First Lines, Series Limits and Ratios

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. The Lyman series of hydrogen has infinitely many lines: two to one, three to one, four to one, and so on, forever. And every new line has a shorter wavelength than the one before. So if the lines never stop, the wavelength should keep shrinking, all the way down to zero, right? But no, it doesn't! There is a wall. Below a certain wavelength, not one single line. Today we find that wall, and it takes just one line of maths. And look, in JEE, first lines, series limits and ratios are quick, sure marks, if you know exactly which jump to take. So, let's begin.

0:46The longest and the shortest line of a series

So, first a quick recall: the Rydberg formula. Nu bar, the Greek nu with a bar on top, is the wavenumber: one over lambda, the wavelength. It equals R H, the Rydberg constant, the R with a small H below, times Z squared, where Z is the atomic number, times one over n one squared, minus one over n two squared. Here n one is the lower level, where the electron lands, and n two is the upper level it falls from. Okay? Now look at the picture. Every arrow is one line of the spectrum. The blue arrows all land on n equals one: that is the Lyman series. The orange ones land on n equals two, the Balmer series, and the green ones land on n equals three, the Paschen series. So, one landing level, one series. Easy, isn't it? Now tell me. In one series, which line has the longest wavelength? The longest arrow, right? Biggest jump, longest wavelength? No, no, wait! Think energy. Long wavelength means a small nu bar, and that means the smallest energy, so the smallest jump. That is n one plus one, down to n one; the arrow means the jump. For Lyman, two to one: the shortest blue arrow in the picture. We call it the first line of the series. Got it? Alright, now the other end. What is the biggest possible jump that can land on n one? Yes, from n two equals infinity! The sideways eight on screen means infinity. In the picture, that is the dashed line at the very top, where the energy is zero. This jump has the most energy, so it gives the shortest wavelength of the series. We call it the series limit. Every line of a series lies between these two: the first line and the limit, okay?

2:43Series limits of hydrogen: put n2 = infinity

Now, let's find this limit. Put n two equal to infinity. What happens to one over n two squared? Yes, it goes to zero! One over a huge number squared is practically nothing. Here the arrow means tends to. So that whole second term just drops out. So the bracket is only one over n one squared. Nu bar limit, the small word limit written below, equals R H times Z squared, over n one squared. And flip it for the wavelength: lambda limit equals n one squared, over R H times Z squared. As simple as that. Now hydrogen, so Z equals one. Lyman lands on n one equals one. So lambda is one over R H: ninety one point two nanometres; the n m on screen is nanometres. See, this is the wall from our puzzle! Infinitely many Lyman lines, and not one of them goes below ninety one point two nanometres. Puzzle solved! Balmer lands on n one equals two, so the top becomes two squared, four. Four over R H: three hundred and sixty four point six nanometres. And Paschen lands on n one equals three: nine over R H, that is eight hundred and twenty point four nanometres. Keep these three limits handy. Clear?

4:08Topper's trick: cancel RH, compare the brackets

Now a trick that toppers use. To compare two lines, you never need the value of R H, okay? It is the same for both lines, so in a ratio, it cancels. Just compare the brackets: Z squared, times one over n one squared, minus one over n two squared. The sign like a stretched alpha means proportional to, okay? But careful here. If one line has the bigger bracket, is its wavelength longer, or shorter? Shorter, yes! The bracket gives nu bar, and lambda is one over nu bar. So lambda one over lambda two equals nu bar two over nu bar one. See, the small one and two swap places. The bracket ratio, flipped. Forget the flip, and you land on the wrong option, which the examiner has kept ready for you! And one more. If two lines, even from two different species, have the same Z squared times bracket, then they have exactly the same wavelength. The double arrow on screen means implies. Keep this ready for the practice problems. With me?

5:14Example 1: ratio of first Lyman and Balmer lines

Right, let's do example one. Find the ratio of the wavelengths of the first lines of the Lyman and Balmer series. Pause the video, and try it first. Done? Okay, let's check. First line means the smallest jump, right? Lyman lands on one, so its first line is two to one. R H cancels, so just the bracket: one minus one fourth, three by four. Balmer lands on two, so its first line is three to two. Bracket: one fourth minus one ninth, which is five by thirty six. So lambda L, with a small L for Lyman, to lambda B, with a small B for Balmer, is three by four, to five by thirty six, right? No, no, that is the trap! These brackets give nu bar, not lambda. Lambda is one over nu bar, so flip! Lambda L over lambda B equals Balmer's bracket over Lyman's: five by thirty six, divided by three by four. That is five by thirty six, times four by three: five by twenty seven. So the ratio is five is to twenty seven. Quick sanity check: the Lyman line comes out shorter. Makes sense, right? Its jump is bigger, so more energy, and a shorter wavelength. Exactly!

6:37Example 2: first Lyman line from the Balmer limit

Now a harder one, JEE level. The series limit of the Balmer series of hydrogen has wavelength lambda. Find the wavelength of the first Lyman line. Pause the video, and try it. Hint: write R H in terms of lambda. Done? Okay, let's check. First, the Balmer series limit. Which jump is that? Yes, from infinity, down to two! So one over lambda equals R H, times one fourth minus zero. That is R H by four. Turn it around: R H equals four over lambda, okay? So we never need the value of R H; lambda does the job. Now the first Lyman line. We just used infinity, so it is infinity to one again, right? No, no, careful! Infinity to one is the Lyman limit. The first line is the smallest jump: two to one. Lambda prime, lambda with the little dash, is its wavelength. One over lambda prime equals R H, times one minus one fourth: three by four R H. Now put R H equals four over lambda. Three by four, times four over lambda: the fours cancel, and we get three over lambda. So lambda prime equals lambda by three. One third of the Balmer limit, and not a single constant needed. Look, that is the power of the trick: cancel R H, and the question finishes itself.

8:12Quick revision: first line, series limit and ratios

So, quick revision. First line, the longest wavelength: the smallest jump, n one plus one, down to n one. Series limit, the shortest wavelength: infinity down to n one, and nu bar is R H times Z squared, over n one squared. And for ratios: cancel R H, compare the brackets, and flip for the wavelength. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!