Magnetic Confinement Fusion: Mass Defect and Particle Motion

Problem Statement

Deuterium-tritium (D-T) fusion is expected as a next-generation energy source. In this problem, we consider a single reaction model inside a magnetic confinement fusion reactor.

In a vacuum coordinate space, a uniform magnetic field of magnetic flux density $B$B exists in the positive $z$z-direction.

At the coordinate origin $(0,0,0)$(0,0,0), a deuterium nucleus (mass $m_D$mD​, charge $+e$+e) and a tritium nucleus (mass $m_T$mT​, charge $+e$+e) that were initially at rest undergo a fusion reaction, producing an alpha particle (mass $m_{\alpha}$mα​, charge $+2e$+2e) and a neutron (mass $m_n$mn​, charge $0$0).

In this reaction, the total mass after the reaction is less than before — this mass defect $\Delta m$Δm is completely converted into the kinetic energy of the products (alpha particle and neutron) according to Einstein's mass-energy equivalence.

The alpha particle and neutron are emitted in opposite directions in the $xy$xy-plane, following conservation of momentum. The alpha particle undergoes uniform circular motion in the $xy$xy-plane due to the Lorentz force from the magnetic field. The neutron, having no charge, is unaffected by the magnetic field and travels in a straight line toward the reactor wall.

The speed of light in vacuum is $c$c. The kinetic energy of the reactants before the reaction is negligible compared to the released energy; treat particles as starting completely at rest.

Also, the speeds of the products are sufficiently small compared to $c$c — use non-relativistic (classical) mechanics for kinetic energy and momentum calculations.

Find the radius $R$R of the uniform circular motion of the alpha particle.

Constraints

• $m_D = 3.30 \times 10^{-27}\text{ kg}$mD​=3.30×10−27 kg
• $m_T = 4.95 \times 10^{-27}\text{ kg}$mT​=4.95×10−27 kg
• $m_{\alpha} = 6.40 \times 10^{-27}\text{ kg}$mα​=6.40×10−27 kg
• $m_n = 1.60 \times 10^{-27}\text{ kg}$mn​=1.60×10−27 kg
• $e = 1.60 \times 10^{-19}\text{ C}$e=1.60×10−19 C
• $c = 3.0 \times 10^8\text{ m/s}$c=3.0×108 m/s
• $B = 0.60\text{ T}$B=0.60 T

Input Format

Find the radius $R$R of the alpha particle's circular motion in meters ($\text{m}$m) and give the answer as a positive integer equal to the value multiplied by $100$100. (Example: if $R = 2.53\text{ m}$R=2.53 m, answer $253$253.)