If two deuterium nuclei (charge +e, mass 3.34*10 -27 kg) get close enough togeth

Question

If two deuterium nuclei (charge +e, mass 3.34*10-27kg) get close enough together, the attraction of the strong nuclearforce will fuse them to make an isotope of helium, releasing vastamounts of energy. The range of this force is about10-15 is the principle behind the fusion reactor. Thedeuterium nuclei are moving much too fast to be contained byphysical walls, so they are confined magnetically. a) How fast would two nuclei have to move sothat in a head-on collision they would get close enough to fuse?(Treat the nuclei as point charges, and assume that a separationof 1*10-15 is required for fusion.) If two deuterium nuclei (charge +e, mass 3.34*10-27kg) get close enough together, the attraction of the strong nuclearforce will fuse them to make an isotope of helium, releasing vastamounts of energy. The range of this force is about10-15 is the principle behind the fusion reactor. Thedeuterium nuclei are moving much too fast to be contained byphysical walls, so they are confined magnetically. a) How fast would two nuclei have to move sothat in a head-on collision they would get close enough to fuse?(Treat the nuclei as point charges, and assume that a separationof 1*10-15 is required for fusion.) How fast would two nuclei have to move sothat in a head-on collision they would get close enough to fuse?(Treat the nuclei as point charges, and assume that a separationof 1*10-15 is required for fusion.)

Explanation / Answer

Assume two deuterium nuclei come in from infinity so thatinitial electrostatic potential is zero and all energy iskinetic. Conservation of Energy: Ei = Ef KEi + Ui = KEf +Uf KEi + 0 = 0 + Uf (1/2)mv2 = kq2/r v2 = 2ke2/mr = 2ke2/mr =2(8.99x109N*m2C-2)(1.60x10-19C)2/[(3.34x10-27kg)(10-15m)] v2 = 13.8 x 109-38+27+15m2/s2 = 1.38 x 1014m2/s2 v = 1.17 x 107 m/s = 11.7Mm/s v = 1.17 x 107 m/s = 11.7Mm/s

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