JEE Advanced Questions (2004–2018): Spectra

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. We shine light of just one wavelength, nine hundred and seventy angstrom, on hydrogen atoms in the ground state. How many lines come out in the emission spectrum? Just one, right? Well, it is not so simple. JEE Advanced asked exactly this in two thousand sixteen. So today, let us solve the JEE Advanced questions on spectra, from two thousand four to two thousand eighteen. Questions like these come up often, so let's begin.

0:34Jumps between levels give spectral lines

So, first, our toolkit; the diagram shows the energy levels of hydrogen. E n, with the small n below, is the energy of level n: minus thirteen point six times Z squared over n squared, in electron volts, the e V on screen. Here n is the principal quantum number, and capital Z is the atomic number. This holds only for atoms with just one electron, like hydrogen and hydrogen-like ions. Okay? Now, when the electron falls from an upper level n two to a lower level n one, a photon comes out; the arrow shows that jump. Tell me, how much energy does that photon carry? Yes, exactly, the difference between the two levels. h nu equals thirteen point six times Z squared, times bracket one over n one squared, minus one over n two squared, in electron volts. Here h is Planck's constant, and nu, the Greek letter like the letter v, is the frequency. And n one is always the lower level. Clear? Now the same jump in wavelengths. One over lambda, the Greek letter for wavelength, equals capital R times Z squared, times the same bracket, where R is the Rydberg constant. And for ions, do not drop the Z squared. Got it? Now the series. The lower level n one names it: lines ending on level one form the Lyman series, in the ultraviolet; on level two, the Balmer series, mostly in the visible; and on level three, the Paschen series, in the infrared. The diagram shows all three groups. With me?

2:11Absorbing a photon, then counting the lines

Now, what happens when light falls on an atom in the ground state? The photon is absorbed only if its energy, h nu, equals the gap between two levels exactly, E n two minus E n one. The one exception: a photon with at least the ionisation energy is also absorbed; it frees the electron, and the extra energy goes into the electron's kinetic energy. Okay? Then the excited atoms fall back, and every jump between two levels gives one line. So, if many atoms are raised to level n, how many different lines can come out? Every pair of levels gives one line. From n levels we can make n times bracket n minus one, over two, pairs, so that is the number of different lines. Right? Next, hydrogen-like ions, with one electron. Every energy is multiplied by Z squared, and every wavelength is divided by Z squared; those are the times sign and the division sign. Clear? And two Bohr model results for example two. Capital L, the angular momentum, equals n times h over two pi. And r n, the radius of orbit n, equals a naught times n squared over Z; a naught, the small a with a zero below, is the Bohr radius. With me?

3:28Example 1: lines emitted after 970 angstrom light (2016)

Now, remember our puzzle? Here it is, example one, from JEE Advanced two thousand sixteen: a hydrogen atom in its ground state is irradiated by light of wavelength nine hundred and seventy angstrom; the capital A with a little circle on top is the angstrom symbol. Take h c as one point two three seven times ten to the power minus six electron volt metre, and the ground-state energy as minus thirteen point six electron volts. Find the number of lines in the emission spectrum. Pause the video and try it first. Done? Okay, let's check. The energy E of one photon is h c over lambda: one point two three seven times ten to the power minus six, over nine hundred and seventy times ten to the power minus ten metre. That gives twelve point seven five electron volts. Okay? Now, will the atom absorb this photon? The energy must match a gap exactly. So, which level can it reach? We add the photon energy to the ground-state energy. Minus thirteen point six, plus twelve point seven five, gives minus zero point eight five electron volts. Right? But is that a real level? Minus thirteen point six over n squared equals minus zero point eight five, so n squared is sixteen, and n is four; the double arrow means implies. Yes, level four is a real level on the diagram, so the photon is absorbed, and the atom goes up to it. Clear? Now it falls back down. Many students write four to three, three to two, two to one, so three lines. Is that the answer? Oh no, careful, that is exactly the trap, because four, three, two, one is only one route down. The emission spectrum comes from many atoms, and different atoms take different routes, so every pair of levels gives a line. n times n minus one over two, that is four times three over two, gives six lines. And our puzzle is solved: one wavelength goes in, and six lines come out!

5:34Example 2: wavelengths of a hydrogen-like atom (2013)

Now a harder one, example two, from JEE Advanced two thousand thirteen. The radius of the orbit of an electron in a hydrogen-like atom is four point five times a naught, and its angular momentum is three h over two pi. Find the wavelengths it can emit on de-excitation, that is, while falling back down, in terms of R, the Rydberg constant. Pause the video and try it first. Done? Okay, let's check. We know neither the orbit nor the atom. So which fact tells us the orbit straight away? Yes, exactly, the angular momentum. Capital L equals n times h over two pi, and here it is three h over two pi, so n equals three. Okay? Many students simply take Z as one here. Is that really safe to do? No, it is not, because the atom is hydrogen-like, not hydrogen, so we must find Z from the radius. The radius is a naught times n squared over Z, which is nine a naught over Z, and this equals four point five a naught. So Z equals two, not one. Right? With Z equal to two, Z squared is four, so one over lambda equals four times R, times the bracket. From level three, the jumps are three to one, three to two, and two to one; the arrows show each jump. That is three lines, as n times n minus one over two says. Clear? First, three to one. The bracket is one minus one ninth, which is eight ninths; the dot means times. Four R times eight over nine gives thirty two R over nine. Flip it, and lambda equals nine over thirty two R, okay? Next, three to two. The bracket is one quarter minus one ninth, which is five over thirty six. Four R times five over thirty six gives five R over nine, so lambda equals nine over five R, right? And last, two to one. The bracket is one minus one quarter, which is three quarters, and four R times three quarters gives three R, so lambda equals one over three R, got it? The usual slip here is forgetting the four from Z squared, so watch for it. You did really well to get through this one!

7:48Quick revision: Rydberg formula and n(n-1)/2 lines rule

So, quick revision. One over lambda equals R times Z squared, times bracket one over n one squared, minus one over n two squared, and the lower level n one names the series. Next, a photon is absorbed only if its energy matches a gap exactly, or is enough to ionise the atom. And many atoms at level n give n times n minus one over two lines, so always find n and Z first. Now try the other JEE Advanced questions on this page. You can do them, so all the best!