Question

what would be the radius of 2nd excited State in Li+2 ion ?

Answer 1

we know that the radius of any orbit of any hydrogen like species and hydrogen is given by 52.9 pm × n²/Z

where Z is the Atomic number and  and n is the respective orbit

Li²⁺ is hydrogen like species and the Atomic number of Lithium is 3

In the 1st excited state of Li²⁺ the electron jumps from n= 1 to n = 2 and the in the 2nd excited state of Li²⁺ the electron jumps from n = 2 to n = 3. so in the 2nd excited state the electron is present in the orbit, n = 3

substituting the above values in the radius equation we get:

r₂ = 52.9 pm × 3²/3 = 158.7 pm = 1.587 angstroms

where pm is picometres = 10⁻¹² metre and one angstrom = 10⁻¹⁰ metre