If the transition in a hydrogen atom is from an excited state (n = 6) to a lower state (p = 3) and Rydberg's constant is 1.0974 x 10^7 m-I, then the wavelength of the spectral line is:

Explanation

To calculate the wavelength of a spectral line resulting from a transition in a hydrogen atom, we can use the Rydberg formula:1/λ = R_H * (1/n_f^2 - 1/n_i^2)where λ is the wavelength of the spectral line, R_H is the Rydberg constant, n_f is the final state (lower energy level), and n_i is the initial state (higher energy level).In this case, the transition is from the excited state (n = 6) to the lower state (p = 3). Plugging these values into the Rydberg formula, we have:1/λ = R_H * (1/3^2 - 1/6^2)Simplifying the equation:1/λ = R_H * (1/9 - 1/36)1/λ = R_H * (4/36 - 1/36)1/λ = R_H * 3/361/λ = R_H/12Now, substituting the value of the Rydberg constant (R_H = 1.0974 x 10^7 m^-1), we have:1/λ = (1.0974 x 10^7 m^-1)/121/λ = 0.09145 x 10^7 m^-1Taking the reciprocal of both sides of the equation:λ = 1/(0.09145 x 10^7 m^-1)λ ≈ 1.093 x 10^-7 mSo, the correct answer is A: 1.093 x 10^-7 m. This is the wavelength of the spectral line resulting from the transition in the hydrogen atom.