The Hydrogen Spectrum

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. Excite hydrogen atoms so that they glow, and pass their light through a prism. You expect a full rainbow, right? Instead, you see just four sharp coloured lines on a dark background: one red, one blue-green and two violet. So why only four, and why these colours? One formula explains every one of them. Questions on this come up often in JEE, so let's begin.

0:31One jump down, one photon

So, look at the figure. Each horizontal line is an energy level, labelled by n, the principal quantum number, with its energy in electron volts, e V, on the right. The dashed line at zero is n equals infinity, the sideways eight. Each arrow is the electron falling to a lower level. So where does the lost energy go? Yes, exactly, it leaves as one photon. Its energy is h nu, where h is Planck's constant, and nu, the Greek letter like a v, is the frequency. So h nu equals the energy at level n two, minus the energy at level n one; the small numbers below are labels, and n two is greater than n one. Okay? Now put in Bohr's energy from the figure: E n, the energy of level n, equals minus thirteen point six times Z squared over n squared electron volts, with capital Z the atomic number, for hydrogen-like atoms with one electron only. So h nu equals thirteen point six times Z squared, times the bracket, one over n one squared, minus one over n two squared, in electron volts. Clear? In terms of the wavelength, lambda, the Greek letter on screen, this becomes the Rydberg formula, also in the figure. One over lambda equals capital R, times Z squared, times the same bracket, and capital R is the Rydberg constant. Got it? Its value is one point zero nine seven times ten to the power seven per metre, written m to the power minus one. And one over R is ninety one point two nanometres, n m on screen. Keep this number handy. With me? And here is a handy topper's trick. Lambda in nanometres equals one thousand two hundred and forty, over delta E in electron volts, where delta, the little triangle, means the change in energy. Energy to wavelength in one step, okay?

2:22A series is named by the level it lands on

Now, these jumps come in families called series, and the figure marks three: Lyman, Balmer and Paschen. So tell me, is a series named by the level the electron starts from, or the one it lands on? It is named by the level it lands on, the lower level, n one. Jumps ending on n equals one form the Lyman series. These are long drops, so the photons are in the ultraviolet. Right? Jumps ending on n equals two form the Balmer series, in the visible and near ultraviolet, okay? Ending on n equals three gives the Paschen series, in the infrared: the short arrows near the top. Clear? The figure leaves out two more: Brackett, ending on four, infrared, and Pfund, ending on five, far infrared. Got it? Now, inside one series, which jump gives the longest wavelength: the smallest jump, or the biggest? The smallest jump, from the very next level up, so n two equals n one plus one, gives the first line of the series. Less energy means a longer wavelength, so picking the biggest jump here is a common trap. With me? And the biggest jump, from n equals infinity, gives the shortest wavelength, the series limit. One over infinity is zero, so lambda limit, with limit written small below, equals n one squared over R times Z squared. Okay?

3:44Back to the four colours

Now, back to our puzzle. Lyman is ultraviolet and Paschen is infrared, so which series gives the four lines we can see? Yes, exactly, only Balmer. Visible light runs from four hundred to seven hundred nanometres, and only four jumps land there: from n equals three, four, five and six, down to two; the arrow means goes to. Right? H alpha, six hundred and fifty six point three nanometres, is red, and H beta, four hundred and eighty six point one, is blue-green. H gamma, four hundred and thirty four point zero, and H delta, four hundred and ten point two, are violet. Alpha, beta, gamma and delta are Greek letters, written small below the H, okay? So the puzzle is solved: the other Balmer lines are in the near ultraviolet. What we saw is an emission spectrum: bright lines on a dark background, from excited atoms. Clear? Pass white light, a continuous spectrum, through the gas instead, and you get an absorption spectrum: dark lines at the same wavelengths. Got it? Now, white light passes through hydrogen at room temperature. Which dark lines appear? Only the Lyman lines, in the ultraviolet. At ordinary temperatures all the atoms are in the ground state, n equals one, so every jump up starts from there. With me?

5:07Example 1: series and wavelength of the n = 4 to 3 line

Now our first example. To which series does the transition from n equals four to n equals three of hydrogen belong? And find its wavelength. Pause the video and try it first. Done? Okay, let's check. It starts at level four, and Brackett is the series of level four, so this is Brackett, right? That is exactly the trap, because a series is named by the lower level, where the electron lands. Here n one is three, so this is the Paschen series. Okay? For hydrogen, Z equals one. So one over lambda equals R times the bracket, one ninth minus one sixteenth, which gives seven times R, over one hundred and forty four. Clear? Flip it over, and lambda equals one hundred and forty four over seven times R, okay? Put in R, one point zero nine seven times ten to the power seven, and the fraction gives the wavelength in metres, the m on screen: about one thousand eight hundred and seventy five nanometres, where the wavy equals sign means approximately. That is infrared, as a Paschen line should be.

6:13Example 2: 3 to 1 line from the 2 to 1 and 3 to 2 lines

Now a harder one. The lines two to one and three to two of hydrogen have wavelengths one hundred and twenty one point six nanometres, and six hundred and fifty six point three nanometres. Without the Rydberg formula, find the wavelength of the line three to one. Pause the video and try it first. Done? Okay, let's check. Three to one is just three to two, then two to one. So add the wavelengths, and you get seven hundred and seventy seven point nine nanometres, right? That sum is the trap, because wavelengths never add; photon energies do. Each energy is h c over lambda, with c the speed of light, so h c over lambda three one equals h c over lambda three two, plus h c over lambda two one. The small numbers below name the jump. Okay? Cancel h c, and one over lambda three one equals one over six hundred and fifty six point three, plus one over one hundred and twenty one point six. Right? So lambda three one is six hundred and fifty six point three times one hundred and twenty one point six, over their sum. That gives one hundred and two point six nanometres, shorter than both, as a bigger jump should be.

7:27Quick revision: Rydberg formula, series names, energies add

Clear? So, quick revision. One over lambda equals R times Z squared, times the bracket, and lambda in nanometres is one thousand two hundred and forty over the energy in electron volts. Each series is named by the lower level: Lyman ends on one, Balmer on two, Paschen on three. Energies add, wavelengths do not, and cool hydrogen absorbs only the Lyman lines. Now go and try the practice problems on this page. All the best!