The de-Broglie wavelength of an electron in an excited state of a Li ²⁺ ion is 6 a₀ (where a₀ is the Bohr radius). The principal quantum number of this state is

Step-by-step solution

According to Bohr's quantization condition, the circumference of the n^ th orbit is equal to n times the de-Broglie wavelength ( _d ) of the electron: 2 r_n = n _d The radius of the n^ th Bohr orbit is given by: r_n = a₀ n^2 Z Substituting r_n into the first equation gives: 2 (a₀ n^2 Z ) = n _d _d = 2 a₀ n Z For a Li ²⁺ ion, the atomic number Z = 3 . We are given that _d = 6 a₀ . Substituting these values: 6 a₀ = 2 a₀ n 3 6 = 2n 3 18 = 2n n = 9 Answer: 9