Calculate the wavelength of the light emitted when an electron in...

Question

Calculate the wavelength of the light emitted when an electron in a
hydrogen atom makes each transition and indicate the region of the
electromagnetic spectrum (infrared, visible, ultraviolet, etc.) where
the light is found.
a. n = 2¡n = 1
b. n = 3¡n = 1
c. n = 4¡n = 2
d. n = 5¡n = 2

Accepted Answer

hey everyone in this example, we need to determine the transitions of each wavelength observed from the atomic emission spectrum of hydrogen. So what we should initially recognize is that because they use the term emission, we want to recall what that means in regards to our electrons transitioning. So we want to recall that this end value here that we see in our answer choices represents energy levels specifically energy levels within our atom. So we can just say that keep that singular. So this is an energy level or shell within our atom. And we also want to recognize that the term emission is describing how our electron travels through these energy levels. And so we would recall that an emission electron transition is describing our electron traveling from a higher energy level to a lower energy level. We also want to recall our term absorption. Sorry, that should be. So when we think of an electron absorption transition, we would imagine our electron and our atom traveling from a lower energy level to a higher energy level. Now again, the question that we're given is asking in regards to a mission. So we want to look at the answer choices that start at a higher energy level and are transitioning to a lower energy level. So we can actually already rule out choices A and C. Due to the fact that that is acting the opposite from an emission are electrons and choices A. N. C. Start out in an absorption transition where they start out at the lower energy levels one And move up to energy levels eight and a and then energy level seven in c. And the other transitions in the spectrum follow the same pattern in A. N. C. So we can say that and see our electron absorption and I keep smelling it with a P but a B. So yeah, these are electron absorption. And so we can say therefore we rule out since this question is in concern about an electron emission. So we can rule out and see. And now we're just going to consider whether B and D. Are correct answers and which one is the most correct answer. And so what we're going to do next is think of two methods that we can use to solve this question. And the first is what we may have recalled from our notes known as the Bohr model. Now, in my opinion, I think that that is a bit more complex. Sometimes mathematically. So I think what we're going to review is actually our Rydberg equation to solve this example. Now, as you can see, they gave us three different wavelengths for our electron emission. And what we're going to do is we're going to utilize our Rydberg equation, which we should recall is going to be one over lambda, which is our wavelength, the denominator is our wavelength set equal to our Rensburg constant, which we recall is 1.97 times 10 to the seventh power inverse meters. And then this is multiplied by Z. We should recall that Z represents our atomic number of our element. And because this question mentions hydrogen, we're assuming that the atom were talking about is hydrogen. And so when we refer to our periodic tables, we see that hydrogen has atomic number one. So one is going to be in our numerator. And this is going to be placed over our denominator, which is going to be always our final electron transition represented by N F squared. So this is going to be in a subtraction where we're going to subtract another fraction beginning with Z and the numerator again, because we're still working with hydrogen, we're only working with hydrogen in this example. And then in the denominator we want to finish off with our initial electron transition and sorry I wrote final again. So initial is N I squared. And it's very important that you are attentive to set it up this way because we want to ensure that our results is positive. And so we should recall that infractions when we have an exponents or when we're doing a subtraction between two different fractions. We want to go ahead and start off with the fraction that is greater in magnitude meaning the denominator is a smaller number. So that is why we begin with our final value minus initial And so because we're comparing choices B to D to see whether either one is most correct To answer this question, we're going to just start off with testing for B. So in testing for choice B according to the Rydberg equation, what we're going to want to do is assume that our initial value for our electron emission is going to be equal to eight because B tells us that we go from energy level eight down to energy level one in the emission. And so we're going to assume that yes, eight is definitely going to be the initial emission level. So In assuming that our initial electron energy level is eight, we can also test for our wavelengths whether that's going to be equal to around the given wavelength that were these three, given wavelength up here, but we can just pick one of them. So we can just go with the 1st 1, 92.63 nanometers. We're going to test to see whether lambda, if we use eight as our initial energy level equals exactly or about To that value 92.63 nm. So let's go ahead and get started with the setup. So we're going to have the inverse fraction of our wavelength lambda. And this is set equal to r reed Zberg constant. We should recall our reads berg constant is again 1. 97 times 10 to the seventh power inverse meters. So this is our here and this is again multiplied by the difference between Z, which is our atomic number of our element hydrogen. That's one. And then in our denominator we want to include our final electron energy level, which according to our emission in B Is going to be energy level one. So that's going to be one squared in our denominator. This is then subtracted from our assumed initial energy level in our mission from B, which again is one for hydrogen, the numerator, the atomic number Divided by our initial assumed value of eight squared. And so now that we have this set up, we're going to go ahead and in our calculators find the quantity of this difference here and it should be a positive number because we've set it up the right way. We know that our first fraction has a greater magnitude than our second fraction over here because our second fraction has a higher denominator. So what we're going to have is inverse wavelength is equal to our reads berg constant in times two to the inverse meters, times the difference, which gives us a value of 20.98 43 75. And so our next step is to now go ahead and take the product of these two quantities are Rydberg constant times 20.98 43 75. And so what that's going to give us is one over lambda Equal to a value of 1.7, 86 Times 10 to the negative or times 10 to the positive seven, inverse meters. And so what we want to recognize is we have a diagonal here and we should recall that in algebra when we have diagonals, we can switch places with those two terms. So what we're going to do is take lambda and set that equal to the inverse value of our product here that we found above. So one point oh 7 86 times 10 to the seventh power inverse meters. And actually know our meters are no longer going to be inverse meters. They're just going to be standard meters. Now due to the fact that we took this quantity and made it into an inverse fraction. And so now meters is no longer in the denominator Or over under one. And so it's the opposite. Now it's just a standard meter value. So our next step is to find this quotient in our calculators to get our proposed wavelength. And so our wavelength is going to equal a value According to this calculation of 9.2604, 7 Times 10 to the negative 8th power meters. Now we want to recognize that our given wavelengths above our units of nanometers. And so this means we're going to want to follow a conversion factor to go from meters to nanometers. And we should recall that one m gives us 10 to the ninth power nanometers because our prefix nano is 10 to the ninth. This allows us to cancel our units of meters, leaving us with the proper units for this wavelength as nanometers. And so what we would have is that our wavelength, If our initial energy level is eight as proposed in choice B is equal to a value of 92. nm and so we're going to compare this to that first wavelength from the question Which is 92.63 nm and what we can conclude is that therefore this is actually because this is fairly close to 92.63 nm which we were given in the question, we can say therefore our initial energy level Is most likely eight. So let's go ahead and consider choice D. Before we just confirm B as our final answer. So this was for B. Now we're going to assume Indy that our initial energy level starts at According to D. They're saying that it should start at seven. So and I squared would equal seven. We're assuming So again, we're going to do the same test for the wavelength value. And let's just continue to compare it to this first wave length so that we can have a direct comparison to choice B. So what we're going to have is one over our wavelength equal to the reeds, berg constant 1.0 97 times 10 to the seventh. Power inverse meters, multiplied by our final energy level Where in the numerator we should have our atomic number of hydrogen from the periodic table. That's one over our final energy level according to choice D. Which is also one just like in B. So 1/1 squared. This is subtracted again from our initial assumed energy level, which according to choice D is going to be at the energy level seven. So that would be our atomic number for hydrogen. One in the numerator divided by Our assumed initial energy level seven squared. Now we're going to go ahead and take the difference between these two fractions. Again, our results should be positive because we are subtracting by the fraction that has the larger number in the denominator. And so what we're going to get is our inverse wavelength is equal to the Rydberg constant, Multiplied by the value of zero points 97, 95, 92. And so our next step is to take the product of these two quantities here being multiplied to one another. That gives us our inverse wavelength equal to 1.0 74 61 times 10 to the seventh power, inverse meters. And so again, we have a diagonal here so we can switch places with our wavelength. And that product we just found. And what we would have is that our wavelength is equal to the inverse value of 1.0 74 61 times 10 to the seventh power And now are inverse meters become standard meters because we have put them in another inverse fraction. And so what we're going to have next is the product of this quotient here, Which gives us a wavelength equal to a value of 9.3, 0 57 times 10 to the negative eighth power meters. And yet again, we want to end up in nanometers because we are given wavelengths in nanometers from the question. So we're going to do the conversion factor where one m gives us 10 to the ninth power nanometers. We cancel out meters, leaving us with nanometers. And this gives us wavelength equal to 93.057 nm. And so we're going to compare this to our first given wavelength. And we would say that this is not really as close to our given wavelength 92.63 nanometers as choice B was. And so we can say that therefore rule out D. And so this can conclude that our only correct choice to complete this example would be choice B. Due to the fact that it was our Rydberg equation in our assumed initial value of our Electron emission being eight, gave us a wavelength that was Pretty much almost approximate. Two are given wavelength in the example, 92.63 nm. And so this will complete this example as our final answer here. So I hope that everything we reviewed was clear. But if you have any questions, please leave them down below. And I will see everyone in the next practice video

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