Question

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?

Short answer

The least center-to-center distance of the alpha particle from the target nucleus is $\mathrm{d}=88\mathrm{fm}$.

Step-by-step solution

Step 1: Given data:

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 $\mathrm{d}$.

Determining the concept:

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 $9.0\times {10}^9{\mathrm{Nm}}^2/{\mathrm{c}}^2$.

The least center-to-center distance of the alpha particle:

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.