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Suppose a hypothetical object has just four quantum states, with the following energies:

-0.3 eV (third excited state)
-1.0 eV (second excited state)
-2.4 eV (first excited state)
-4.7 eV (ground state)

suppose that the beam of electrons is shut off so that all of the objects are in the ground state almost all the time. If electromagnetic radiation with a wide range of energies is passed through the material, what will be the three energies of photons corresponding to missing ("dark") lines in the spectrum? Remember that there is hardly any absorption from excited states, because emission from an excited state happens very quickly, so there is never a significant number of objects in an excited state. Assume that the detector is sensitive to a wide range of photon energies, not just energies in the visible region. List the dark-line energies in order from largest to smallest.
eV (largest)
eV
eV (smallest)

Respuesta :

Answer:

Explanation:

In this problem we can use Bohr's atomic model, to deal with the electronic transition, so we can have transitions between two given states.

           ΔE = [tex]E_{f}[/tex] -E₀

Lower state     final state                     energy

Fundamental      first excited state          ΔE₁ = -2.4 - (-4.7)  = 2.3 eV

Fundamental      second excited state   ΔE₂ = -1.0 - (-4.7) = 3.7 eV

Fundamental      third excited state        ΔE₃ = -0.3 - (-4.7) = 4.4 eV

As they indicate that there are no electrons in the excited states these are the only possible transitions.

When a wide range of light strikes, the frequencies of these energies are absorbed and observed as black bands (absence of radiation) in the spectrum.

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