Question

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 ________. [nearest integer

Answer 1

Number of electrons in $\\\\_{22}^{48}X^{3−}$ is 25.
Number of neutrons = 48 – 22 = 26.
Percentage increase in the number of neutrons over electrons

=$(\frac{26–25}{25})$100=4%
∴ a = 4