An electron in a Li ²⁺ ion is in an excited state where its de-Broglie wavelength is 4 a₀ (where a₀ is the Bohr radius). When this electron makes a transition to the ground state, the maximum number of spectral lines observed in the emission spectrum is

Step-by-step solution

The condition for standing waves in a Bohr orbit is: 2 r_n = n _d The radius of the n^ th orbit for a hydrogen-like species is: r_n = a₀ n^2 Z Substituting the expression for r_n into the standing wave condition: 2 (a₀ n^2 Z ) = n _d _d = 2 a₀ n Z For the Li ²⁺ ion, Z = 3 . Given _d = 4 a₀ , we can find the principal quantum number n of the excited state: 4 a₀ = 2 a₀ n 3 4 = 2n 3 n = 6 The electron is in the 6^ th orbit. When it transitions to the ground state ( n = 1 ), the maximum number of spectral lines produce