Question

Trying to pull two quarks apart would produce more quarks in groups or hadrons. Suppose that when the separation reaches 1 fm ( the approximate radius of a nucleon), the lightest hadron a ${\pi }^0$is created.

(a) Roughly how much force is involved?

(b) Compare this with the electrostatic force between two fundamentalcharges the same distance apart. Does your results agree with the strengths in table 12.1 ?

Short answer

(a) The force is $21600N$.

(b) The strong force is$21600N$ , about 100 times as much as the electrostatic force $\left(F_e\right)$, just as expected.

Step-by-step solution

 Step 1: Given data

The given data : separation =$1fm$ .

 Step 2: Concept of Electrostatic force

Coulomb's law: magnitude of electric force$F_e$ between two protons is given by:

$$\begin{gathered} F_e=k\frac{(+e)(+e)}{r^2} \\ {F}_e=k\frac{e^2}{r^2} \end{gathered}$$

HereKrepresents proportionality constant and is equal to$8.99\times {10}^{9}N·m^2/C^2$ , the charge of the electron$1.6\times {10}^{-19}C$ .

The electrostatic force is given by,

$F_e=\frac{e^2}{4\pi {\epsilon }_0{r}^2}$

Calculation of the force required to pull the quarks 

(a)

The strong force is roughly constant at the short-range when it is active, so we have the work done by the force is equal to the mass-energy of a pion, which is $135MeV$, so the force is given as,

$F=\frac{E}{d}$

Substitute $135MeV$ for E and $1.0fm$ for d in the above equation, and we get,

$$\begin{gathered} F=\frac{(135MeV)}{1fm}\frac{\left(1.6\times {10}^{-13}J\right)}{1MeV}\frac{(1.0fm)}{{10}^{-15}m} \\ F=21600N \end{gathered}$$

The force is .$21600N$

Comparison of the electrostatic force between two fundamental charges

(b)

Substitute $1.6\times {10}^{-19}C$ for the charge of the electron $\left(e\right),8.85\times {10}^{-12}F/m$for ${\epsilon }_0$and ${10}^{-15}m$for r in the above equation.

$$\begin{gathered} F_e=\frac{{\left(1.6\times {10}^{-19}\text{C}\right)}^2}{4\pi \left(8.85\times {10}^{-12}\text{F/m}\right){\left({10}^{-15}\text{m}\right)}^2} \\ {F}_e=230\text{N} \end{gathered}$$

The strong force is $21600N$, about 100 times as much as the electrostatic force $\left(F_e\right)$, which is 230 N, just as expected.