Answer to Question #84408 in Mechanics | Relativity for Ra

Question

In a nuclear reactor one deuterium atom(D) and one triton atom(T) at rest decay into one alpha particle(He) and one neutron(n). Obtain an expression for the momentum of the alpha particle and the neutron and the value of the kinetic energy of the alpha particle given that the values of the rest masses are: m_D=2.01474amu; m_T=3.017amu; m_He=4.00387amu and 1 amu = 931 MeV.

Answer 1

$$\text{D + T} \to _2^4\text{He + n}^0$$ $$\Delta m=(m_D+m_T-m_{He}-m_n)=$$ $$=(2.01474+3.017-4.00387-1.0086648)=0.0192052\text{ amu}$$

or 17.88 MeV.

$$T_\alpha=\frac{Q}{1+m_\alpha/m_n}=\frac{17.88}{1+4.00387/1.0086648}=3.59 \text{ MeV}.$$

Momenta:

$$p_\alpha=\sqrt{2m_\alpha T_\alpha},$$ $$p_\text{n}=-\sqrt{2m_n T_n}.$$