p44_035

p44_035 - encountering other deuterons in order to produce...

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35. (a) Our calculation is identical to that in Sample Problem 44-4 except that we are now using R ap- propriate to two deuterons coming into “contact,” as opposed to the R =1 . 0 fm value used in the Sample Problem. If we use R =2 . 1 fm for the deuterons (this is the value given in problem 33), then our K is simply the K calculated in Sample Problem 44-4, divided by 2 . 1: K d + d = K p + p 2 . 1 = 360 keV 2 . 1 170 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
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Unformatted text preview: 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 ux 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 conFnement schemes more cost-eective....
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