(a) Our calculation is identical to that in Sample Problem — “Fusion in a gas of protons and required temperature” except that we are now using R appropriate to two deuterons coming into “contact,” as opposed to the R=1.0fm value used in the Sample Problem. If we use R=2.1fm for the deuterons, then our K is simply the K calculated in the Sample Problem, divided by 2.1:
kd+d=kp+p2.1=360keV2.1≈170keV
Consequently, the voltage needed to accelerate each deuteron from rest to that value of K is 170 kV.
(b) Not all deuterons that are accelerated toward each other will come into “contact” and not all of those that do so will undergo nuclear fusion. Thus, a great many deuterons must be repeatedly encountering other deuterons in order to produce a macroscopic energy release. An accelerator needs a fairly good vacuum in its beam pipe, and a very large number flux is either impractical and/or very expensive. Regarding expense, there are other factors that have dissuaded researchers from using accelerators to build a controlled fusion “reactor,” but those factors may become less important in the future — making the feasibility of accelerator “add-ons” to magnetic and inertial confinement schemes more cost-effective.