A singly ionized helium atom ( He ⁺ ) in an excited state with principal quantum number n transitions to the first excited state, emitting a photon of wavelength . The value of n is : ( R is the Rydberg constant)

Step-by-step solution

For a hydrogen-like ion, the Rydberg formula is given by: 1 = R Z^2 ( 1 n_f^2 - 1 n_i^2 ) For a singly ionized helium atom ( He ⁺ ), the atomic number Z = 2 . The transition is to the first excited state, so the final state is n_f = 2 . The initial state is n_i = n . Substituting these values into the formula: 1 = R (2)^2 ( 1 2^2 - 1 n^2 ) 1 = 4R ( 1 4 - 1 n^2 ) 1 = R - 4R n^2 Rearranging to solve for n : 4R n^2 = R - 1 = R - 1 n^2 4R = R - 1 n^2 = 4 R R - 1 n = 2 R R - 1 Answer: 2 R R - 1