题目内容
一个静止的氡核
Rn,放出一个α粒子后衰变为钋核
Po,同时放出能量为E=0.26 MeV的光子.假设放出的核能完全转变为钋核与α粒子的动能,不计光子的动量.已知M氡=222.08663 u、mα=4.0026 u、M钋=218.0766 u,1 u相当于931.5 MeV的能量.
(1)写出上述核反应方程;
(2)求出发生上述核反应放出的能量;
(3)确定钋核与α粒子的动能.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141321671390.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141321734404.png)
(1)写出上述核反应方程;
(2)求出发生上述核反应放出的能量;
(3)确定钋核与α粒子的动能.
见解析
(1)
Rn→
Po+
He+γ
(2)质量亏损Δm=222.08663 u-4.0026 u-218.0766 u=0.00743 u
ΔE=Δmc2≈6.92 MeV
(3)设α粒子、钋核的动能分别为Ekα、Ek钋,动量分别为pα、p钋,由能量守恒定律得:ΔE=Ekα+Ek钋+E
不计光子的动量,由动量守恒定律得:
0=pα+p钋
又Ek=![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141321983577.png)
故Ekα∶Ek钋=218∶4
联立上面式子解得
Ek钋=0.21 MeV,Ekα=6.54 MeV.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141321671390.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141321734404.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141321937292.png)
(2)质量亏损Δm=222.08663 u-4.0026 u-218.0766 u=0.00743 u
ΔE=Δmc2≈6.92 MeV
(3)设α粒子、钋核的动能分别为Ekα、Ek钋,动量分别为pα、p钋,由能量守恒定律得:ΔE=Ekα+Ek钋+E
不计光子的动量,由动量守恒定律得:
0=pα+p钋
又Ek=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141321983577.png)
故Ekα∶Ek钋=218∶4
联立上面式子解得
Ek钋=0.21 MeV,Ekα=6.54 MeV.
![](http://thumb2018.1010pic.com/images/loading.gif)
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