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प्रश्न
- 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.
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उत्तर
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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