（2009•南通二模）一个静止的质量为M的放射性原子核发生衰变，放出一个质量为m、速度大小为v的α粒子，则衰变后新原子核速度大小为mvM−mmvM−m；设衰变过程中释放的核能全部转化为新

Question

（2009•南通二模）一个静止的质量为M的放射性原子核发生衰变，放出一个质量为m、速度大小为v的α粒子，则衰变后新原子核速度大小为

(mv)/(M−m)

(mv)/(M−m)

；设衰变过程中释放的核能全部转化为新原子核和α粒子的动能，真空中光速为c，则衰变过程中质量亏损为

(Mmv2)/(2(M−m)c2)

(Mmv2)/(2(M−m)c2)

．

(mv)/(M−m)

(mv)/(M−m)(mv)/(M−m)mvmvM−mM−m

(mv)/(M−m)

(mv)/(M−m)(mv)/(M−m)mvmvM−mM−m

(Mmv2)/(2(M−m)c2)

(Mmv2)/(2(M−m)c2)(Mmv2)/(2(M−m)c2)Mmv2Mmv2v2v222(M−m)c22(M−m)c2c2c22

(Mmv2)/(2(M−m)c2)

(Mmv2)/(2(M−m)c2)(Mmv2)/(2(M−m)c2)Mmv2Mmv2v2v222(M−m)c22(M−m)c2c2c22

Answer 1

根据动量守恒定律得，0=mv-（M-m）v′，解得v′＝(mv)/(M−m)．
△E＝(1)/(2)mv2+(1)/(2)(M−m)v′2=(Mmv2)/(2(M−m))
根据爱因斯坦质能方程得，△m＝(△E)/(c2)＝(Mmv2)/(2(M−m)c2)．
故答案为：(mv)/(M−m)   (Mmv2)/(2(M−m)c2) v′＝(mv)/(M−m)mvmvmvM−mM−mM−m．
△E＝(1)/(2)mv2+(1)/(2)(M−m)v′2=(Mmv2)/(2(M−m))
根据爱因斯坦质能方程得，△m＝(△E)/(c2)＝(Mmv2)/(2(M−m)c2)．
故答案为：(mv)/(M−m)   (Mmv2)/(2(M−m)c2) △E＝(1)/(2)111222mv2+(1)/(2)(M−m)v′2=(Mmv2)/(2(M−m))
根据爱因斯坦质能方程得，△m＝(△E)/(c2)＝(Mmv2)/(2(M−m)c2)．
故答案为：(mv)/(M−m)   (Mmv2)/(2(M−m)c2) 2+(1)/(2)111222(M−m)v′2=(Mmv2)/(2(M−m))
根据爱因斯坦质能方程得，△m＝(△E)/(c2)＝(Mmv2)/(2(M−m)c2)．
故答案为：(mv)/(M−m)   (Mmv2)/(2(M−m)c2) 2=(Mmv2)/(2(M−m))
根据爱因斯坦质能方程得，△m＝(△E)/(c2)＝(Mmv2)/(2(M−m)c2)．
故答案为：(mv)/(M−m)   (Mmv2)/(2(M−m)c2) (Mmv2)/(2(M−m))Mmv2Mmv2Mmv222(M−m)2(M−m)2(M−m)
根据爱因斯坦质能方程得，△m＝(△E)/(c2)＝(Mmv2)/(2(M−m)c2)．
故答案为：(mv)/(M−m)   (Mmv2)/(2(M−m)c2) △m＝(△E)/(c2)△E△E△Ec2c2c22＝(Mmv2)/(2(M−m)c2)Mmv2Mmv2Mmv222(M−m)c22(M−m)c22(M−m)c22．
故答案为：(mv)/(M−m)   (Mmv2)/(2(M−m)c2) (mv)/(M−m)mvmvmvM−mM−mM−m   (Mmv2)/(2(M−m)c2) (Mmv2)/(2(M−m)c2)Mmv2Mmv2Mmv222(M−m)c22(M−m)c22(M−m)c22