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3D【解析】粒子在加速电场中加速,由动能定理得q=2mv2=0,粒子在磁场中做匀速团周运动,洛伦兹力提供向心力,由牛顿第二定律得q0B=m,解得U=-,故为使a粒子9He)经同一匀强磁场偏转后仍从同一出口离开磁场,需将加速电压变为原来的方,故D项正确A、BC项错误

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