The de-Broglie wavelength of an electron in a particular Bohr orbit of a He ⁺ ion is equal to 2 a₀ , where a₀ is the Bohr radius. The principal quantum number of this orbit is:

Step-by-step solution

According to Bohr's quantization condition, 2 r = n The radius of the n^ th orbit for a hydrogen-like species is given by r = a₀ n^2 Z Substituting r into the first equation: 2 ( a₀ n^2 Z ) = n = 2 a₀ n Z For the He ⁺ ion, the atomic number Z = 2 . We are given = 2 a₀ . Therefore, 2 a₀ = 2 a₀ n 2 Solving for n , we get n = 2 . Answer: 2