An atomic emission spectrum of hydrogen shows three
wavelengths:
...

Question

An atomic emission spectrum of hydrogen shows three
wavelengths:
121.5 nm, 102.6 nm, and 97.23 nm. Assign these
wavelengths to transitions in the hydrogen atom.

Accepted Answer

Hey everyone, welcome back. So let's get started with this video. So here we're told three wavelengths are observed from an atomic emission spectrum of hydrogen. There are 97.23 nm. 102.6 nm and 94.96 nm. What is their corresponding electronic transition? Okay, so here We are dealing with wavelengths that are all around 100 nm. Therefore we're talking about the Lyman series and this is from And that's greater than one. Going to end Is equal to one. So then we know the final And so what we're going to have to figure out is the initial end. And then here we have a mission. So that means it's going to go from high energy level too low. So then it makes sense that our final end is one because we're starting off From a high one. So then we're going to use to write the Rydberg equation, which is one over wavelength is equal to R which is Rydberg constant times e squared, which is the atomic number times one over and final squared -1 over an initial squared. Okay, so then the Rydberg constant has a unit in meters. So the wavelengths were given use nanometers so that our first step is going to have to convert those nanometers two m. So then let's start with the first one, which was 97 point 23 nanometers. And then one nanometer Is equal to 10 to the - meters. And then the meters cancel out. And we get nine points 7, 2 times 10 To the -8 m. Okay, our 2nd 1 is 102 point six nanometers times one nanometer is equal to 10 to the negative nine m nanometers cancel out. And we get 1.03 times To the -7 m. And then the last one is 94 . nm times one nanometer, it's equal to 10 to the negative nine m nanometers canceled out. And we get 9.50 times to the negative eight meters. Okay, So then now we're able to plug in our values to the equation then it color. Okay, So then let's start with one. So we're going to have one overweight blinks, Which is going to be 9.7.2 Times 10 to the negative eight m is equal to R. Which is the Rydberg constant of one point 097 Times 10 to the seven meters. Universe Times E. Which is going to be one because as the atomic number of hydrogen and it's going to be squared over one. We said we know and final, which is one because it's right here minus one over and in the show squared and this is what we're going to be solving for. So then whenever we solve for an initial Continak it that it's going to be four. Okay, so then let's do it for the 2nd 1. one over wavelength. The wavelength is one 03 times 10 to the -7 meters Is equal to R. Which is 1.097 Times 10 to the seven meters in verse. Once again atomic number of hydrogen is one squared one over and final, which is one -1 over an initial squared we saw for an initial and we get three. Okay. And then finally for the 3rd 1 We have one over wavelength which is 9.50 Times 10 to the - meters Is equal to R. Which is 1.097 times 10 to the seven meters in verse Times. Atomic number of one squared over one and finals one squared minus one over an initial squared we saw for an initial And we get five. So then for number one it's going to be going from N 4 to end one. So then this one is 4-1. This one's 1-4. So it's not going to be big one. no, 321. No. Okay. So then the second one is going to go from And is equal to 32 and is equal to one. So 321. Okay. And then finally And is equal to 5 to N is equal to one. So then answer choice A is going to be our answer. Thank you for watching. I hope this helped. And I'll see you in the next video

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