Question

Calculate the height of the Coulomb barrier for the head-on collision of two deuterons, with effective radius2.1 fm.

Short answer

The height of the Coulomb barrier for the head-on collision is 170KeV.

Step-by-step solution

Write the given data

1. Head-on collision of two deuterons.
2. Effective radius of the barrier, $R=2.1\mathrm{fm}$

Determine the formulas for potential barrier

The potential energy of the two charged system is as f $\mathbf{U}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}\mathbf{r}}$ …… (i)

Here, the distance rbetween the protons when they stop are their center-to-center distance, 2R, and their charges $q_1\mathrm{and}{q}_2$are both e.

Calculate the height of the Coulomb barrier

Now, consider the conservation of energy for the two-deuteron system, determine the total energy using equation (i) as follows:

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

Simplify the equation as:

$$\begin{gathered} \mathrm{k}=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\left(4\mathrm{R}\right)} \\ =\frac{\left(9\times {10}^9{\displaystyle \frac{\mathrm{V}.\mathrm{m}}{\mathrm{C}}}\right){\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2}{4(2.1\times {10}^{-15}\mathrm{m})} \\ =2.74\times {10}^{-14}\mathrm{J} \\ =170\mathrm{KeV} \end{gathered}$$

This kinetic energy is the required potential energy to cross the barrier.

Hence, the potential height of the Coulomb barrier is 170 KeV.