It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves built out of more elementary units called quarks.

Solution:

A proton has three quarks. Let there be $n$ up quarks in a proton, each having a charge of $+\frac{2}{3} e$.

Charge due to $n$ up quarks $=\left(\frac{2}{3} e\right) n$

Number of down quarks in a proton = 3 − n

Each down quark has a charge of $-\frac{1}{3} e$.

Charge due to $(3-n)$ down quarks $=\left(-\frac{1}{3} e\right)(3-n)$

Total charge on a proton = + e

$\therefore e=\left(\frac{2}{3} e\right) n+\left(-\frac{1}{3} e\right)(3-n)$

$e=\left(\frac{2 n e}{3}\right)-e+\frac{n e}{3}$

$2 e=n e$

$n=2$

Number of up quarks in a proton, n = 2

Number of down quarks in a proton = 3 − n = 3 − 2 = 1

Therefore, a proton can be represented as ‘uud’.

A neutron also has three quarks. Let there be $n$ up quarks in a neutron, each having a charge of $+\frac{3}{2} e$.

Charge on a neutron due to $n$ up quarks $=\left(+\frac{3}{2} e\right) n$

Number of down quarks is $3-n$, each having a charge of $\left(-\frac{1}{3}\right) e$.

Charge on a neutron due to $(3-n)$ down quarks $=\left(-\frac{1}{3} e\right)(3-n)$

Total charge on a neutron = 0

$0=\left(\frac{2}{3} e\right) n+\left(-\frac{1}{3} e\right)(3-n)$

$0=\frac{2}{3} e n-e+\frac{n e}{3}$

$e=n e$

$n=1$

Number of up quarks in a neutron, n = 1

Number of down quarks in a neutron = 3 − n = 2

Therefore, a neutron can be represented as ‘udd’.