A hydrogen atom in a state with a binding energy of 0.85 eV undergoes a transition to a state having an excitation energy of 10.2 eV . Calculate the wavelength of the emitted radiation.

Step-by-step solution

The binding energy of an electron in the n^ th state of a hydrogen atom is given by B.E. = 13.6 n^2 eV . For the initial state, B.E. = 0.85 eV : 13.6 n_i^2 = 0.85 n_i^2 = 16 n_i = 4 The excitation energy of a state is the energy required to excite an electron from the ground state ( n=1 ) to that state. For the final state, the excitation energy is 10.2 eV : E_ n_f - E₁ = 10.2 eV - 13.6 n_f^2 - (-13.6) = 10.2 13.6 n_f^2 = 13.6 - 10.2 = 3.4 n_f^2 = 4 n_f = 2 The energy of the emitted photon during the transition fro