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Two particle accelerators were used to accelerate two beams of deuterons directly towards each other to have a head-on collision.

A particle accelerator is used to speed up charged particles and channel them into a beam. For the condition of a head-on collision with the target nuclei, the accelerator can be used to direct the used particle to reach the target material.

Formula:

The potential energy of the two charged system is as follows:

$U=\frac{q_1{q}_2}{4{\mathrm{蟺蔚}}_0\mathrm{r}}$ 鈥︹�� (i)

The calculation is identical to that in Sample Problem 鈥� 鈥淔usion in a gas of protons and required temperature鈥� except that using Rappropriate to two deuterons coming into 鈥渃ontact,鈥� as opposed to the R = 1.0 fm value used in the Sample Problem.Here, the distance rbetween the protons when they stop are their center-to-center distance, 2R, and their charges$q_1$and$q_2$are both e.Thus, the total potential barrier to be overcome by each accelerator for the two-deuteron system to come into contact considering the energy conservation concept is given using equation (i) as:

$$\begin{gathered} 2K_{p+p}=U \\ =\frac{e^2}{4{\mathrm{蟺蔚}}_0\left(2R\right)} \end{gathered}$$

Substitute the value and solve as:

$K_{p+p}=\frac{e^2}{4{\mathrm{蟺蔚}}_0\left(4R\right)}$

Substitute the values and solve as:

$$\begin{gathered} 2K_{p+p}=\frac{\left(9脳{10}^9{\displaystyle \frac{\mathrm{V}.\mathrm{m}}{\mathrm{C}}}\right){\left(1.6脳{10}^{-19}\mathrm{C}\right)}^2}{4(1.0脳{10}^{-15}\mathrm{m})} \\ =5.75脳{10}^{-14}\mathrm{J} \\ =360\mathrm{KeV} \end{gathered}$$

If the radius is considered to be R = 2.1, fm then the above barrier height with differ by 2.1 term to the required voltage value (or above energy value) required by the accelerator becomes the following considering equation (i) for$K=U伪\frac{1}{r}$:

$K_{d+d}=\frac{{\mathrm{K}}_{\mathrm{d}+\mathrm{d}}}{2.1}$

Substitute the values and solve as:

$$\begin{gathered} K_{d+d}=\frac{360\mathrm{KeV}}{2.1} \\ =170\mathrm{KeV} \end{gathered}$$

Consequently, the voltage needed to accelerate each deuteron from rest to that value of Kis 170 KeV.

Not all deuterons that are accelerated toward each other will come into 鈥渃ontact鈥� 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 鈥渞eactor,鈥� but those factors may become less important in the future 鈥� making the feasibility of accelerator 鈥渁dd-ons鈥� to magnetic and inertial confinement schemes more cost-effective.