質(zhì)子（
 
1
1
H）和α （
 
4
2
He）粒子的電荷量之比是1：2，質(zhì)量之比是1：4，以相同的速率在同一勻強磁場中做勻速圓周運動，軌道半徑分別為RH和RHe，周期分別為TH和THe，則（　�。�
A、RH：RHe=1：2，TH：THe=1：1
B、RH：RHe=1：1，TH：THe=1：1
C、RH：RHe=1：1，TH：THe=1：2
D、RH：RHe=1：2，TH：THe=1：2
分析：質(zhì)子H和α粒子以相同的速率在同一勻強磁場中作勻速圓周運動，均由洛侖茲力提供向心力，由牛頓第二定律和圓周運動的規(guī)律，可求得v及T的表達式，根據(jù)表達式可以得到半徑以及周期之比．
解答：解：質(zhì)子p和α粒子在勻強磁場中作勻速圓周運動，均由洛侖茲力提供向心力，由牛頓第二定律得：
 qvB=m
v2
R
，得軌道半徑：R=
mv
qB

則運動周期為：T=
2πR
v
=
2πm
qB

根據(jù)質(zhì)子（
 
1
1
H）和α （
 
4
2
He）粒子的電荷量之比是1：2，質(zhì)量之比是1：4，
則得：RH：RHe=1：2，TH：THe=1：2．
故選：D
點評：本題主要考查了帶電粒子在磁場場中運動的問題，知道洛倫茲力充當(dāng)向心力，熟練掌握圓周運動的基本公式即可．
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