Video Transcript
The diagram shows an atom in the target material of a Coolidge tube that generates X-rays. An electron in the electron beam used in the tube ejects an electron in the K-shell of the atom and is scattered. Either the high-energy-level electron or the L-shell electron can transition to the K-shell. Which electron would produce a photon that would be part of a characteristic line of the spectrum with an energy closer to the maximum energy value of the spectrum?
Before we review the answers, let’s clear some space by only including the last sentence of this question. Now then we’re going to look at the four electrons in this diagram that could possibly produce a photon. These are (A) the ejected electron, (B) the scattered electron, (C) the high-energy-level electron, and (D) the electron in the L-shell. We also have the option of (E) all of the electrons.
Now all of these electrons can produce a photon, but the methods by which they do it will differ. The ejected electron and scattered electron are moving outside of the atom, meaning that they will produce Bremsstrahlung or braking radiation as they slow down. The photons produced through such a method vary heavily in their energy. Though some energies are more common. On a spectrum of X-ray energy versus intensity, like this one here, Bremsstrahlung produces a smooth curve with middling energy values having higher intensity, meaning that they are more common.
But we are not looking for a smooth curve. We’re looking for a characteristic line of the spectrum, which means that we are looking for distinct sharp peaks in the spectrum like this, which are not produced by Bremsstrahlung. These peaks are only produced when an electron in a higher energy level transitions downwards to a lower energy level, meaning that only the electron in the L-shell or the high-energy-level electron will be able to produce these peaks, which they do in a process called energy level transition.
The reason that this method produces such sharp peaks is because the energy of the photons produced using this method have the exact same energy as the difference in electron energy levels between the electron shells of this atom, which on a spectrum looks like a peak because we only have a very specific value of energy rather than a potential range of values. We can see then that since the ejected electron and the scattered electron use Bremsstrahlung rather than an energy level transition, they will not produce a characteristic line, meaning that they cannot be the correct answers. And by extension, (E) cannot be it either.
To determine now whether it is (C) or (D), we have to look at the other part of the question, which is “Which characteristic line will have an energy closer to the maximum energy value of the spectrum?” which means on the spectrum that we have over here, which characteristic line will be further to the right, indicating a higher X-ray energy?
As previously noted, the energy of a photon produced using an energy level transition is equal to the difference in electron energy levels that the electron transitions through, which means that an electron transitioning downwards from a higher energy level will produce a photon with higher energy. Therefore, when we look at the characteristic lines of a spectrum, the one that has the higher X-ray energy closer to the right of the spectrum, and thus closer to the maximum energy value of the spectrum, would be produced by the high-energy-level electron since it is in a higher energy level than the L-shell and they’re both transitioning down to the same place, the K-shell.
Therefore, the electron that would produce a photon that would be part of a characteristic line of the spectrum with an energy closer to the maximum energy value of the spectrum would be (C), the high-energy-level electron.
