Question

In a quark model of elementary particles, a neutron is made of one up quarks [charge $\left(2/3\right)e$] and two down quarks [charges $-\left(1/3\right)e$]. Assume that they have a triangle configuration with side length of the order of ${10}^{–15}m$.

A) Calculate electrostatic potential energy of neutron and compare it with its mass $939MeV$.
B) Calculate electrostatic potential energy of proton and compare it with its mass $939MeV$.

Answer 1

1) Step 1: Draw a rough diagram.

Step 2: Find potential energy.

Charge on an up quark

$q_{up}=\frac{2}{3e}$

Charge on a down quark

$q_{down}=-\frac{1}{3e}$

Distance between each quark

$d={10}^{-15}m$

The potential energy of the system is given by,

$U=\frac{1}{4\pi {ϵ}_{0}d}[q_{down}{q}_{down}+q_{down}{q}_{up}+q_{down}{q}_{up}]$

$(e=1.6\times {10}^{-19}C)$

$U=\frac{1}{4\pi (8.854\times {10}^{-12})\left({10}^{-15}\right)}(1.6\times {10}^{-19}{)}^2[\frac{1}{9}-\frac{4}{9}]$

$U=-76.81\times {10}^{-15}$ $J=\frac{-76.81\times {10}^{-15}}{1.6\times {10}^{-19}eV}$

$(\because 1eV=1.6\times {10}^{-19}J)$

$U=-48\times {10}^{4}eV=-0.48MeV$

By comparing this energy with the neutrons mass $\left(939MeV\right)$.

potential energy is equivalent to,

$\frac{-0.48}{939}=5.11\times {10}^{-4}$ neutron masses

Final Answer: $5.11\times {10}^{-4}$ neutron masses



2) Step 1: Draw a rough diagram.

Step 2: Find potential energy.

Charge on an up quark

$=\frac{2}{3}e$

Charge on a down quark

$=-\frac{1}{3}e$

Distance between each quark $={10}^{-15}m$ The potential energy of the system is given by,

$U=\frac{1}{4\pi {ϵ}_{0}r}[q_{up}{q}_{up}+q_{down}{q}_{up}+q_{down}{q}_{up}]$

Substitute the values, we get

$U=\frac{1}{4\pi (8.854\times {10}^{-12})\left({10}^{-15}\right)}$

$[\frac{2}{3}e\left(\frac{2}{3}e\right)+(-\frac{1}{3}e)\left(\frac{2}{3}e\right)+(-\frac{1}{3}e\left)\left(\frac{2}{3}e\right)\right]$

$(e=1.6\times {10}^{-19}C)$

$U=\frac{1}{4\pi (8.854\times {10}^{-12})\left({10}^{-15}\right)}(1.6\times {10}^{-19}{)}^2[\frac{4}{9}-\frac{4}{9}]$

$=0eV$