AIEEE and Earlier JEE Main Questions: Spectra

Transcript

0:00Introduction

Hello everyone! Let us start with a small puzzle. Hydrogen atoms falling from level two to one give out ultraviolet photons of ten point two electron volts, and this light falls on helium ions. But a helium ion in its ground state needs forty point eight electron volts for even its first jump, four times as much. So the helium ions can never absorb this light, right? Hold on to your answer. Today we solve older JEE Main questions on spectra, and this puzzle is one of them. Let's begin.

0:33Levels and the Rydberg formula

First, 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 Bohr result holds only for one electron: hydrogen and hydrogen-like ions. Okay? Now the electron falls from an upper level, n two, to a lower level, n one. Tell me, what decides the wavelength of the photon that comes out? Yes, exactly, it is the energy gap between the two levels. One over lambda equals capital R times Z squared, times bracket one over n one squared, minus one over n two squared. Lambda, the Greek letter, is the wavelength, R is the Rydberg constant, and n one and n two, with small numbers below, are the lower and upper levels. Clear? Now, the lower level, n one, names the series. Lines ending on level one form the Lyman series, on level two, the Balmer series, and on level three, the Paschen series. The diagram shows these groups of arrows, labelled ultraviolet, visible and infrared. Got it? In every series, the first line, with the longest wavelength, has n two equal to n one plus one. The series limit, with the shortest wavelength, has n two equal to infinity; that sideways eight is the infinity sign. With me?

2:08Ratios: let the constants cancel

Now, many of these questions give one wavelength and ask for another. For the same atom, capital R and Z squared are the same for both lines, so in a ratio they simply cancel. Okay? So the ratio depends only on the two brackets. But which way up? Is lambda two over lambda one equal to bracket two over bracket one, or bracket one over bracket two? It is the bracket of line one over the bracket of line two. Lambda sits in the denominator of the Rydberg formula, so the wavelength goes as one over the bracket; lambda one and lambda two carry small numbers below. Right? Now, a quick check. Lambda in nanometres, the n m, equals one thousand two hundred and forty over delta E in electron volts, where capital delta, the Greek triangle, means the gap. So a bigger gap gives a shorter wavelength. If your answer breaks this, you flipped the ratio. Clear?

3:01Absorption and hydrogen-like ions

Now, absorption. The photon is absorbed only if its energy, h nu, equals a gap exactly; h is Planck's constant, and nu, the Greek letter that looks like a v, is the frequency. The exception is a photon big enough to free the electron completely, which is ionisation. Okay? Next, hydrogen-like ions, with just one electron left. Every energy is multiplied by Z squared, and every wavelength is divided by Z squared; those are the times and division signs. Got it? Now, helium plus has Z equal to two, so its energies are four times hydrogen's. Can any of its lines match a hydrogen line exactly? Yes, it can, and here is a trick toppers use. The symbol H e with a small plus sign at the top is helium plus, helium that has lost one electron. Its line from two n two to two n one equals the hydrogen line from n two to n one; the arrow shows the jump. Why? Z squared gives a factor of four, and doubling both levels makes each n squared four times bigger, so the fours cancel. With me?

4:08Example 1: second Balmer line from the first (JEE Main 2020)

Now, example one, from JEE Main two thousand twenty. The first member of the Balmer series of hydrogen has wavelength six thousand five hundred and sixty one angstrom; the capital A with a little circle on top is the angstrom symbol. Find the wavelength of the second member in nanometres, the n m. Pause the video and try it first. Done? Okay, let's check. Balmer means the lower level is two, so which jumps give the first and the second members? The first member has n two equal to n one plus one, so it is three to two. Its bracket is one quarter minus one ninth, which is five over thirty six. Okay? The second member is the next line, four to two. Its bracket is one quarter minus one sixteenth, which is three over sixteen. Right? Many students now put the second bracket on top: three over sixteen, divided by five over thirty six. Is that correct? That one is upside down, so be careful, because this is the trap here. Lambda two over lambda one is five over thirty six, divided by three over sixteen. That works out to twenty over twenty seven. Clear? So lambda two is six thousand five hundred and sixty one, times twenty over twenty seven. Dividing by twenty seven gives two hundred and forty three, and times twenty gives four thousand eight hundred and sixty angstrom, which is four hundred and eighty six nanometres. It is shorter than the first line, as a bigger gap should give. Well done!

5:46Example 2: which He+ jump absorbs 10.2 eV (JEE Main 2019)

Now, remember our puzzle? Here it is, example two, from JEE Main two thousand nineteen. Radiation from the transition n equals two to one of hydrogen atoms falls on helium plus ions in the states n equals one and n equals two. Which transition of the helium ions can absorb it? Pause the video and try it first. Done? Okay, let's check. The hydrogen photon has energy thirteen point six times bracket one minus one quarter, which is ten point two electron volts. Okay? Helium plus has Z equal to two, so E n equals minus fifty four point four over n squared, in electron volts. For n from one to four, the levels are minus fifty four point four, minus thirteen point six, minus six point zero four, and minus three point four. Right? Now the ions in the ground state, n equal to one. Can they absorb a photon of ten point two electron volts? No, the ions in level one cannot absorb it. Even the smallest jump, one to two, needs fifty four point four minus thirteen point six, which is forty point eight electron volts. That is far too big, and the gap must match exactly. Clear? But some ions sit in level two. From two to three, the gap is thirteen point six minus six point zero four, which is seven point five six electron volts, not ten point two. With me? Now try two to four. Thirteen point six minus three point four gives ten point two electron volts, an exact match! Got it? So the answer is n equals two to n equals four: hydrogen's two to one line with both levels doubled, our topper's trick. Puzzle solved: ions in the ground state cannot absorb this light, but ions in level two can. Well done!

7:35Quick revision: Rydberg ratios, exact-gap absorption, He+

So, quick revision. One over lambda equals capital R times Z squared, times the bracket, and the lower level, n one, names the series. In a ratio, R and Z squared cancel, and the wavelength goes as one over the bracket. And a photon is absorbed only when it matches a gap exactly; helium plus, with both levels doubled, matches hydrogen. Now try the other JEE Main questions on this page. All the best!