Let E₀ be the magnitude of the ground state energy of a hydrogen atom. If the angular momentum of the electron in a particular Bohr orbit is 3h , where h is Planck's constant, the energy of the electron in this orbit is:

Step-by-step solution

According to Bohr's quantization rule, the angular momentum of an electron in the n^ th orbit is given by: L = nh 2 Given that L = 3h , we can equate the two expressions: nh 2 = 3h n = 6 The energy of an electron in the n^ th Bohr orbit of a hydrogen atom is given by: E_n = - E₀ n^2 where E₀ is the magnitude of the ground state energy. Substituting n = 6 : E₆ = - E₀ 6^2 = - E₀ 36 Answer: - E₀ 36