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An electron in chromium moves from the $n=2$ state to the $n=1$ state without emitting a photon. Instead, the excess energy is transferred to an outer electron (one in the $n=4$ state), which is then ejected by the atom. In this Auger (pronounced "ohjay") process, the ejected electron is referred to as an Auger electron. (a) Find the change in energy associated with the transition from $n=2$ into the vacant $n=1$ state using Bohr theory. Assume only one electron in the $\mathrm{K}$ shell is shielding part of the nuclear charge. (b) Find the energy needed to ionize an $n=4$ electron, assuming 22 electrons shield the nucleus. (c) Find the kinetic energy of the ejected (Auger) electron. (All answers should be in electron volts.)

a) 5400 $\mathrm{eV}$

b) 3.40 $\mathrm{eV}$

c) 5400 $\mathrm{cV}$

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Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

University of Washington

Hope College

In this exercise, we have a chromium Adam that has, Ah, an atomic number Z 0 24 And in the Sadam, we have an Elektrim that is originally in the second energy level and transitions to the 1st 1 But instead of admitting a Fulton in this process, the energy, um, the energy that the F D election lost goes to Ah, an electron that is in an outer shell in this case, in the show and equals four and iron eyes is the be Adam. Ah. With this in mind, we have in question and you have to answer what is the energy associate ID with the transition from the second the from the second energy level to the 1st 1 and we have to consider that Ah, for the second energy level for the l show, there's only one electron shooting. Be the show. Okay. Ah, So if there is only one electron shooting, then the effective atomic number Z f z e f f is equal to the atomic number minus one. So this is just 23 and the energy off the second level is minus ZF squared times 13.6, like from votes divided by to square, which is four. Okay, so this is Ah, minus 23 squared times 13.6, divided by four. So this is just minus 1800 election votes, and the energy of the first level is minus is you have square. So that's minus 23 squared times 13.6 silicon votes. And this is minus 7000 194 electron votes. Okay, So the energy associated with the position is the difference in energy between be energy of the second level in the energy of the 1st 1 And this is approximately approximately 500. I'm sorry. Uh, 5400 electoral votes. Okay, now to the second. In the second question question be, they have to estimate the ionization energy for the fourth energy level. So in order to do that, I'm gonna under estimate the energy off the fourth energy level. And we have to suppose here that there are 22 electrons that are shooting the the fourth shell. So the effective atomic number is Z minus 22. So this is 24 minutes 22. Which is it with it, too. So the energy of the fourth level ISS minus Z F square times 13.6 electoral votes divided by foursquare, which is 16. So this is minus four squared minus two squared, which is four times 13.6, divided by 16 and this is just minus 3.4 electoral votes. So if the energy of the fourth level is minus, people are going for that promotes the ionizing energy. The energy required to remove in electric is 3.4 electron books. This is the ionizing energy and in question. See, you have to find what is the kinetic energy of the electric that is omitted from the fourth level, when the fourth level is ionized and from conservation of energy, we know that the kinetic energy is just don t. That is the difference between the second level, the energy of a second level in the 1st 1 That's the energy that the electric lost when the transition from the second level to the first minus the energy required to lionize the fourth level. So that's e I four. So the kinetic energy is 500. Yeah, I'm sort of 5400 electoral votes minus 3.4 the vote. Okay. And this is still approximately 5400 election votes. Cake? Uh, I'm making this approximation because I did an approximation similar to this before. So for consistency, we can consider that the kinetic energy of the electron nous approximately approximately 5 5400

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