JEE Main Chemistry Question (2026) — Solution

Step-by-step solution

The radius of the n-th Bohr orbit for a hydrogen-like species is given by the formula: r_n = a_0 n^2 Z where a_0 is the Bohr radius of the hydrogen atom (52.9 pm), n is the principal quantum number, and Z is the atomic number. Given r_n = 70.53 pm, we can write: 70.53 = 52.9 n^2 Z n^2 Z = 70.53 52.9 = 1.333 = 4 3 Checking the given options: For Li ^ 2+ and n = 3, Z = 3 n^2 Z = 9 3 = 3 For He ^+ and n = 3, Z = 2 n^2 Z = 9 2 = 4.5 For He ^+ and n = 2, Z = 2 n^2 Z = 4 2 = 2 For Li ^ 2+ and n = 2, Z = 3 n^2 Z = 4 3 The ratio matches for Li ^ 2+ and n = 2. Answer: Li ^ 2+ , 2