An electron in a hydrogen atom has a de-Broglie wavelength of 6 a₀ , where a₀ is the Bohr radius. If this electron undergoes a transition to the ground state, the energy of the emitted photon is approximately:

Step-by-step solution

The de-Broglie wavelength of an electron in the n^ th Bohr orbit is given by = 2 r n . Since r = a₀ n^2 Z , the wavelength can be written as = 2 a₀ n Z . For a hydrogen atom, Z = 1 . Given = 6 a₀ : 6 a₀ = 2 a₀ n 1 n = 3 The electron is initially in the 3^ rd orbit and transitions to the ground state ( n = 1 ). The energy of the emitted photon is given by Rydberg's formula: E = 13.6 ( 1 n₁^2 - 1 n₂^2 ) eV E = 13.6 ( 1 1^2 - 1 3^2 ) eV E = 13.6 ( 1 - 1 9 ) eV = 13.6 8 9 eV E = 12.088 eV 12.09 eV . Answer: 12.09 eV