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$(a)$ In a quark model of elementary particles, a neutron is made of one up quarks [ charge $\frac{2}{3}e$ ] and two down quarks [ charges $ - \frac{1}{3}e$ ]. Assume that they have a triangle configuration with side length of the order of ${10^{ - 15}}$ $m$. Calculate electrostatic potential energy of neutron and compare it with its mass $939$ $Me\,V$. $(b)$ Repeat above exercise for a proton which is made of two up and one down quark.
Solution
$(a)$ Electrostatic potential energy $\mathrm{U}=\frac{k q_{1} q_{2}}{r}$ The system of three charges is as below,
Electrostatic potential energy,
$\mathrm{U}=k\left(\frac{q_{d} q_{d}}{r}+\frac{q_{u} q_{d}}{r}+\frac{q_{u} q_{d}}{r}\right)$
$\mathrm{U}=k\left[\frac{\left[\left(-\frac{1}{3} e\right)\left(-\frac{1}{3} e\right)\right.}{r}+\frac{\left(\frac{2}{3} e\right)\left(-\frac{1}{3} e\right)}{r}+\frac{\left(\frac{2}{3} e\right)\left(-\frac{1}{3} e\right)}{r}\right]$
$\mathrm{U}=\frac{k}{r}\left[\frac{1}{9} e^{2}-\frac{2}{9} e^{2}-\frac{2}{9} e^{2}\right]$
$=\frac{k e^{2}}{q r}[1-2-2]$
$=\frac{k e^{2}}{q r} \times(-3)$
$=\frac{9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2} \times(-3)}{9 \times 10^{-15}}$
$=-7.68 \times 10^{-14} \mathrm{~J}$
$\therefore \quad \mathrm{U}=\frac{-7.68 \times 10^{-14}}{-1.6 \times 10^{-19}} \mathrm{eV} $
$\therefore \quad\mathrm{U}=4.8 \times 10^{5} \mathrm{eV}$
$=0.48 \times 10^{6} \mathrm{eV}=0.48 \mathrm{MeV}$
