Question

An energy-level diagram for a hypothetical atom is shown above.

a. Determine the frequency of the lowest energy photon that could ionize the atom, initially in its ground state.

b. Assume the atom has been excited to the state at -1.0 electron volt.

i. Determine the wavelength of the photon for each possible spontaneous transition.

ii. Which, if any, of these wavelengths are in the visible range?

c. Assume the atom is initially in the ground state. Show on the following diagram the possible transitions from the ground state when the atom is irradiated with electromagnetic radiation of wavelengths ranging continuously from 2.5 x 10–7 meter to 10.0 x 10–7 meter.

Answer 1

(a) E = hf … 5.5 = (4.14x10–15 eV-*) f … f = 1.33x1015 Hz

(b) i. From the –1eV state, the following transitions could happen –1ev → –3eV → –5.5 →, or –1ev → –5.5eV, three different possible energy differences: 2eV, 2.5 eV and 4.5 eV Using E = hc / λ … with hc = 1240 eV-nm, and the energies above give the following wavelengths λ1 = 621 nm, λ2 = 497 nm λ1 = 276 nm

ii. The visible range is 400–700 nm, so the 1st and 2nd wavelengths are in that range.

(c) The energies corresponding to the given wavelength are found with E = hc / λ and are 1.24–4.97 eV. This range allows a transition to the first and second excited state but does not allow ionization because its not enough energy for ionization.