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Contests/Gnit Sunday Open 002 (GSO002)/Problem 9 Magnetic Confinement Fusion: Mass Defect and Particle Motion
Problem 9

Magnetic Confinement Fusion: Mass Defect and Particle Motion

Finished
300 ptsLv.5 ElementaryQuantum
2026/03/22 19:30〜2026/03/22 21:00
Author: admin

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 BBB exists in the positive zzz-direction.

At the coordinate origin (0,0,0)(0,0,0)(0,0,0), a deuterium nucleus (mass mDm_DmD​, charge +e+e+e) and a tritium nucleus (mass mTm_TmT​, charge +e+e+e) that were initially at rest undergo a fusion reaction, producing an alpha particle (mass mαm_{\alpha}mα​, charge +2e+2e+2e) and a neutron (mass mnm_nmn​, charge 000).

In this reaction, the total mass after the reaction is less than before — this mass defect Δm\Delta mΔm 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 xyxyxy-plane, following conservation of momentum. The alpha particle undergoes uniform circular motion in the xyxyxy-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 ccc. 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 ccc — use non-relativistic (classical) mechanics for kinetic energy and momentum calculations.

Find the radius RRR of the uniform circular motion of the alpha particle.

Constraints

  • mD=3.30×10−27 kgm_D = 3.30 \times 10^{-27}\text{ kg}mD​=3.30×10−27 kg
  • mT=4.95×10−27 kgm_T = 4.95 \times 10^{-27}\text{ kg}mT​=4.95×10−27 kg
  • mα=6.40×10−27 kgm_{\alpha} = 6.40 \times 10^{-27}\text{ kg}mα​=6.40×10−27 kg
  • mn=1.60×10−27 kgm_n = 1.60 \times 10^{-27}\text{ kg}mn​=1.60×10−27 kg
  • e=1.60×10−19 Ce = 1.60 \times 10^{-19}\text{ C}e=1.60×10−19 C
  • c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}c=3.0×108 m/s
  • B=0.60 TB = 0.60\text{ T}B=0.60 T

Input Format

Find the radius RRR of the alpha particle's circular motion in meters (m\text{m}m) and give the answer as a positive integer equal to the value multiplied by 100100100. (Example: if R=2.53 mR = 2.53\text{ m}R=2.53 m, answer 253253253.)

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