91Ó°ÊÓ

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:

$\begin{array}{rcl}∆\mathrm{U}& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\\ & =& \mathrm{k}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\end{array}$

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

Write the equation for the potential energy as below.

$\begin{array}{rcl}∆\mathrm{U}& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{d}}\end{array}$

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.

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

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