Answer to Question #113171 in Classical Mechanics for Lizwi

Question

A uranium-238 atom can break up into a thorium-234 atom and a particle called an alpha particle, α-4. The numbers indicate the inertias of the atoms and the alpha particle in atomic mass units (1 amu = 1.66 × 10−27 kg). When an uranium atom initially at rest breaks up, the thorium atom is observed to recoil with an x component of velocity of -2.5 × 105 m/s.
How much of the uranium atom's internal energy is released in the breakup?

Answer 1

The initial momentum of the system was zero, the final momentum consists of the momentum of the thorium atom and the momentum of the alpha particle. Thus, according to momentum conservation principle, we can write

$$0=m_Tv_T+m_\alpha v_\alpha,\\ m_Tv_T=-m_\alpha v_\alpha,\\ v_\alpha=-\frac{m_Tv_T}{m_\alpha}.$$

According to the law of conservation of energy, the energy released in the breakup was spent to the increase of the kinetic energy of thorium and alpha-particle:

$$E=E_T+E_\alpha=\frac{1}{2}m_Tv_T^2+\frac{1}{2}m_\alpha v_\alpha^2=\\ \space\\ =\frac{1}{2}m_Tv_T^2+\frac{1}{2}m_\alpha \bigg(-\frac{m_Tv_T}{m_\alpha}\bigg)^2=\\ \space\\ =\frac{1}{2}m_Tv_T^2\bigg[1+\frac{m_T}{m_\alpha}\bigg]=\\ \space\\ =\frac{1}{2}(234\cdot1.66\cdot10^{−27})(-2.5\cdot10^5)^2\bigg(1+\frac{234}{4}\bigg)=\\=7.22\cdot10^{-13}\text{ J}.$$

If we divide this value by the charge of an electron, we will convert the energy into electron-volts:

$$E=4.51\text{ MeV}.$$