The electron, in a hydrogen atom, initially in a state of quantum number n1 makes a transition to a state whose excitation energy, with respect to the ground state, is 10.2 eV. If the wavelength, associated with the photon emitted in this transition, is 487.5 mm, find the (i) energy in ev, and (ii) value of the quantum number, n1 of the electron in its initial state.

According to question the orbit in which the electron is transiting has an excitation energy of 10.2 eV with respect to ground state energy as -13.6eVi.e., E2 - E1 = 10.2eV E2 -(-13.6) =10.2 E2 = -3.4eVi.e., the electron is transiting into n = 2 from a initial orbit n1(1) As given the wavelength emitted during the transition is 487.5nm which when converted into energy comes out to be 1237.5/487.5 ≈ 2.55eVit means that excitation energy for transiting an electron from n1 to n = 2 is 2.55eVi.e., En2 - En1 = 2.55eV here, En1 -(-3.4)= 2.55 En1 = -0.85eV(2) As we know in hydrogen spectrum the energy in 4th orbit is -0.85eV ( -13.6/n^2 = 0.85eV )Therefore, n = 4