Consider an imaginary ion X 3 - 22 48 . The nucleus contains ' a ' % more neutrons than the number of electrons in the ion. The value of ' a ' is

Step-by-step solution

X 22 48 3 - In an atomic representation X Z A 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 = 26 - 25 25 × 100 = 4   %   =   a % So, the value of a=4.