恒、质量数守恒,以及掌握爱因斯坦质能方程.
解答:解:根据电荷数守恒、质量数守恒有:
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_DA/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_DA/1.png)
释放的核能△E=2Ek+hv-Ek0.
根据爱因斯坦质能方程得,△E=△mc2,
解得
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_DA/2.png)
答:核反应方程为
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_DA/3.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_DA/4.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028203419780683857/SYS201310282034197806838008_DA/5.png)
点评:解决本题的关键知道在核反应过程中电荷数守恒、质量数守恒,以及掌握爱因斯坦质能方程.
![](http://thumb2018.1010pic.com/images/loading.gif)
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