10.1.6. Using a cloud chamber placed in a magnetic field with induction $B$, the elastic scattering of $\alpha$-particles on deuterium nuclei is observed. Find the initial energy of the $\alpha$-particle if the radius of curvature of the initial sections of the trajectories of the nucleus and the $\alpha$-particle after scattering turned out to be equal to $R$. Both trajectories lie in a plane perpendicular to the magnetic field induction.
Solution
After the collision, the particles move along circular trajectories in a plane perpendicular to the magnetic field $B$ under the action of the Lorentz force. Let us write Newton's second law: $$qvB = m\frac{v^2}{r}$$
From here, the trajectory radius is: $$r = \frac{mv}{qB}$$
The magnitude of the particle's momentum $p = mv$ can be expressed from the last formula: $$p = qBr$$
In an elastic scattering on a stationary deuterium nucleus, both momentum and energy are conserved. Let us write down the masses and charges of the $\alpha$-particle and the deuterium nucleus. An $\alpha$-particle consists of two protons and two neutrons. Since the masses of a proton and a neutron are practically the same, the mass of the $\alpha$-particle is equal to four proton masses, and its charge is equal to double the elementary charge: $m_\alpha = 4m_p$ and $q_\alpha = 2e$.
The deuterium nucleus consists of one proton and one neutron, so for it we get: $m_D = 2m_p$ and $q_D = e$.
According to the problem statement, after scattering, both particles have the same trajectory radius $R$ in the magnetic field, so the momentum of the particles will be: $$p_\alpha = q_\alpha BR = 2eBR$$ $$p_D = q_D BR = eBR$$
Let us write the law of conservation of energy: $$K = \frac{p_\alpha^2}{2m_\alpha} + \frac{p_D^2}{2m_D}$$