An alpha particle (which has two protons) is sent directly toward a target nucleus containing$\mathbf{92}\mathbf{}\mathbf{protons}$. The alpha particle has an initial kinetic energy of$\mathbf{0}\mathbf{.}\mathbf{48}\mathbf{}\mathbf{pJ}$. What is the least center-to-center distance the alpha particle will be from the target nucleus, assuming the nucleus does not move?

After reading the question, Figure drawn below

The initial kinetic energy of the particle is, ${\mathrm{K}}_{\mathrm{i}}=0.48\mathrm{pJ}$

The final kinetic energy is,${\mathrm{K}}_{\mathrm{f}}=0$

The charge,${\mathrm{q}}_1=+2\mathrm{e}$

The charge, ${\mathrm{q}}_2=+92\mathrm{e}$

Let the center-to-center separation be $\mathrm{d}$.

Use the law of conservation of energy.

The law of conservation of energy states that energy can neither be created nor destroyed - only converted from one form of energy to another.

Formulae:

$\begin{array}{rcl}∆\mathrm{U}& =& \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\\ & =& \mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\end{array}$

Where, U is the potential energy, q is charge of the particle, t is center-to-center separation, ${\mathrm{\epsilon }}_0$is the permittivity of free space, and k is the Coulomb’s constant having a value $9.0\times {10}^9{\mathrm{Nm}}^2/{\mathrm{c}}^2$.

Write the equation for the potential energy as below.

$\begin{array}{rcl}∆\mathrm{U}& =& \frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\end{array}$

From the Law of Conservation of Energy,

$\begin{array}{rcl}{\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}& =& -∆\mathrm{U}\\ & =& -\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\end{array}$

Solve the above equation for distance as below.

$\mathrm{d}=-\mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{K}}_{\mathrm{f}}-{\mathrm{K}}_{\mathrm{i}}}$

Substitute known vales in the above equation.

$\begin{array}{rcl}\mathrm{d}& =& \frac{9.0\times {10}^9\times \left(2\mathrm{e}\right)\left(92\mathrm{e}\right)}{0-0.48\times {10}^{-12}}\\ & =& \frac{9.0\times {10}^9\times 2\times \left(1.6\times {10}^{-19}\right)\times 92\times \left(1.6\times {10}^{-19}\right)}{0.48\times {10}^{-12}}\\ & =& 8.8\times {10}^{-14}\mathrm{m}\\ & =& 8.8\mathrm{fm}\end{array}$

Therefore, using the law of conservation of energy we can determine the distance of alpha particle from nucleus.