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

Question

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

Suppose a hypothetical object has just four quantum states, with the following energies: -1.4 eV (third excited state) -1.9 eV (second excited state) -3.0 eV (first excited state) -4.8 eV (ground state) Suppose that material containing many such objects is hit with a beam of energetic electrons that ensures that there are always some objects in all of these states. What are the six energies of photons that could be strongly emitted by the material? (In actual quantum objects there are often "selection rules" that forbid certain emissions even though there is enough energy; assume that there are no such restrictions here.) List the photon emission energies in order from largest to smallest. If two different transitions would produce photons of the same energy, list that energy twice: Next, 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.

Explanation / Answer

Hi,

In this case we must remember some concepts from quantum processes.

- When an electron passes from a level with certain energy to a level with less energy, photons are emitted.

- When a photon collides with matter it gives energy to it. This energy can be used by electrons to pass from a fundamental state to a excited one.

- Excited states have more energy than the ground one.

- The relation between the energy of the photons emitted and the energy levels of the electrons is:

Ep = E2 - E1 ; where Ep is the photon's energy, while E2 and E1 are the energy levels (in this case, E2 is bigger than E1)

(a) The six energies correspond to the following transitions between levels:

Third excited state to ground one: E = -1.4 + 4.8 = 3.4 eV

Second excited state to ground one: E = -1.9 + 4.8 = 2.9 eV

First excited state to ground one: E = -3.0 + 4.8 = 1.8 eV

Third excited state to first excited state: E = -1.4 + 3.0 = 1.6 eV

Second excited state to first excited state: E = -1.9 + 3.0 = 1.1 eV

Third excited state to second excited state : E = -1.4 + 1.9 = 0.5 eV

(b) If all the objects are in the ground state and the absortion of exited states is not consider, then the only energies into account are the ones from the ground state to each excited state.

Ground state to Third excited state: E = 3.4 eV

Ground state to Second excited state: E = 2.9 eV

Ground state to First excited state: E = 1.8 eV

I hope it helps.

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