Mixed Problems: Set 2 (Bohr Model, Spectra and de Broglie)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Take a sample of hydrogen atoms, all in the ground state. We shine photons of twelve point seven five electron volt on them, and the atoms absorb them. So only one energy goes in, okay? Now the electrons come back down. How many different lines can the sample emit? Just one line, the same energy back again? Look, JEE Main often asks such questions: Bohr model, spectra and de Broglie, all mixed in one set. Today we pick up three quick tools from this set, and by the end, you will crack this puzzle in thirty seconds. So, let's begin.

0:45Tool 1: for ratios, keep only n and Z

So, tool one: ratios. For a ratio, forget all the constants. Keep only n, the orbit number, and capital Z, the nuclear charge. The radius, r, goes as n squared over Z; the sign like a stretched alpha means proportional to, okay? Try this one from our set. Find the ratio of the radius of the third orbit of L i two plus, the lithium ion, to that of the second orbit of H e plus, the helium ion. Quickly, what do you get? Right, n squared over Z for each. The brackets on screen tell you which ion and which orbit. L i two plus: n is three, Z is three, so nine over three on top. H e plus: n is two, Z is two, so four over two at the bottom. That gives three over two: the ratio is three is to two. As simple as that! Now a twist. The wavelength of the first Lyman line of H e plus is lambda. Find it for L i two plus. Same line in both ions, so only Z matters, as the screen says: lambda goes as one over Z squared. Z goes from two to three, so multiply by nine over four, right? No, no, careful! This is the JEE trap. Z squared sits at the bottom, so a bigger Z means a shorter wavelength. The arrow on screen means going from H e plus, where Z is two, to L i two plus, where Z is three. So multiply lambda by two squared over three squared. Four lambda by nine. Clear?

2:28Tool 2: energy tells you the level

Next up, tool two: energy. E n, with the small n below, is the energy of level n. It equals minus thirteen point six, times Z squared over n squared, in electron volts, e V on screen. And the kinetic energy, capital K, is just minus E n: plus thirteen point six, times Z squared over n squared, okay? Quick one from our set. Find the kinetic energy of the electron in the third orbit of L i two plus. Just look at Z squared over n squared. That's it! Z is three and n is three, so nine over nine, exactly one. So K is simply thirteen point six electron volt. See? No big calculation at all. Now a classic trap. Find the least energy needed to ionise an H e plus ion that is already in its first excited state. The ionisation energy of H e plus is fifty four point four electron volt, so the answer is fifty four point four, right? No, no, careful! Fifty four point four is from the ground state. First excited state means the first level above the ground state: n equals two, not n equals one, okay? For H e plus, thirteen point six times two squared is fifty four point four. So E two, with the small two below, is minus fifty four point four over two squared: minus thirteen point six electron volt. So the least energy to ionise is minus E two: just thirteen point six electron volt. Always read the words carefully. Got it?

4:09Tool 3: the electron's de Broglie wavelength

Alright, now de Broglie. Every moving electron has a wavelength, lambda. For an electron accelerated through a potential difference of V volts, lambda equals twelve point two seven over root V, in angstrom, the capital A with a small ring on top, okay? From our set: find the de Broglie wavelength of an electron of kinetic energy one hundred and fifty electron volt. But the formula wants volts. So what number goes in for V here? Just one fifty! An electron accelerated through one fifty volts gains exactly one fifty electron volt of kinetic energy. So lambda is twelve point two seven over root one fifty. Root one fifty is about twelve point two five, so lambda is about one point zero angstrom. The wavy equals sign means approximately. And JEE also asks it backwards: the potential needed for a wavelength of one angstrom. Twelve point two seven over root V equals one, so root V is twelve point two seven, and V is about one fifty volts. The same pair of numbers, just turned around. With me?

5:18Example 1: angular momentum when E = -1.51 eV in hydrogen

On to example one. The energy of the electron in a hydrogen atom is minus one point five one electron volt. Find its angular momentum. On the right, you can see the hydrogen energy levels, with the Lyman, Balmer and Paschen series; keep an eye on it. Pause the video, and try it first. Done? Okay, let's check. Tool two: energy tells you the level. Hydrogen, so Z is one, and minus thirteen point six over n squared equals minus one point five one. So n squared is thirteen point six over one point five one, which is about nine. Simple division, isn't it? Now Bohr's condition. The angular momentum, m v r, equals n h over two pi. Here m is the electron's mass, v its speed, r its orbit radius, and h is Planck's constant. We got nine, so nine h over two pi, right? No, no, careful! Nine is n squared, not n. So n is three, and the angular momentum is three h over two pi. And look at the picture: the level n equals three sits at minus one point five one, exactly our energy. Perfect!

6:34Example 2: 12.75 eV absorbed, how many lines are emitted

Now our puzzle from the start, JEE level. Hydrogen atoms in the ground state absorb photons of twelve point seven five electron volt. To which level are they excited, and how many lines can the sample then emit? Pause and try. Hint: energy tells you the level. Done? Okay, let's check. The energy absorbed lifts the electron from level one to level n, so it equals E n minus E one. For hydrogen, that is thirteen point six, times one minus one over n squared. Put it equal to twelve point seven five. So one over n squared is one minus twelve point seven five over thirteen point six. That ratio is zero point nine three seven five, so one over n squared is one over sixteen. n squared is sixteen, so n is four. The electrons go up to level four. Correct! Now the lines. The arrows on screen mean falls to. Coming down one step at a time: four to three, three to two, two to one. Three steps, so three lines, right? No, no, careful! This is the JEE trap. One atom takes one route. But a sample has many atoms, and different atoms take different routes, so every pair of levels gives a line. Count them: n times n minus one, over two. Four times three over two: six lines! And there you go, puzzle solved. One energy goes in, but six different lines come out.

8:11Quick revision: ratios, first excited state, n(n-1)/2 lines

So, quick revision. One: for ratios, keep only n and Z, nothing else. Radius goes as n squared over Z, and for the same line, lambda goes as one over Z squared. Two: energy tells you the level, and first excited state means n equals two. Three: for an electron, lambda is twelve point two seven over root V, in angstrom. Four: a sample falling from level n to the ground state gives n times n minus one, over two lines. Now go and try the three practice problems on this page. All the best!