GSO002 Problem 9
Problem Statement
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 exists in the positive z-direction.
At the coordinate origin (0,0,0), a deuterium nucleus (mass mD, charge +e) and a tritium nucleus (mass mT, charge +e) that were initially at rest undergo a fusion reaction, producing an alpha particle (mass mα, charge +2e) and a neutron (mass mn, charge 0).
In this reaction, the total mass after the reaction is less than before — this mass defect 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-plane, following conservation of momentum. The alpha particle undergoes uniform circular motion in the 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. 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 — use non-relativistic (classical) mechanics for kinetic energy and momentum calculations.
Find the radius R of the uniform circular motion of the alpha particle.
Constraints
- mD=3.30×10−27 kg
- mT=4.95×10−27 kg
- mα=6.40×10−27 kg
- mn=1.60×10−27 kg
- e=1.60×10−19 C
- c=3.0×108 m/s
- B=0.60 T
Input Format
Find the radius R of the alpha particle's circular motion in meters (m) and give the answer as a positive integer equal to the value multiplied by 100. (Example: if R=2.53 m, answer 253.)
Solution
1. Mass Defect and Released Energy
Δm=(mD+mT)−(mα+mn)=(3.30+4.95)×10−27−(6.40+1.60)×10−27=8.25×10−27−8.00×10−27=0.25×10−27 kgThe total kinetic energy released:
Q=Δmc2=(0.25×10−27)×(3.0×108)2=2.25×10−11 J2. Momentum Conservation and Energy Distribution
Since the reactants are at rest, the total momentum is 0. The momenta of the alpha particle and neutron are equal in magnitude and opposite in direction; let p=∣pα∣=∣pn∣.
Using non-relativistic mechanics (K=2mp2):
2mαp2+2mnp2=Q p2=mα+mn2mαmnQNumerically:
- mα+mn=8.00×10−27 kg
- mαmn=10.24×10−54 kg2
3. Radius of the Alpha Particle's Circular Motion
The alpha particle (charge q=+2e) undergoes circular motion in the xy-plane with the Lorentz force as centripetal force:
mαRv2=qvB R=qBmαv=2eBpSubstituting:
R=2×(1.60×10−19)×0.602.40×10−19=1.922.40=1.25 mMultiplied by 100: 125.
Final Answer: 125