Consider the deuterium–tritium fusion reaction with the tritium nucleus at rest:
2/1H + 3/1H → 4/2He + 1/0n
(a) Suppose the reactant nuclei will spontaneously fuse if their surfaces touch. From Equation 43.1, determine the required distance of closest approach between their centers. (b) What is the electric potential energy (in electron volts) at this distance? (c) Suppose the deuteron is fired straight
at an originally stationary tritium nucleus with just enough energy to reach the required distance of closest approach. What is the common speed of the deuterium and tritium nuclei, in terms of the initial deuteron speed υ_i, as they touch? (d) Use energy methods to find the minimum initial deuteron energy required to achieve fusion. (e) Why does the fusion reaction actually occur at much lower deuteron energies than the energy calculated in part (d)?

Question

Accepted Answer

(a) distance of closest approach

r= aA!/3 (43.1)