The visible emission lines observed by Balmer all involved nf = 2. (b) Calculate the wavelengths of the first three lines in the Balmer series—those for which ni = 3, 4, and 5—and identify these lines in the emission spectrum shown in Figure 6.11.

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Hey everyone in this example, we're given the following Balmer series where the final energy level is the second energy level and we need to calculate wavelength in nanometers for the given emission line. So what we want to recall our first two different formulas for our first formula, we would recall the calculation for inverse wavelength, which is found from taking our Rydberg constant and multiplying by one over our final energy level squared minus one over our initial energy level squared. We then want to calculate the or recall our formula for the energy of a photon and we would recall that this is equal to the speed of light, multiplied by or sorry, Plank's constant, which is H multiplied by the speed of light and divided by our wavelength. We should recognize that because we're given a Balmer series, our wavelength occurs at visible light on our electromagnetic spectrum. And so what we can do now that we have recalled these two formulas is reinterpret for one formula. So that wavelength is no longer in the inverse and we're going to have that wavelength is equal to Planck's constant, multiplied by the speed of light divided by the negative value of our Rydberg constant. This is now then multiplied by one over because we had to take our wavelength out of the inverse. So 1/1 Over our final energy level squared - over our initial energy level squared. So we're going to solve for our wavelength at the initial energy level being equal to six. And what we should recognize is that because it's equal to six, we can say our initial energy level is greater than our final energy level. And so when we're plugging in for our this part of our equation, we're just going to put the initial energy level first subtracted from the final energy level here. So these are going to be switched and you'll see here below. So what we should have is that our wavelength is equal to Planck's constant, which we would recall is equal to a value of 6.626 times 10 to the negative 34th power jules times seconds. We're then recalling our speed of light which is three times 10 to the eighth. Power meters per second. And then we're going to divide by negative one times our rights burg constant, which we recall is 2.18 Times 10 to the -18 Power Jewels. We're then going to multiply this by 1/1 over our initial energy level which is given in the prompt as six squared. Then we're going to subtract this from one over our final energy level given in the prompt as to and we're going to square this as well. So what we're going to have when we simplify this is that our wavelength is equal to in our numerator. We're going to get the product of the speed of light. Times are plank's constant, which is 1.987, 8 times 10 to the negative 25th power meters, we're left with meters because we're able to cancel out jewels as well as our units of seconds. And then in our denominator we're going to have negative one times our Rydberg constant, which gives us negative 2.18 times 10 to the negative 18th power. And then this is multiplied by the value of our fraction on the right hand side, which should give us a value of -4.5. And so what we're going to simplify too is now that the wavelength is equal to a value of negative or positive for rather 0.103 to six times 10 to the negative seventh power. And we're left with units of meters. But the prompt says to give wavelength in nanometers. So we're going to do a conversion factor where we would recall that our prefix nano tells the stuff for one nanometer. We have 10 to the negative ninth power meters. And so now we're able to cancel our units of meters. Leaving us with nanometers. And what we're going to get for our wavelength is a value of 410.32 nanometers at the initial energy level being six in our emission line. And when we look at our electromagnetic spectrum we would see that this is going to be corresponding or this wavelength is going to correspond to the visible light spectrum where we would probably see violet light emitting from our photons. So moving on to the next emission line, we have that our initial energy level is now equal to seven, which is still greater than the final energy level at two. So what we're going to have is that our wavelength is equal to in our numerator, we have the product of our plank's constant times the speed of light, which above was 1.9878 times 10 to the negative 25th power meters. And in our denominator we have our negative value of our Rydberg constant 2.18 times 10 to the negative 18th power. And then now we're just going to multiply by a different fraction here so that we have one over 1/7 squared minus 1/2 squared, which was our final energy level here. So this is what we're going to be calculating for now. And what we should have is that our wavelength is equal to 3.97155 times 10 to the negative seventh power. We have units of meters and we're going to convert two nanometers again. So we would recall again, one nanometer is 10 to the negative nine power meters. We cancel out meters were left with nanometers for wavelength and what we should get here is a wavelength of 397 nm. So this would be our second answer when our initial energy level is being at seven in our mission line. Our first answer was 410 nm when our initial energy level was six. And now we're going to calculate for the last emission line where our initial energy level is now equal to eight. So we're going to follow the same steps. We would have that wavelength lambda is equal to. We still have the same fraction here. So 1.9878 times 10 to the negative 25th power meters. When we multiply the speed of light by our plank's constant. And then in our denominator we have negative one times Rydberg constant, giving negative 2.18 times 10 to the negative 18th power. Now we're multiplying by 1/1 over eight squared initial energy level here minus 1/2 squared our final energy level. And what we should get is that our wavelength is equal to 3.8905 times 10 to the negative seventh power meters. Which will then convert two nanometers just like before. Where one nanometer is equal to 10 to the negative ninth power meters. And this gives us our final wave length Equal to a value of 389 nm. And so this would be our final answer Where our initial energy level and our mission line stops at the value eight or the energy level eight. So everything boxed in here represents our final answers for this example. So I hope that everything I went through is clear. If you have any questions, please leave them down below, and I will see everyone in the next practice video.

The visible emission lines observed by Balmer all involved nf = 2. (b) Calculate the wavelengths of the first three lines in the Balmer series—those for which ni = 3, 4, and 5—and identify these lines in the emission spectrum shown in Figure 6.11.

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Key Concepts

Balmer Series

Rydberg Formula

Emission Spectrum