The figure shows the energy levels of a mercury atom. The atom is initially in the 2nd excited state. Which one of the following photon energies cannot be emitted by a mercury atom which is initially in its 2nd excited state?

Question

BlackTom AI · Accepted Answer

Photon energies equal the differences between permitted atomic energy levels, so only transitions from a higher level to a lower level (including cascades) can produce emitted photons.

1. Relevant knowledge points and concepts
- Energy levels given are measured above ground: ground = 0.0 eV, 1st excited = 4.9 eV, 2nd excited = 6.7 eV, 3rd excited = 8.8 eV.
- A photon emitted when an electron falls from energy E_high to E_low has energy E_photon = E_high − E_low (conservation of energy).
- Starting in the 2nd excited state (6.7 eV), the atom may emit photons by direct downward transitions or by cascades through intermediate levels.

2. Complete reasoning process step by step
- List all possible downward transitions available from the initial 2nd excited level (6.7 eV) and any subsequent cascades:
  a) Direct: 2nd (6.7 eV) → ground (0.0 eV)
  b) Direct/cascade: 2nd (6.7 eV) → 1st (4.9 eV), then 1st (4.9 eV) → ground (0.0 eV)

3. Calculations with intermediate steps
- For 2nd → 1st: E = 6.7 − 4.9 = 1.8 eV
- For 2nd → ground: E = 6.7 − 0.0 = 6.7 eV
- For 1st → ground (after cascade): E = 4.9 − 0.0 = 4.9 eV

4. Why the correct answer is correct
- The only photon energies that can be emitted starting from the 2nd excited state are 1.8 eV (direct fall to 1st), 6.7 eV (direct fall to ground), and 4.9 eV (emitted later if the atom first drops to 1st and then to ground).
- 8.8 eV would require a transition from the 3rd excited level (8.8 eV) down to ground, or some net drop of 8.8 eV. Starting at 6.7 eV there is no single allowed downward transition that yields 8.8 eV, nor can you emit a single photon of greater energy than the initial excitation. Therefore 8.8 eV cannot be emitted.

5. Why the other options are incorrect
- a) 1.8 eV: Correctly produced by 6.7 → 4.9 (calculated above).
- c) 4.9 eV: Produced by 1st → ground after a cascade (4.9 − 0 = 4.9 eV).
- d) 6.7 eV: Produced by direct 2nd → ground (6.7 − 0 = 6.7 eV).
- b) 8.8 eV: Impossible from an initial 6.7 eV state (no downward transition of 8.8 eV available).

6. Conclusion confirming final answer
- Therefore the photon energy that cannot be emitted when the atom starts in the 2nd excited state is 8.8 eV, matching the provided correct answer (b. 8.8 eV).

The figure shows the energy levels of a mercury atom. The atom is initially in the 2nd excited state. Which one of the following photon energies cannot be emitted by a mercury atom which is initially in its 2nd excited state?Single choice

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