Consider an imaginary ion $$_{22}^{48}X^{3-}$$. The nucleus contains '$$a$$' $$\%$$ more neutrons than the number of electrons in the ion. The value of '$$a$$' is ______.

We are given an imaginary ion $$_{22}^{48}X^{3-}$$ and need to find the percentage ‘a’ by which neutrons exceed the number of electrons.

Since the atomic number (Z) = 22, the number of protons is 22. Furthermore, the mass number (A) = 48 implies that the number of neutrons is A − Z = 48 − 22 = 26. Because the ion carries a charge of 3−, the total number of electrons is 22 + 3 = 25.

Since the nucleus contains ‘a’ % more neutrons than the number of electrons, we set up $$\text{Number of neutrons} = \text{Number of electrons} + \frac{a}{100} \times \text{Number of electrons}$$.

Substituting the known values yields $$26 = 25 + \frac{a}{100} \times 25$$.

Rearranging gives $$26 - 25 = \frac{25a}{100}$$, which simplifies to $$1 = \frac{25a}{100}$$.

Solving for a, we find $$a = \frac{100}{25} = 4$$.

Therefore, the value of $$a$$ is 4.