Consider an imaginary ion X 3 - 2248 . The nucleus contains ' a ' %...

${X 2248}^{3 -}$

In an atomic representation $X ZA$

Z is atomic number = 22

A is atomic mass number = 48.

No. of protons $= Z = 22$

No. of neutrons $= A - Z = 48 - 22 = 26$

No. of electrons $= P - charge = 22 + 3 = 25$

$$ of extra neutrons than electrons $= \dfrac{26 - 25}{25} \times 100 = 4$

So, the value of a=4.