Question

In a quark model of elementary particles, a neutron is made of one up quark of charge ${\displaystyle \frac{2}{3}}e$ and two down quarks of charges $\left(-{\displaystyle \frac{1}{3}}e\right)$. If they have a triangle configuration with side length of order ${10}^{-15}m$. The electrostatic potential energy of neutron in $MeV$ is:

Answer 1

Hint: The numerical problem given above can be solved easily by assuming the quarks to be the same as charged particles such as electrons. The three charged particles are arranged in a triangle arrangement.

Formula Used:
The mathematical formula for calculating electric potential energy is given as:
$U$ = ${\displaystyle \frac{1}{4\pi {\epsilon }_{0}r}}\left[q_u{q}_d+q_d{q}_d+q_d{q}_u\right]$
In this mathematical expression, $r=$ the distance between two quarks, $q_u=$charge in an up quark and $q_d=$charge in a down quark.

Complete step by step solution:
In the numerical problem given above the quarks are arranged in the form of an equilateral triangle as shown in the figure drawn below:







From the given numerical problem we know that the separation between the two charges is $r={10}^{-15}m$.
Also the charge of an Up quark is given as $q_u={\displaystyle \frac{2}{3}}e$
Similarly, the charge of a Down quark is given as $q_d=\left(-{\displaystyle \frac{1}{3}}e\right)$
Now we can substitute these values in the mathematical expression given above. Substituting these values we get:
$U=\left[{\displaystyle \frac{1}{4\pi {\epsilon }_{\circ }r}}\right]\times \left\{\left({\displaystyle \frac{2}{3}}e\right)\left({\displaystyle \frac{-1}{3}}e\right)+{\left({\displaystyle \frac{-1}{3}}e\right)}^2+\left({\displaystyle \frac{-1}{3}}e\right)\left({\displaystyle \frac{2}{3}}e\right)\right\}$
When we calculate the value of this mathematical equation we get:
$U=\left[{\displaystyle \frac{1}{4\pi {\epsilon }_{\circ }r}}\right]\times \left\{{\displaystyle \frac{-e^2}{3}}\right\}$
Now we know that ${\displaystyle \frac{1}{4\pi {\epsilon }_{\circ }}}=9\times {10}^9$, $r={10}^{-15}$and $e=1.6\times {10}^{-19}$. We can put these values in the above equation. Thus, we get:
$U={\displaystyle \frac{9\times {10}^9}{{10}^{-15}}}\left[{\displaystyle \frac{-{\left(1.6\times {10}^{-19}\right)}^2}{3}}\right]$
Solving this we get:
$U=-7.68\times {10}^{-14}J$
But this value of electrostatic potential energy of the neutron is given in Joules but we have to calculate the value in Mega-electron Volts. Thus we have to convert the unit to $MeV$
We know that $1J=6.4\times {10}^{12}MeV$. Using this to convert the equation.
Thus, $U=-0.48MeV$

Hence, we find that the electrostatic potential energy of the neutron in $MeV$ is equal to $-0.48MeV$.

Note: It is important to be very careful while solving numerical problems of this type as the calculations tend to get very tedious and time consuming. Even if a minor mistake is made in any step it could affect the final result.