ch43-p034 - 34. (a) Our calculation is identical to that in...

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K K dd pp + + == 21 360 170 .. keV keV. 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 towards 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
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