Q31P - Fundamentals Of Physics

Question

Calculate the height of the Coulomb barrier for the head-on collision of two deuterons, with effective radius 2.1 fm.

Step-by-Step Solution

Answer

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

Step 1: Write the given data

a)   Head-on collision of two deuterons.

b)   Effective radius of the barrier, R = 2.1 fm

Step 2: Determine the formulas for potential barrier

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

$\mathbf{U}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{4}{\mathbf{\tau \tau \epsilon }}_{\mathbf{0}}\mathbf{r}}$                                                                  …… (i)

 

Here, the distance r between the protons when they stop are their center-to-center distance, 2R, and their charges $q_1 \mathrm{and} q_2$ are both .

Step 3: 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:

 $$\begin{gathered} 2K=U \\ =\frac{e^2}{4\tau \tau {\epsilon }_0\left(2R\right)} \\ \mathrm{Simplify} \mathrm{the} \mathrm{equation} \mathrm{as}: \\ \mathrm{K}=\frac{{\mathrm{e}}^2}{4{\mathrm{\pi \epsilon }}_0\left(4\mathrm{R}\right)} \\ =\frac{\left(9\times {10}^9{\displaystyle \frac{\mathrm{V}.\mathrm{m}}{\mathrm{C}}}\right){\left(1.6\times {10}^{-19} \mathrm{C}\right)}^2}{4\left(2.1\times {10}^{-15} \mathrm{m}\right)} \\ =2.74\times {10}^{-14 }\mathrm{J} \\ =170 \mathrm{keV} \end{gathered}$$

 

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

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