An electron in an excited state of a hydrogen atom has a de-Broglie wavelength of 6 a₀ , where a₀ is the Bohr radius. If this electron makes a transition to the ground state, what is the energy of the emitted photon? (Given: Ground state energy of H-atom is -13.6 eV )

Step-by-step solution

The de-Broglie wavelength of an electron in the n^ th Bohr orbit is related to the orbit's circumference by: 2 r_n = n For a hydrogen atom ( Z=1 ), the radius of the n^ th orbit is r_n = a₀ n^2 . Substituting this into the equation gives: 2 (a₀ n^2) = n = 2 a₀ n We are given that the de-Broglie wavelength in the excited state is = 6 a₀ . Equating the two expressions: 2 a₀ n = 6 a₀ n = 3 The electron is initially in the n = 3 state and transitions to the ground state ( n = 1 ). The energy of the emitted photon is gi