1.让氦核(4 2 He) ,氚核(2 1 H) 的混合物沿着与电场垂直的方向进入匀强电场偏转,要使它们最后偏转角相同,这些粒子必须有相同的( )A.B.动能 C.电量 D.质量老师讲过速度tan的公式 是 tan=UQ

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1.让氦核(42He),氚核(21H)的混合物沿着与电场垂直的方向进入匀强电场偏转,要使它们最后偏转角相同,这些粒子必须有相同的()A.B.动能C.电量D.质量老师讲过速度tan的公式是tan=UQ1

1.让氦核(4 2 He) ,氚核(2 1 H) 的混合物沿着与电场垂直的方向进入匀强电场偏转,要使它们最后偏转角相同,这些粒子必须有相同的( )A.B.动能 C.电量 D.质量老师讲过速度tan的公式 是 tan=UQ
1.让氦核(4 2 He) ,氚核(2 1 H) 的混合物沿着与电场垂直的方向进入匀强电场偏转,要使它们最后偏转角相同,这些粒子必须有相同的( )
A.B.动能 C.电量 D.质量
老师讲过速度tan的公式 是 tan=UQL/mdv2 U是电场的电势差,Q是粒子的电荷量,m是粒子的质量,d是极板的宽度,L是极板的长度,偏转相同就是tan相同,U L d都是固定不边的
2.如果速度和力不在一条直线上,轨迹是趋向速度还是力?

1.让氦核(4 2 He) ,氚核(2 1 H) 的混合物沿着与电场垂直的方向进入匀强电场偏转,要使它们最后偏转角相同,这些粒子必须有相同的( )A.B.动能 C.电量 D.质量老师讲过速度tan的公式 是 tan=UQ
1、你可能打错字了,氚核(2 1 H)是错的,只能是氘核(2 1 H),氚核(3 1 H).
既然你打错字,答案我就不说了,可告诉你想要的分析过程.
可以象你们老师那样推导,也可象我这样推导(因匀强电场E=U / d):设板长为L,电场强度是E,电荷电量是q,进入电场时的初速是 V0,则用类平抛运动知识得
电荷通过两板间的电场所用时间是 t =L / V0 (在平行极板方向是匀速的)
在垂直极板方向是初速为0的匀加速直线运动,
该方向的末速度是 Vy=a*t =(qE / m)*(L / V0)=qEL / (m*V0)
所以偏转角θ的正切值是 tanθ=Vy / V0=qEL / (m*V0^2)
可见,E、L相同的,要使它们离开电场时的偏转角相同,必须要满足 q / (m*V0^2)相同.
在本题中,如果原题是给氦核(4 2 He) ,氘核(2 1 H),因它们的荷质比相等,所以答案是A对.
2、如果速度和力不在一条直线上,物体是做曲线运动.
当这个力(合力)是恒力时,轨迹是趋近力.

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