For overcoming the Coulomb barrier for fusion, methods other than heating the fusible material have been suggested. For example, if you were to use two particle accelerators to accelerate two beams of deuterons directly toward each other so as to collide head-on, (a) what voltage would each accelerator require in order for the colliding deuterons to overcome the Coulomb barrier? (b) Why do you suppose this method is not presently used?

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:

$U=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0\mathrm{r}}$ …… (i)

The calculation is identical to that in Sample Problem — “Fusion in a gas of protons and required temperature” except that using Rappropriate to two deuterons coming into “contact,” 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:

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

Substitute the value and solve as:

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

Substitute the values and solve as:

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

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$K=U\alpha \frac{1}{r}$:

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

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 “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.