An electron in a hydrogen atom is in a state with a total energy of -1.51 eV . If the Bohr radius of the ground state is 0.53 , what is the radius of this orbit?

Step-by-step solution

The total energy of an electron in the n^ th orbit of a hydrogen atom is given by: E_n = - 13.6 n^2 eV Given E_n = -1.51 eV : -1.51 = - 13.6 n^2 n^2 = 13.6 1.51 = 9 n = 3 The radius of the n^ th orbit is given by: r_n = a₀ n^2 where a₀ is the ground state radius. Substituting n = 3 and a₀ = 0.53 : r₃ = 0.53 3^2 r₃ = 0.53 9 = 4.77 Answer: 4.77