题目内容
1998年6月2日,我国科学家研究的阿尔法磁谱仪由“发现号”航天飞机搭载升空,用于探测宇宙中是否有反物质和暗物质,所谓反物质的原子(反原子)是由带负电的反原子核和核外的正电子组成,反原子核由反质子和反中子组成.与![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/0.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/1.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/2.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/3.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/4.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/5.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/6.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/7.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/8.png)
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_ST/images9.png)
【答案】分析:根据左手定则,及运动的方向,可画出运动轨迹图;根据牛顿第二定律,由洛伦兹力提供向心力,并由几何关系,即可求解.
解答:解:轨迹如图![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/images0.png)
设质子的质量为m,电荷量为q,
据题意有:
qvB=
(r为质子的半径)
解得:
=2x
又由于反氢核(反质子)质量为m,电荷量为-q,
故其轨道半径r1=r,但偏转方向相反.即x1=-x
由于反氦核(反α粒子)的质量为4m,电量为-2q,
同理,轨道半径r2=
=2x
由图中几何关系可知:
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/3.png)
答:
x1=-x
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/4.png)
点评:考查左手定则与牛顿第二定律的应用,学会画出运动轨迹图,注意几何关系的运用.
解答:解:轨迹如图
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/images0.png)
设质子的质量为m,电荷量为q,
据题意有:
qvB=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/0.png)
解得:
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/1.png)
又由于反氢核(反质子)质量为m,电荷量为-q,
故其轨道半径r1=r,但偏转方向相反.即x1=-x
由于反氦核(反α粒子)的质量为4m,电量为-2q,
同理,轨道半径r2=
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/2.png)
由图中几何关系可知:
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/3.png)
答:
x1=-x
![](http://thumb.1010pic.com/pic6/res/gzwl/web/STSource/20131028204511060856617/SYS201310282045110608566016_DA/4.png)
点评:考查左手定则与牛顿第二定律的应用,学会画出运动轨迹图,注意几何关系的运用.
![](http://thumb2018.1010pic.com/images/loading.gif)
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