Line Spectra and the Rydberg Formula

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Hydrogen has just one electron. Excite hydrogen gas in an electric discharge, and pass its light through a prism. One line? No! We see a whole set of sharp lines: in the ultraviolet, in the visible, and in the infrared. One electron, so many lines? But how? And one visible line is a bright red line. Can you find its wavelength, with just pen and paper? By the end of this video, you will. And look, almost every hydrogen spectrum question in JEE runs on one formula. So, let's begin.

0:43Line spectra: every element's fingerprint

So, first, what is a line spectrum? When the atoms of an element are excited, in a flame or in an electric discharge, they emit light, okay? Now pass this light through a prism. What will you see? Every colour, smoothly spread out, or only some? Only some, yes! Bright lines, at definite wavelengths. Lambda, the Greek letter on screen, is the wavelength. This is called an emission line spectrum. Okay? Now pass white light through the cold gas. Dark lines appear, at the same wavelengths! This is the absorption spectrum. So the gas absorbs at the very wavelengths where it can emit, right? And here is the beautiful part. Each element has its own set of lines, like a fingerprint. Completely unique! Now hydrogen. In a discharge through hydrogen gas, H two on screen, the molecules first break up into atoms: they dissociate. Then the excited H atoms emit the light. So the hydrogen spectrum comes from hydrogen atoms, not molecules, okay?

1:51The Rydberg formula

Alright, so where exactly do these lines sit? One formula gives all of them: the Rydberg formula. On the left, nu bar, the Greek nu with a bar on top: the wavenumber, one over lambda. It equals R H, R with a small H below, times Z squared, times the bracket: one over n one squared, minus one over n two squared. n one and n two, with the small one and two below, are two energy levels. Clear? Now look at the bracket. Tell me: why must n two be bigger than n one? Yes, exactly! Only then is the bracket positive. So n two is greater than n one, the sign on screen, and the line comes from a jump down, from level n two to level n one. So every line is one jump down. Got it? Next, R H, the Rydberg constant: one point zero nine seven times ten to the power seven per metre, written m to the minus one. Or one zero nine six seven seven per centimetre, c m to the minus one. Careful: with R H per metre, lambda comes out in metres, okay? And capital Z is the atomic number. For hydrogen, Z is one. The formula also works for hydrogen-like species, with one electron only: for the helium ion, H e plus, the small plus on top means it has lost one electron, Z is two. And Z comes in squared, so don't forget it, especially in H e plus questions, right?

3:25Five series: fix the lower level n1

Now, the picture. The horizontal lines are hydrogen's energy levels: n equals one at the bottom, at minus thirteen point six electron volts, up to n equals infinity, the dashed line at zero. Each arrow is one jump down, and each jump gives one line. So, our puzzle: one electron, but many levels and many possible jumps. That is why so many lines! Now look at the blue arrows on the left. Where do they all end? On n equals one, yes! Lines ending on the same lower level form one series. This is the Lyman series: n one is one, n two is two, three, and so on; the three dots mean the list continues. All in the ultraviolet, the U V on screen. The orange arrows end on n equals two: the Balmer series, with n two from three upwards. These are the visible lines. Our red line is one of them! Clear? The green arrows end on n equals three: the Paschen series, in the infrared, the I R on screen. Two more series are not drawn in the picture. Brackett: n one equals four, also in the infrared. And Pfund: n one equals five, in the far infrared. So look, the whole table is one rule: n one fixes the series, and n two runs above it. That's it!

4:52Topper's trick: the energy version

Now a trick that toppers use. Where does this formula come from? Bohr's energies! R H equals thirteen point six electron volts, the e V on screen, divided by h c, okay? Here h is Planck's constant, and c is the speed of light. So multiply the Rydberg formula by h c. On the left, h c nu bar is the photon energy, and on the right: thirteen point six, times Z squared, times the same bracket, in electron volts. Question in electron volts? Use this line, skip the wavelength. Much faster, you see?

5:30Example 1: first Balmer line of hydrogen, 656 nm

Now, example one, and it finishes our puzzle. Find the wavelength of the first line of the Balmer series of hydrogen. Pause the video, and try it first. Done? Okay, let's check. First line, so the smallest numbers: n one is one, and n two is two, right? No, no, careful! That jump is the first line of Lyman, not Balmer. Hydrogen, so Z is one, and in the Balmer series n one is always two. The first line is the smallest jump onto it: from n two equals three. The arrow means three down to two, okay? So nu bar equals R H, times one over two squared, that is one quarter, minus one over three squared, one ninth. One quarter minus one ninth is five over thirty six. So nu bar is five R H over thirty six. Now flip it: lambda is thirty six, over five times one point zero nine seven times ten to the power seven, in metres. Approximately six point five six times ten to the power minus seven metre; the wavy equals sign means approximately. In nanometres, the n m on screen: about six hundred and fifty six nanometres. Red light! See, this is the bright red line from our puzzle, called H alpha: H with a small Greek alpha. Just pen and paper, and we found it. Perfect!

7:00Example 2: wavelength of the second Lyman line

Now a harder one, JEE level. Find the wavenumber and the wavelength of the second line of the Lyman series of hydrogen. Pause the video, and try it. Hint: first decide the jump. Done? Okay, let's check. Lyman, so n one is one. Second line, so n two is two, right? No, no, careful! Two down to one is the first Lyman line, the smallest jump onto n equals one. The second line comes from the next level up: three down to one. And hydrogen, so Z is one. Okay? So nu bar equals R H, times one minus one ninth; one over one squared is just one, right? That gives eight ninths of R H. Put in R H: eight ninths, times one point zero nine seven times ten to the power seven. Approximately nine point seven five times ten to the power six, per metre. That is the wavenumber. And the wavelength is just one over nu bar: about one point zero two six times ten to the power minus seven metre, which is one hundred and two point six nanometres. Ultraviolet, U V, exactly as the Lyman table said. Perfect!

8:19Quick revision: Rydberg formula and the five series

So, quick revision. Line spectra: bright lines in emission, dark lines in absorption, a fingerprint for every element. Nu bar equals one over lambda equals RHZ squared, times one over n one squared, minus one over n two squared, with n two bigger than n one. And n one from one to five gives Lyman, Balmer, Paschen, Brackett and Pfund. Now go and try the practice problems on this page. All the best!