If two deuterium nuclei (charge +e, mass 3.34×1027kg) get close enough together,
ID: 1351509 • Letter: I
Question
If two deuterium nuclei (charge +e, mass 3.34×1027kg) get close enough together, the attraction of the strong nuclear force will fuse them to make an isotope of helium, releasing vast amounts of energy. The range of this force is about 1015m. This is the principle behind the fusion reactor. The deuterium nuclei are moving much too fast to be contained by physical walls, so they are confined magnetically.
Part A How fast would two nuclei have to move so that in a head-on collision they would get close enough to fuse? (Treat the nuclei as point charges, and assume that a separation of 1.0×1015m is required for fusion.)
Part B What strength magnetic field is needed to make deuterium nuclei with this speed travel in a circle of diameter 2.30 m ?
Explanation / Answer
a) Find the electrical PE gained by the deuterium pair at the required separation and equate this to the KE the pair had before 'impact'. From KE obtain velocity.
Electrical potential (V) of a point charge at a distance d from the point is ..
V = kQ/r (J/C)
Q = 1.60×10^-19 C
r = 1.0×10^-15 m ( (the required separation)
V = (9×10^9)(1.60×10^-19)/ (1×10^-15) = 1.44×10^6 J/C
so another point charge (e) brought to this position would give the system an elec.PE, E = eV (J)
E = 1.44×10^6 J/C x 1.60×10^-19 C
= 2.30×10^-13 J
this is transferred from the KE's of the two moving nuclei as they collide.
KE possessed by each nucleus = E/2
½ (2.30×10^-13) = ½ mv²
> v = (2.30×10^-13 / 3.34×10^-27)
v = 8.30×10^6 m/s
b) In the B- field centripetal force (mv²/R) is provided by magnetic force (BQv)
mv²/R = BQv
B = mv/RQ (T)
B = (3.34×10^-27)(8.30×10^6) / (2.30/2)(1.60×10^-19)
B = 0.151 T