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Question
- In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges –(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 MeV.
- Repeat above exercise for a proton which is made of two up and one down quark.
Solution
This system is made up of three charges. The potential energy of the system is equal to the algebraic sum of the PE of each pair. So,
a. `q_d = 1/3 e` charge on down quark
`q_u = + 2/3 e` charge on up quark
Potential energy U = `(kq_1q_2)/r`
`k = 1/(4piε_0)`
U = `(kq_1q_2)/r + (kq_1q_3)/r + (kq_2q_3)/r`
∴ Un = `1/(4piε_0) ((-q_d)(-q_d))/r + ((-q_d)q_u)/(4piε_0) + (q_u (-q_d))/(4piε_0r)`
= `(q_d)/(4piε_0r) [+q_d - q_u - q_u]` Talking sign of charge
= `(q_d)/(4piε_0r) [q_d - 2q_u]`
= `(9 xx 10^9 xx 1/3 e)/(10^-15) [1/3 e - 2 * 2/3 e]` nature sign of charge taken already
= `(9 xx 10^9 xx e)/(3 xx 10^-15) * e/3 [1 - 4]` Joule
= `(-3 xx 9 xx 10^9 xx 1.6 xx 10^-19)/(9 xx 10^-15)` e Joule
= `- 4.8 xx 10^(9 - 19 - 15) eV`
= `4.8 xx 10^5 eV`
= `- 0.4.8 xx 10^6 eV`
U = – 0.48 MeV
So, charges inside neutron [1qu and 2qd] are attracted by energy of 0.48 MeV.
Energy released by a neutron when converted into energy is 939 MeV.
∴ Required ratio = `(1 - 0.481 MeV)/(939 MeV)` = 0.0005111 = 5.11 × 10–4
b. P.E. of proton consists of 2 up and 1 down quark
r = 10–15 m
qd = `- 1/3 e, q_u = 2/3 e`
Up = `1/(4 piε_0) (q_u xx q_u)/r + (q_u (-q_d))/(4 piε_0r) + (q_u (-q_d))/(4 piε_0r)`
= `q/(4 piε_0r) [q_u - q_d - q_d]`
= `q/(4 piε_0r) [q_u - 2q_d]`
= `(9 xx 10^9)/10^-13 2/3 e [2/3 e - 2 1/3 e]` = 0
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