Find the longest and shortest wavelengths in the Lyman and Paschen series for hydrogen. In what region of the electromagnetic spectrum does each series lie?

Video transcript

Welcome back, everyone. We are making observations about a doubly ionized lithium atom here. And we are tasked with finding what is going to be the shortest wavelength and the longest wavelengths in both the Lyman and Pasin series here. And we are also told to specify the region of the electromagnetic spectrum in which the wavelengths appear. I'm gonna scroll down here just a little bit to give us some room for our calculations. Let's go and start off with the Lyman series here. So for the Liman series, we have the following equation, we have one over our wavelength is equal to our lithium Ryberg constant times 1/1 squared. Since N equals one is going to be our final excited state minus one over N squared. But before using our equation here, let's go ahead and calculate our lithium ride beg constant. We have that our lithium rider constant is going to be equal to our atomic number times our hydrogen ride be constant. What this gives us is three squared times one point oh times 10 to the seventh which when we plug into our calculator gives us 9.873 times 10 to the seventh here. Now let's go ahead and dive into finding our shortest wavelength, our shortest wavelength in our Liman series. For our lemon series, we are going to start from an excited state of N equals infinity and go all the way down to N equals one following our equation here. What we get is one over lambda is equal to 9.873 times 10 to the seventh times 1/1 squared minus one over infinity squared. This is equal to 9.927 times 10 to the seventh. And solving for Lambda, we get 10.1 nanometers. Wonderful. So now let's go ahead in our lineman series, solve for our longest wavelength delta L or Lambda L. And for this, we are going to go from N equals two, the closest excited state to N equals one. Following the same equation, we have one over Lambda is equal to 9.873 times 10 to the seventh times 1/1 squared minus 1/2 squared. Plugging this into our calculator. What we get is 7.4 oh five times 10 to the seventh meters to the negative first power which solving for Lambda gives us 13.5 nanometers. Wonderful. So now we are going to switch over to the posh series. So for the Poin series, we have the following formula here. It's very similar. We have one over Lambda is equal to our lithium rider constant, our lithium rider constant times 1/3 squared. Since our final excited state will be three minus one over N squared. Let's once again, start off with our shortest wavelength here, shortest wavelength denoted by Lambda S. This and we are still going from N equals infinity all the way down to N equals three. We have one over Lambda is equal to 9.873 times 10 to the seventh meters to the negative first power times one divided by three squared minus one over infinity squared. Plugging this into our calculator. What we get is one point oh 97 times 10 to the seventh meters to the negative first power which solving for Lambda gives us 91.2 nanometers. Wonderful. And finally, we are moving on to the longest wavelength in the peine series going from the closest excited state of N equals four to N equals three. We have one over Lambda is equal two. And let me scroll down here just a little bit one over Lambda is equal to 9.873 times 10 to the seventh times meters to the negative first power times 1/3 squared minus 1/4 squared. What this gives us when we plug it into our calculator is 4.799 times to the six meters to the negative first power which solving for Lambda gives us two oh eight nanometers Now, if you observe all of our wavelengths here, you can see that all wavelengths will be in the U V region here. So with that in mind, we have found for both the Lyman and pushing series, the shortest and longest wavelengths corresponding to our final answer. Choice of a. Thank you all so much for watching. I hope this video helped. We will see you all in the next one.