Part A: JEE Main Level

Transcript

0:00Introduction

Hello everyone! First, one small puzzle before we start. In a hydrogen atom, the electron drops from level four to level three, and a photon of light comes out. So tell me: what colour will you see? Red? Violet? Hmm, everybody has a guess, right? Hold on to yours; we will settle it in the last problem today. This is the JEE Main part of the atomic structure problem bank. I will give you three quick ideas from it, and then two full problems. So, let's begin.

0:33Idea 1: Bohr orbits, watch the powers

Idea number one: Bohr's formulas. E n, with a small n below, is the energy in the n-th orbit: minus thirteen point six, into Z squared by n squared, in electron volts. Here n is the orbit number, and Z is the atomic number. And r n is the radius: zero point five two nine, into n squared by Z, okay? Its unit is the angstrom, the capital A with a small ring on top. Look, energy has Z squared, but radius has just Z, not Z squared. Clear? First, H, hydrogen, so Z is one, in the level n equals four. E four is minus thirteen point six by four squared, that is by sixteen: minus zero point eight five electron volt. Easy, isn't it? Now the radius of the third orbit of He plus, the helium ion, where Z is two. So n squared is nine, and Z squared is four: nine by four, right? No, no, careful! This is the JEE trap. Radius has Z, not Z squared. So r three is zero point five two nine, into three squared by two, that is into nine by two: two point three eight angstrom. Radius goes with n squared by Z; energy goes with Z squared by n squared. Never mix the powers! One more. Find the energy needed to remove the electron from n equals three of L i two plus, the lithium ion, where Z is three. Thirteen point six, into three squared, over three squared. One twenty two point four by nine: thirteen point six electron volt. See, Z squared and n squared are both nine, so they simply cancel. That's it!

2:10Idea 2: quantum numbers and nodes

Alright, idea number two: quantum numbers. Each set in brackets gives four of them, in this order: n, then l, then m l, and m s, the spin. The small l and s below the m are just labels, okay? And our question: which of these sets is not allowed? Here are four sets. The last number, the spin, is plus half or minus half. Now tell me, which one breaks a rule? The second one, yes! l runs from zero only up to n minus one; the three dots mean, and so on. In two, two, zero, plus half, l equals n, so it is not allowed. The others are all fine. Next: how many electrons at most can have n equal to three and l equal to two? l equal to two means a d subshell, so this is three d, with five orbitals. Two electrons in each, so ten electrons. Now the nodes of a three d orbital. Radial nodes are n minus l, so three minus two, one, right? Oh no, careful! n minus l counts the maxima, not the nodes. Radial nodes are n minus l minus one: three minus two minus one, zero. And angular nodes are just l: two. So a three d orbital has zero radial nodes and two angular nodes. Got it?

3:31Idea 3: ions lose 4s electrons first

Idea number three. Write the configuration of C u plus. Copper is the argon core, the A r in square brackets, then three d ten, four s one. The small raised numbers count the electrons in each subshell. Three d was filled last, so copper loses a three d electron first, right? No, no! Ions lose their four s electrons first. So C u plus is the argon core, then three d ten, a completely full d subshell. Nice and neat. Same rule for F e two plus: the four s electrons leave first, and three d six remains. Five d orbitals, six electrons. By Hund's rule, one goes into each orbital first, and only the sixth pairs up. So, four unpaired electrons. Now the spin-only magnetic moment. Mu equals the square root of n into n plus two, and here n is the number of unpaired electrons, not the shell. Root of four into six, root twenty four: four point nine zero Bohr magnetons, written B M on screen. With me?

4:32Example 1: spectral lines from n = 6 to the ground state

Now, example one. How many lines can a sample of hydrogen give when its electrons fall from n equals six to the ground state? Pause the video, and try it first. Done? Okay, let's check. Six to five, five to four, and so on, down to one. Five jumps, so five lines, right? No, no! Five is the most that one single atom can give. But a sample has a huge number of atoms, and they take different routes. From six, an electron can fall to five, four, three, two or one; the arrow means, falls to. Five lines already. Then, from five, it can fall to four lower levels; from four, to three; from three, to two; and from two, to just one. Total: five plus four plus three plus two plus one, fifteen lines. And the shortcut: n into n minus one, by two. Six into five by two: fifteen. Exactly! Remember: one atom gives at most five, but the sample gives fifteen.

5:35Example 2: wavelength of the 4 to 3 line of hydrogen

Now the harder one, and it is our opening puzzle. Find the wavelength of the four to three line of hydrogen. Pause the video and try it. Hint: start from the Rydberg formula. Done? Okay, let's check. Nu bar, the nu with a bar on top, is the wavenumber: one over lambda. It equals R H, the R with a small H below, the Rydberg constant, into one by three squared, minus one by four squared. The lower level goes first, so the bracket stays positive. That is one by nine, minus one by sixteen. Take one forty four as the common denominator: sixteen minus nine is seven. So nu bar is seven R H by one forty four. Clear? So the wavelength is seven R H by one forty four, right? Wait, careful! That is nu bar, one over lambda, not lambda itself. Flip it. Lambda is one forty four by seven R H, and R H is one point zero nine seven into ten to the power seven, so the answer comes out in metres. That is about one point eight seven five into ten to the power minus six metre: one thousand eight hundred and seventy five nanometres. The wavy equals sign means, approximately. Now, our puzzle. What colour is this light? It's a trick question! Visible light runs only from four hundred to seven hundred and fifty nanometres. One thousand eight hundred and seventy five is far beyond that: infrared, in the Paschen series, the lines that end at level three. So light does come out, but your eyes see nothing at all. Puzzle solved!

7:13Quick revision: Bohr powers, nodes, 4s first, line count

So, quick revision. One: radius varies as n squared by Z, and that sign on screen means, varies as. But energy varies as Z squared by n squared. Never mix the powers. Two: radial nodes are n minus l minus one, and ions lose their four s electrons before three d. Three: a sample falling from level n gives n into n minus one by two lines, and nu bar is one over lambda, so flip it at the end. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!