Consider the three electronic transitions in a hydrogen atom shown here, labeled A, B, and C. (c) Calculate the wavelength of the photon emitted for each transition. Do any of these transitions lead to the emission of visible light? If so which one(s)?

Calculate the wavelength of the photon emitted for transition B.

Video transcript

Welcome back everyone labeled A, B, C and D. In the diagram below are four electronic transitions in a hydrogen atom. Find the wavelength in nanometers of the photon emitted for A. B. C and D, identify which transitions leads or lead to the emission of visible light. So based on transition A we can see that our initial energy level is an equals three and our final energy level is an equals two. Because the arrow points down to n equals two. We want to recall that for this example, we're going to need to utilize our Rydberg equation, in which we have our inverse wavelength represented by one over lambda set equal to our Rydberg constant. Our sub lambda, which is multiplied by the following bracket where we have one over our final energy level squared subtracted from one over our initial energy level squared and it should always be final minus initial so that the value in our bracket is positive. Recall that for our unit of wavelength we should have units in meters. So now with us outlined, let's go to transition A. So for calculating our wavelength for transition A. We have our Rydberg equation one over lambda equal to our Rydberg constant, which we should recall is the value 1.0 97 times 10 to the seventh Power inverse meters from our textbooks. This is multiplied by our brackets where we have one over our final energy level, which we agreed is the second energy level squared subtracted from one over our initial energy level, which we agreed is three and this is squared in the denominator. So that finishes off our brackets. And we can simplify this in the next line to our Rydberg constant, which is multiplied by the resultant of our brackets being 1/4 subtracted from 1 9th. And so in our next line we'll have the product of our Rydberg constant multiplied by 0. 8889 as the result of our brackets. And from this product here we would get a value of 1.52361 times 10 to the sixth power. Notice that our units are still inverse meters. Now we can isolate wavelength fully by having one divided by our result from above 1.52361 times 10 to the six power inverse meters. And this results in our wavelength equal to 6.56336 times 10 to this negative seventh power. And our units are meters but I'd like to make a correction. So in our notes for the Rydberg equation, our units for wavelength should be in nanometers. And so we're going to need to convert from meters in the denominator two nanometers in the numerator, recall that our prefix nano tells us that for one m we have an equivalent of 10 to the ninth power nanometers. So canceling out meters were left with nanometers as our final unit for wavelength. And this will result in our wavelength equal to a value in our next line of 656.3 nanometers. As our wavelength of the photon emitted from the transition defined by a. So this would be our first answer so far. Moving on to our next transition defined by transition B. Which is here we have the initial energy level being an I equal four And our final energy level at N. equals three. So let's do our work for transition be below. We have again the same equation, one over r inverse wavelength equal to our Rydberg constant, which is the same as before. This is multiplied by our brackets. Where we have now won over our final energy level for transition be now being the third energy level squared, subtracted from one over our initial energy level for transition B being the fourth energy level squared. So this finishes off our brackets. In our next line we're going to simplify our brackets so that we have. So in our brackets we have 1/9 subtracted from 1/16. This is going to give us our product between our Rydberg constant, Multiplied by the result of our difference in our brackets being a value of 0.048611. So taking the product between our Rydberg constant and that term we have a value of (533-63. inverse meters. Where we can isolate wavelength now by taking one divided by that value 533 to 63.8889 inverse meters. This gives us our wavelength equal to 1.875244 times 10 to the negative six power meters. Which we will convert to nanometers in the numerator by putting meters in the denominator where one m is equivalent to 10 to the ninth Power nanometers, canceling out meters were left with nanometers. We get our final wavelength for transition be equal to and I'll actually write that in the next line so that it's clear, 1875 nm. So this is our second answer so far for our wavelength of our photon for transition. B Our first answer was for transition. A Now let's move on to transition. See so going to our chart, we have transition. See here in blue where our initial energy level is at an equals five and our final energy level is at n equals four. So we'll do the work below. Following the same process. We have our inverse wavelength equal to Rydberg constant. Which all right in quickly, one point oh 97 times 10 to the seventh Power inverse meters multiplied by our brackets where we have one over our initial energy level for transition. See now being and sorry correction, we have our final energy level which were writing in first, which according to our diagram was four squared. This is subtracted from one over our initial energy level, which from the chart is given as five. and this is squared. So finishing off our brackets in our next line, we have our Rydberg constant still Multiplied by the following fractions. Where we have 1/16 subtracted from 1/25. And so now we would just take the difference in our brackets. So we still have our Rydberg constant, Multiplied by the resulting difference of the brackets being a value of 0.2, sorry, 0.0 2 to 5. So taking the product between this number and our Rydberg constant, we get a value of 246825 inverse meters. As our units. We can now isolate wavelength by taking one divided by this value 2468 to 5 inverse meters. And this results in a wavelength equal to a value of 4.51453 times 10 to the negative. Six power inverse meters. Or sorry, no longer inverse meters. Since we're not in a fraction, we have standard meters converting from our conversion factor. Where meters are in the denominator, nanometers are in the numerator. We have 10 to the ninth power nanometers equivalent to one m canceling out meters were left with nanometers as our unit of wavelength and we get a wavelength equal to about 4051 nanometers. This is our third answer. So far as our wavelength of our photon for transition C. And now we just have our last transition being transitioned D. Given in green. So for transition D our initial energy level is at the sixth energy level and our final energy level is at the fifth energy level. Let's do the work below for our last transition. So for a transition dino we have again the same formula, our Rydberg equation, one over our inverse wavelength equal to our Rydberg constant, 1.97 times 10 to the seventh power inverse meters multiplied by our brackets where we have for our or rather our final energy level squared in our denominator for transition D. That is the fifth energy level squared as our final energy level, subtracted from one over our initial energy level six, which is squared for transition D. That completes our brackets. We would simplify to the following where we have 1/25 subtracted from 1/36. And so taking the difference between our fractions and the brackets, we would have in our next line the following results, which we're multiplying by as 0.1 to to two. And so taking the product here to our Rydberg constant, we get a value of 134075.34. Our units are inverse meters. And we need to isolate wavelength. So we would take one divided by that value 134075.34 inverse meters. This is going to equal a wavelength of 7. 311 times 10 to the negative six power meters, which we multiplied by the conversion factor meters. In the denominator nanometers. In the numerator, one m is equivalent to 10 to the ninth power nanometers. We cancel out the units of meters were left with nanometers and we get a wavelength in nanometers equal to about nanometers. And this would be our fourth final answer as the wavelength of our photon for transition D. And so all of our final answers, highlighted in yellow, will correspond to choice D and the multiple choice as our final answers. So I hope that this made sense. And let us know if you have any questions.