
In a quark model of elementary particles, a neutron is made of one up quark of charge $\dfrac{2}{3}e$ and two down quarks of charges $\left( { - \dfrac{1}{3}e} \right)$. If they have a triangle configuration with side length of order ${10^{ - 15}}m$. The electrostatic potential energy of neutron in $MeV$ is:
Answer
148.8k+ views
Hint: The numerical problem given above can be solved easily by assuming the quarks to be the same as charged particles such as electrons. The three charged particles are arranged in a triangle arrangement.
Formula Used:
The mathematical formula for calculating electric potential energy is given as:
$U$ = $\dfrac{1}{{4\pi {\varepsilon_0}r}}\left[ {{q_u}{q_d} + {q_d}{q_d} + {q_d}{q_u}} \right]$
In this mathematical expression, $r = $ the distance between two quarks, ${q_u} = $charge in an up quark and ${q_d} = $charge in a down quark.
Complete step by step solution:
In the numerical problem given above the quarks are arranged in the form of an equilateral triangle as shown in the figure drawn below:

From the given numerical problem we know that the separation between the two charges is $r = {10^{ - 15}}m$.
Also the charge of an Up quark is given as ${q_u} = \dfrac{2}{3}e$
Similarly, the charge of a Down quark is given as ${q_d} = \left( { - \dfrac{1}{3}e} \right)$
Now we can substitute these values in the mathematical expression given above. Substituting these values we get:
$U = \left[ {\dfrac{1}{{4\pi {\varepsilon _ \circ }r}}} \right] \times \left\{ {\left( {\dfrac{2}{3}e} \right)\left( {\dfrac{{ - 1}}{3}e} \right) + {{\left( {\dfrac{{ - 1}}{3}e} \right)}^2} + \left( {\dfrac{{ - 1}}{3}e} \right)\left( {\dfrac{2}{3}e} \right)} \right\}$
When we calculate the value of this mathematical equation we get:
$U = \left[ {\dfrac{1}{{4\pi {\varepsilon _ \circ }r}}} \right] \times \left\{ {\dfrac{{ - {e^2}}}{3}} \right\}$
Now we know that $\dfrac{1}{{4\pi {\varepsilon _ \circ }}} = 9 \times {10^9}$, $r = {10^{ - 15}}$and $e = 1.6 \times {10^{ - 19}}$. We can put these values in the above equation. Thus, we get:
$U = \dfrac{{9 \times {{10}^9}}}{{{{10}^{ - 15}}}}\left[ {\dfrac{{ - {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{3}} \right]$
Solving this we get:
$U = - 7.68 \times {10^{ - 14}}J$
But this value of electrostatic potential energy of the neutron is given in Joules but we have to calculate the value in Mega-electron Volts. Thus we have to convert the unit to $MeV$
We know that $1J = 6.4 \times {10^{12}}MeV$. Using this to convert the equation.
Thus, $U = - 0.48MeV$
Hence, we find that the electrostatic potential energy of the neutron in $MeV$ is equal to $ - 0.48MeV$.
Note: It is important to be very careful while solving numerical problems of this type as the calculations tend to get very tedious and time consuming. Even if a minor mistake is made in any step it could affect the final result.
Formula Used:
The mathematical formula for calculating electric potential energy is given as:
$U$ = $\dfrac{1}{{4\pi {\varepsilon_0}r}}\left[ {{q_u}{q_d} + {q_d}{q_d} + {q_d}{q_u}} \right]$
In this mathematical expression, $r = $ the distance between two quarks, ${q_u} = $charge in an up quark and ${q_d} = $charge in a down quark.
Complete step by step solution:
In the numerical problem given above the quarks are arranged in the form of an equilateral triangle as shown in the figure drawn below:

From the given numerical problem we know that the separation between the two charges is $r = {10^{ - 15}}m$.
Also the charge of an Up quark is given as ${q_u} = \dfrac{2}{3}e$
Similarly, the charge of a Down quark is given as ${q_d} = \left( { - \dfrac{1}{3}e} \right)$
Now we can substitute these values in the mathematical expression given above. Substituting these values we get:
$U = \left[ {\dfrac{1}{{4\pi {\varepsilon _ \circ }r}}} \right] \times \left\{ {\left( {\dfrac{2}{3}e} \right)\left( {\dfrac{{ - 1}}{3}e} \right) + {{\left( {\dfrac{{ - 1}}{3}e} \right)}^2} + \left( {\dfrac{{ - 1}}{3}e} \right)\left( {\dfrac{2}{3}e} \right)} \right\}$
When we calculate the value of this mathematical equation we get:
$U = \left[ {\dfrac{1}{{4\pi {\varepsilon _ \circ }r}}} \right] \times \left\{ {\dfrac{{ - {e^2}}}{3}} \right\}$
Now we know that $\dfrac{1}{{4\pi {\varepsilon _ \circ }}} = 9 \times {10^9}$, $r = {10^{ - 15}}$and $e = 1.6 \times {10^{ - 19}}$. We can put these values in the above equation. Thus, we get:
$U = \dfrac{{9 \times {{10}^9}}}{{{{10}^{ - 15}}}}\left[ {\dfrac{{ - {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{3}} \right]$
Solving this we get:
$U = - 7.68 \times {10^{ - 14}}J$
But this value of electrostatic potential energy of the neutron is given in Joules but we have to calculate the value in Mega-electron Volts. Thus we have to convert the unit to $MeV$
We know that $1J = 6.4 \times {10^{12}}MeV$. Using this to convert the equation.
Thus, $U = - 0.48MeV$
Hence, we find that the electrostatic potential energy of the neutron in $MeV$ is equal to $ - 0.48MeV$.
Note: It is important to be very careful while solving numerical problems of this type as the calculations tend to get very tedious and time consuming. Even if a minor mistake is made in any step it could affect the final result.
Recently Updated Pages
Wheatstone Bridge - Working Principle, Formula, Derivation, Application

Young's Double Slit Experiment Step by Step Derivation

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Uniform Acceleration

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Degree of Dissociation and Its Formula With Solved Example for JEE

Electrical Field of Charged Spherical Shell - JEE

Charging and Discharging of Capacitor
