An idealized atom has three available energy levels: -9.6 eV, -4.8 eV and -1.6 eV; the lowest is the ground state. An excited atom cascades from the highest to the middle level, then to the ground state. Use h = 6.626 × 10⁻³⁴ J s, c = 3.00 × 10⁸ m s⁻¹ and 1 eV = 1.602 × 10⁻¹⁹ J. Find both emitted photon wavelengths and the wavelength for a direct highest-to-ground transition. Identify the two photon energies that ground-state atoms can absorb within this three-level model, and explain why cascade wavelengths do not add to the direct wavelength.

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Cascade wavelengths 3.878 × 10^-7 m and 2.585 × 10^-7 m; direct wavelength 1.551 × 10^-7 m. Ground-state absorption energies 4.8 eV and 8 eV. A photon carries the difference between the initial and final atomic energies. The cascade energy gaps are E_high − E_middle = 3.2 eV and E_middle − E_ground = 4.8 eV, both positive. Convert each energy to joules and use λ = hc/E_photon. The first wavelength is 3.878 × 10^-7 m and the second is 2.585 × 10^-7 m. The direct gap is 8 eV, so its wavelength is 1.551 × 10^-7 m. This photon has the greatest energy and therefore the shortest wavelength of the three. An atom initially in the ground state can absorb 4.8 eV to reach the middle level or 8 eV to reach the highest level. The 3.2 eV transition requires an atom already in the middle level. Energy is additive through the cascade, but wavelength is inversely proportional to photon energy. Thus E_direct = E_first + E_second, while 1/λ_direct = 1/λ_first + 1/λ_second; wavelengths themselves do not add.

Photon energies in an atomic cascade · set 6

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