(1)設(shè)某速度為
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的粒子從A點(diǎn)入射后到達(dá)AC上的G點(diǎn),因
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與AC成45
0角,其對(duì)應(yīng)圓心角為90
0,即恰好經(jīng)過(guò)四分之一圓周,故到達(dá)G點(diǎn)時(shí)速度仍為
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,方向沿Y軸正向. ----------------------------------------------------------------------(2分)
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粒子在電場(chǎng)中沿Y軸正向加速運(yùn)動(dòng),設(shè)G點(diǎn)坐標(biāo)為G(x,y),剛好穿出電場(chǎng)時(shí)坐標(biāo)為(x,
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),粒子穿出電場(chǎng)時(shí)速度為
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,在電場(chǎng)中運(yùn)動(dòng)的過(guò)程中,由動(dòng)能定理得:
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------------------------------------------ (2分)
而
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又
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-----------------------------------------------(2分)
代入數(shù)據(jù)解得
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,可見(jiàn)粒子穿出電場(chǎng)時(shí)速度大小與x無(wú)關(guān)。---------(3分)
因
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,由
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代入數(shù)據(jù)得:
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----------------------------------------------------------------------(1分)
由數(shù)學(xué)知識(shí)可知,k點(diǎn)坐標(biāo)為k(-0.2m,-0.225m),故從A點(diǎn)射出的所有粒子均從AK之間以20m/s的速度沿Y軸正向射出電場(chǎng),在到達(dá)X軸之前粒子作勻速直線運(yùn)動(dòng),故所有粒子從第三象限穿越X軸時(shí)的速度大小均為20m/s的速度沿Y軸正向。---------------------------------------------------(1分)
(2)因?yàn)?img src="http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824122943485408.gif" style="vertical-align:middle;" />,故離子束射入
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時(shí),離子束寬度剛好與2
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相等,設(shè)粒子在
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中運(yùn)動(dòng)軌道半徑為
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,解得
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=
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---------------------------------------------------(1分)
考察從任一點(diǎn)J進(jìn)入
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的粒子,設(shè)從H穿出
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磁場(chǎng),四邊形
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為菱形,又因?yàn)?img src="http://thumb.zyjl.cn/pic2/upload/papers/20140824/20140824122943688240.gif" style="vertical-align:middle;" /> 水平,而
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,故H應(yīng)與F重合,即所有粒子經(jīng)過(guò)
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后全部從F點(diǎn)離開(kāi)
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進(jìn)入
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磁場(chǎng)。對(duì)
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趨于
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的粒子,圓心角∠
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,故射入
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時(shí)速度趨于Y軸負(fù)向;對(duì)
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趨于0的粒子,圓心角∠
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,故射入
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時(shí)速度趨于Y軸正向,即進(jìn)入
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的所有粒子速度與Y軸正向夾角在0~180
0之間。-- ----------------------(4分)
由于
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=
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,所以
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,由幾何關(guān)系知:
無(wú)限靠近Y軸負(fù)向射入的粒子軌跡如圖所示,最終打在PQ板的右側(cè)
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; ---(1分)
與Y軸負(fù)向成60
0角的粒子剛好經(jīng)過(guò)P點(diǎn)到達(dá)Q點(diǎn);-------- ----------- -(1分)
因此與Y軸正向在0~120
0之間從F點(diǎn)射出的粒子要么打在PQ板的左側(cè),要么打不到板上而穿越Y(jié)軸離開(kāi)
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。由于是“大量”粒子,忽略打在P或Q的臨界情況,所以最終打在擋板PQ右側(cè)的粒子數(shù)
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------------------------------ -----------------------------------------(3分)