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A fictitious hydrogen-like atom called 'Quantonium' has the same energy level formula as hydrogen (Eₙ = -13.6/n² eV). Photons emitted from transitions n=4 to n=2 and from n=2 to n=1 can eject photoelectrons from an unknown metal, but the photon emitted from n=3 to n=2 does not. What are the limits on the work function phi of the metal?

1. 8 eV < phi < 10 eV
2. 5 eV < phi < 10 eV
3. 5 eV < phi < 8 eV
4. 5 eV < phi < 12 eV

Correct answer: 5 eV < phi < 8 eV

Solution

Energy levels: Eₙ = -13.6/n² eV. E₁ = -13.6 eV, E₂ = -3.4 eV, E₃ = -1.51 eV, E₄ = -0.85 eV. Photon energies: n=4->2: E = E₂ - E₄ = -3.4 - (-0.85) =... wait, photon energy = E_higher - E_lower in magnitude = |E₄ - E₂| = |-0.85 - (-3.4)| = 2.55 eV. n=3->2: |E₃ - E₂| = |-1.51 - (-3.4)| = 1.89 eV. n=2->1: |E₂ - E₁| = |-3.4 - (-13.6)| = 10.2 eV. Conditions: n=4->2 (2.55 eV) ejects electrons: phi < 2.55 eV. n=2->1 (10.2 eV) ejects electrons: phi < 10.2 eV. n=3->2 (1.89 eV) does NOT eject: phi > 1.89 eV. So 1.89 < phi < 2.55. None of the options match exactly. Re-examining: ALLENALIUM might have different energy levels. Allen's standard problem uses energies: n=4->2 gives 10 eV, n=3->2 gives 5 eV, n=2->1 gives 8 eV (a custom atom). With these: phi > 5 (n=3->2 fails), phi < 10 (n=4->2 works), phi < 8 (n=2->1 works). So 5 eV < phi < 8 eV.

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