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Consider a large number of hydrogen atoms with electrons randomly distributed in the n = 1, 2, 3, and 4 orbits.(a) How many different wavelengths of light are emitted by these atoms as the electrons fall into lower-energy orbitals?(b) Calculate the lowest and highest energies of light produced by the transitions described in part (a).(c) Calculate the frequencies and wavelengths of the light produced by the transitions described in part (b).

Chemistry 101

Chapter 6

Electronic Structure and Periodic Properties of Elements

Electronic Structure

Periodic Table properties

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for this problem, whereas to consider a pool of hydrogen atoms that have electrons somewhere between the n Igel want an equal for energy level. And we want to consider all of the possible electron transitions that involve going from higher energy level to a lower energy level between the unequal one of the people. For I want to think about which of those transfers going to have the highest energy, the lowest energy and then what frequencies? We're going to be associated with each of those changes. So to do that, I have a hydrogen atom drawn here, and I have tried Thio represent the fact that as you go up on end on these energy levels, the spacing between the energy level starts to decrease the distance between the N Igel one and the angle to energy level is a little bit wider than the distance between the nickel two and three and three and four and so forth. That's something to keep in mind, uh, everything through the logic of this question. So the first thing we need to do is to think about how many different possible transition there there are in this, uh, Adam. So I want to represent that with a rough table here. So I've drawn the four possible energy levels across the top as well as the four possible energy levels. A crossed the column here and across the top will consider that to be our starting point. And the column will consider to be our ending point. So the dots across the diagonal represent the fact that if we go from the unequal one to the nickel one, we haven't made a change at all. We're just going to ignore the lower half of this table. So from this, we can see that the possible energy transfer's going to the antique one. We're gonna be from 23 and four. We can also go from 3 to 2 and 4 to 2, and we can also go from 4 to 3. So using these different energy transfers and looking at the diagram that I've drawn up here, the largest gap in ah distance is going to represent the largest energy change. In other words, we go from the n equal four to the in eagle one that's going to represent our highest difference and energy. So the highest amount of energy is gonna be admitted when we go from the end, Uncle Ford of the ethical one. Conversely, the narrowest energy change is going to be represented by the shortest distance traveled in this case, also starting it for but going to three. So now that we've identified our two energy levels, we can go through and calculate the energy associated with those transfers. We're gonna do that using our Rydberg equation. So in this case, for our high energy change, it's going to be equal to negative 2.18 times 10 to the negative 18 Jules times a relationship of the energy levels. So in this case, we're gonna do one divided by the final end squared in this case is won. We're not subtract from that one divided by the starting ends. Weird. So the internal fraction is gonna be one minus one over 16. When we multiply this through, what we get is that the transfer, the amount of energy released as an electron transitions from the N ical four to the n equal one is negative. 2.4 times 10 to the negative 18th Jules. So to convert this value into a frequency we can use one of our standard energy equation. Z equals planks constant times the frequency. If we rearrange that unsolved for frequency, we're going to take the energy that we just calculated. Divide that by planks, constant. And what we see is that the frequency associated with this, uh, electron transfer between energy levels is 3.8 times 10 to the 15th hurts. So now, using our other well known light equation where the speed of light is equal to the wavelength times the frequency, we can take this frequency that we just calculated and rearrange the speed of light equation to calculate the wavelength associated with that. So for our high energy electron transfer, that's going to be the speed of light divided by the frequency that we just calculated and the value that we're going to get out of. That a 7.89 times, 10 to the negative eight meters. We could do the same principle in the same flow through of equations for our low energy transfer. So call for a lower your insurance. We were going from an equal Florida and equal three. So we're just gonna set up a Rydberg equation the same way that we did before, except this time, our fractions air going to reflect the new ending value, which in this case, we're going to stop it and equal free. So we have one invited by three squared minus one, vital by force. Where or 19 minus 1/16 when we multiply this all the way through. We see that the energy associated with this electron change is negative one point of six times 10 to the negative 19 Jules, which is less than what we had in our high energy change. So that's good. That's kind of a self check that you can do. So now that we have this energy calculated, we can go through the same steps we did four and use our two light equations to calculate frequency and the way life of this light or the light emitted by this change. So same way we did before. We're going to take that energy associated with that transfer. Divide that by planks constant, and we'll find that the frequency of the light emitted by this electron transfers would be 1.6 times 10 to the 14th hurts. We can now take that value and use that to solve our wavelength. The speed of light Divide that by that frequency. What we find is that the wavelength this late and low energy gonna be 1.7 times 10 to the negative six meters.

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