Question

How much work is needed to assemble an atomic nucleus containing three protons such as Be) if we model it as equilateral triangle of side 2.0*10^-15 m with a proton at each vertex? Assume the protons started very far away.

Answer 1

Concepts and reason

The main concepts used to solve the problem are work done and energy.

Initially, find the equation for the work done to assemble two protons. Finally, using this equation to find the work done to assemble three protons.

Fundamentals

The work done to bring two positive charges separated by a distance is equal to the electrostatic energy between the two charges.

$U = \frac{{k{q_1}{q_2}}}{r}$

Here, $k$ is the Coulomb constant, ${q_1}$ is the first positive charge, ${q_2}$ is the second positive charge, and $r$ is the separation between the charges.

The charges are placed in the vertices of the equilateral triangle. So, the distances between the adjacent charges are the same.

The energy between two charges separated by a distance is,

$U = \frac{{k{q_1}{q_2}}}{r}$

Here, $k$ is the Coulomb constant, ${q_1}$ is the first positive charge, ${q_2}$ is the second positive charge, and $r$ is the separation between the charges.

As the charges here are positive, the charges ${q_1}$ and ${q_2}$ can be replaced with the charge of a proton.

Substitute e for ${q_1}$ and ${q_2}$ in $U = \frac{{k{q_1}{q_2}}}{r}$ .

$U = \frac{{k{e^2}}}{r}$

The energy between two protons placed at the vertices of an equilateral triangle is,

$U = \frac{{k{e^2}}}{r}$

When bringing a third charge to the third vertex of the equilateral triangle, the total energy becomes

$U = 3\frac{{k{e^2}}}{r}$

This energy is equal to the work done to assemble three protons such that there are three protons present at the vertex of an equilateral triangle. So, the work done is,

$W = 3\frac{{k{e^2}}}{r}$

Substitute $9 \times {10^9}\,{\rm{N}}{{\rm{m}}^2}{{\rm{C}}^{ - 2}}$ for $k$ , $1.6 \times {10^{ - 19}}\,{\rm{C}}$ for $e$ and $2.0 \times {10^{ - 15}}\,{\rm{m}}$ for $r$ to find $W.$

$\begin{array}{c}\\W = \frac{{3\left( {9 \times {{10}^9}\,{\rm{N}}{{\rm{m}}^2}{{\rm{C}}^{ - 2}}} \right){{\left( {1.6 \times {{10}^{ - 19}}\,{\rm{C}}} \right)}^2}}}{{2.0 \times {{10}^{ - 15}}\,{\rm{m}}}}\\\\ = 3.5 \times {10^{ - 13}}\,{\rm{J}}\\\end{array}$

Ans:

The work done to assemble three protons at the vertices of an equilateral triangle is $3.5 \times {10^{ - 13}}\,{\rm{J}}.$