For a hydrogen atom, the ratio of the total energies of an electron in two different orbits A and B is 1:9 . The ratio of the orbital angular momenta of the electron in these two orbits, L_A : L_B , is

Step-by-step solution

The total energy of an electron in the n^ th orbit of a hydrogen atom is given by E - 1 n^2 . The ratio of the total energies in orbits A and B is: E_A E_B = n_B^2 n_A^2 = 1 9 Taking the square root of both sides gives: n_B n_A = 1 3 n_A n_B = 3 1 According to Bohr's quantization condition, the orbital angular momentum of an electron is L = nh 2 , which implies L n . Therefore, the ratio of their orbital angular momenta is: L_A L_B = n_A n_B = 3 1 The ratio L_A : L_B is 3:1 . Assuming L E directly would incorrectly