91Ó°ÊÓ

Two nuclei are fired at each other with same kinetic energy K.

Formulae:

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

$\mathbf{U}\mathbf{=}\frac{{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{\mathbf{4}{\mathbf{Ï„Ï„Î\mu }}_{\mathbf{0}}\mathbf{r}}$ ….. (i)

The radius of an atom due to its nucleon or mass number as follows:

$\mathbf{r}\mathbf{=}{\mathbf{r}}_{\mathbf{0}}{\mathbf{A}}^{\frac{\mathbf{1}}{\mathbf{3}}}$ …… (ii)

Here,${\mathrm{r}}_0=1.2\mathrm{fm}$

Using the mass number of a lithium nucleus (A = 7) in equation (ii), the radius of the lithium nucleus can be given as follows:

$$\begin{gathered} \mathrm{r}\left(\mathrm{R}\right)=(1.2\mathrm{fm}){\left(7\right)}^{\frac{1}{3}} \\ =2.3\mathrm{fm} \end{gathered}$$

Now, use this value in equation (i) for two lithium nuclei with same energy K, determine the Coulomb barrier height as follows: (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 3e.)

$$\begin{gathered} 2K=U \\ K=\frac{\left(3e\right)\left(3e\right)}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(2R\right)} \\ K=\frac{9e^2}{4\mathrm{Ï€}{\mathrm{Î\mu }}_0\left(4R\right)} \end{gathered}$$

Substitute the values and solve as:

$$\begin{gathered} \mathrm{K}=\frac{9\left(9×{10}^9{\displaystyle \frac{\mathrm{V}.\mathrm{m}}{\mathrm{C}}}\right){\left(1.6×{10}^{-19}\mathrm{C}\right)}^2}{4\left(2.3×{10}^{-15}\mathrm{m}\right)} \\ =2.25×{10}^{-13}\mathrm{J} \\ =1.41\mathrm{MeV} \end{gathered}$$

Hence, the required barrier height is 1.41 MeV.