(a) In a quark model of elementary particles, a neutron is made of one up quarks $$\left[Charge\left(\dfrac{2}{3}\right) e\right]$$ and two down quarks $$\left[Charge\left(-\dfrac{1}{3}\right) e\right] .$$ Assume that they have a triangle configuration with side length of the order of $$10^{-15} \mathrm{m}$$.
Calculate electrostatic potential energy of neutron and compare it with its mass $$939 \mathrm{MeV}$$
(b) Repeat above exercise for a proton which is made of two up and one down quark.
(a) $$q_{d}=-\dfrac{1}{3}
e$$ [charge on down quark]
$$q_{\mathrm{u}}=+\dfrac{2}{3}
e[\text { charge on up quark }]$$
Potential energy $$U=\dfrac{k q_{1} q_{2}}{r}$$
$$k=\dfrac{1}{4
\pi \epsilon_{0}}$$
$$U=\dfrac{kq_1
q_{2}}{r}+\dfrac{kq_1 q_{1} q_{3}}{r}+\dfrac{kq_2 q_{3}}{r}$$
$$\therefore
U_{n}=\dfrac{1}{4 \pi \varepsilon_{0}} \dfrac{\left(-q_{d}\right)\left(-q_{d}\right)}{r}+\dfrac{\left(-q_{d}\right)
q_{\mathrm{u}}}{4 \pi \varepsilon_{0}}+\dfrac{q_{\mathrm{u}}\left(-q_{d}\right)}{4
\pi \varepsilon_{0} r}$$
$$=\dfrac{q_{d}}{4
\pi \varepsilon_{0} r}\left[+q_{d}-q_{u}-q_{u}\right]$$ [Talking sign of charge]
$$=\dfrac{q_d}{4
\pi \varepsilon_{0} r}\left[q_{d}-2 q_{u}\right]=\dfrac{9 \times 10^{9} \times
\dfrac{1}{3} e}{10^{-15}}\left[\dfrac{1}{3} e-2 \cdot \dfrac{2}{3} e\right]$$
[nature sign of charges taken already]
$$=\dfrac{9
\times 10^{8} \times e}{3 \times 10^{-15}} \cdot \dfrac{e}{3}[1-4]$$ joule
$$=\dfrac{-3
\times 9 \times 10^{8} \times 1.6 \times 10^{-19}}{9 \times 10^{-15}}$$ eJoule
$$=-4.8
\times 10^{9-19-15} e V=4.8 \times 10^{5} e V=-0.48 \times 10^{6} e V$$
$$\mathrm{U}=-0.48
\mathrm{MeV}$$
So, charges inside neutron $$\left[1 q_{u} \text { and } 2 q_{d}\right]$$ are attracted by energy of 0.48 Mev.
Energy released by a neutron when converted into energy is 939 Mev.
$$\therefore$$
Required ratio $$=\dfrac{1-0.481 \mathrm{MeV}}{939 \mathrm{MeV}}=0.0005111=5.11
\times 10^{-4}$$
(b) P.E. Of proton consists of 2 up and 1 down quark
$$r=10^{-15}
\mathrm{m}$$
$$q_{d}=-\dfrac{1}{3}
e. q_{u}=\dfrac{2}{3} e$$
$$U_{p}=\dfrac{1}{4
\pi \varepsilon_{0}} \dfrac{q_{\mathrm{u}} \times q_{\mathrm{u}}}{r}+\dfrac{q_{\mathrm{u}}\left(-q_{d}\right)}{4
\pi \varepsilon_{0} r}+\dfrac{q_{\mathrm{u}}\left(-q_{d}\right)}{4 \pi
\varepsilon_{0} r}$$
$$=\dfrac{q_{u}}{4
\pi \varepsilon_{0} r}\left[q_{u}-q_{d}-q_{d}\right]$$
$$=\dfrac{q_{\mathrm{u}}}{4
\pi \varepsilon_{0} r}\left[q_{\mathrm{u}}-2 q_{d}\right]=\dfrac{9 \times 10^{9}}{10^{-15}}
\dfrac{2}{3} e\left[\dfrac{2}{3} e-2 \cdot \dfrac{1}{3} e\right]=0$$
