Hydrogen-like Ions: A Drill

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle to warm up. Hydrogen holds its electron with thirteen point six electron volts: that is the energy you need to pull it out. Now take lithium two plus. It also has just one electron, but three times the nuclear charge. So, three times the energy, right? Hmm, not so fast! Today is a drill on hydrogen-like ions, and you will see the real answer is much bigger. And look, in JEE these are one-line questions, if you know exactly where the nuclear charge goes. So, let's begin.

0:35Step one: find Z, the nuclear charge

So, what is a hydrogen-like ion? Any species with just one electron, like hydrogen itself. Bohr's formulas work only for these one-electron species, okay? And capital Z is the charge on the nucleus. Clear? Now tell me. Lithium two plus: the small two plus up top is its charge. So, is Z equal to two? No, no! Look at the brackets. Hydrogen has Z one, the helium ion, H e plus, has Z two, and L i two plus has Z three. Its nucleus carries a charge of three. So remember: Z is the charge on the nucleus, not on the ion. The two plus only tells you the charge of the ion. Z is what goes into every formula, and that is step one, always. Got it?

1:25Step two: put Z in the right place

Step two: put Z in the right place. The radius of orbit n, r with a small n below, equals zero point five two nine, times n squared over Z, in angstrom: the capital A with a small ring on top. Z sits at the bottom, so more charge pulls the orbit in. The speed, v n, equals two point one eight times ten to the power six, times Z over n, in metres per second, written m s to the minus one. Here Z is on top: more charge, a faster electron. Alright, now the energy, and our puzzle. Radius and speed have just Z in them, okay? So does the energy go as Z, or as Z squared? Z squared, yes! E n equals minus thirteen point six, times Z squared over n squared, in electron volts, e V on screen. And here is the trap: radius and speed have Z, but energy has Z squared. Don't mix up the powers!

2:26Energy goes as Z squared: puzzle solved

So, the ionisation energy: the energy to pull the electron out completely. From the ground state, n equals one, it is thirteen point six times Z squared, electron volt. In the picture, hydrogen is the blue ladder on the left, and its n equals one level sits at minus thirteen point six. Now the helium ion, the orange ladder on the right. Z is two, so Z squared is four. Thirteen point six times four: fifty four point four. See its lowest level, way down at minus fifty four point four. Four times deeper! And now our puzzle. L i two plus: Z is three, so Z squared is nine. Thirteen point six times nine: one hundred and twenty two point four electron volts. Not three times hydrogen, but nine times! Puzzle solved! And if the electron starts higher up, in level n, just divide by n squared: thirteen point six Z squared, over n squared. Keep this line ready; the next drill needs it, okay?

3:28Quick drill: He+ in its first excited state

Now, a quick drill. Find the ionisation energy of the helium ion, in its first excited state. Fifty four point four electron volts, right? No, no, read the words carefully! The ground state is n equals one. The first excited state is the next level up: n equals two. With me? So, thirteen point six Z squared over n squared. For the helium ion, thirteen point six Z squared is fifty four point four, so we get fifty four point four over two squared, that is fifty four point four over four. Thirteen point six electron volts. Same as ground-state hydrogen! Now look at the picture: the helium ion's n equals two level, at minus thirteen point six, sits at exactly the same height as hydrogen's ground state. That is the double arrow marked equal. So this electron is just as hard to pull out as hydrogen's. Simple as that!

4:24Topper's trick: comparisons need only the powers

Now a trick that toppers use. To compare two ions, forget the constants and keep only the powers. The sign like a stretched alpha means proportional to. r goes as n squared over Z, v as Z over n, E as Z squared over n squared, and the time period, capital T, as n cubed over Z squared. Try this one. The electron's speed in n equals two of the helium ion, against n equals three of L i two plus. Which one is faster? Many students say L i two plus, because it has more charge. But look, only Z over n matters for speed, right? Helium ion: two over two, that is one. L i two plus: three over three, also one. Same Z over n, so the speeds are equal! Ten seconds, really easy, and not one constant needed. Well done if you got it!

5:19Example 1: radius of the first orbit of Be3+

Right, example one. Find the radius of the first orbit of B e three plus, the beryllium ion with charge three plus. Pause the video, and try it first. Done? Okay, let's check. Beryllium three plus. So Z is three, right? No, no, careful! Same trap as before. The three plus is the charge of the ion. Z is the nuclear charge, and for beryllium it is four. One electron is left, so it is hydrogen-like. First orbit, so n is one. r one equals zero point five two nine, times one squared over four: zero point five two nine over four. That is approximately zero point one three two angstrom; the wavy equals sign means approximately. More charge, a smaller orbit. Makes sense!

6:11Example 2: identify the -122.4 eV ion and its 2 to 1 line

Now a harder one, JEE level. Some hydrogen-like ion has ground-state energy minus one hundred and twenty two point four electron volts. Identify the ion, and find the wavelength of its two to one line: the arrow means the jump from n equals two down to n equals one. Pause the video, and try it. Hint: find Z first. Done? Okay, let's check. Ground state, so n is one, and the energy is minus thirteen point six Z squared. Put it equal to minus one twenty two point four. The double arrow means so: Z squared is one twenty two point four over thirteen point six, which is nine. So Z is nine, right? No, no! Nine is Z squared, not Z itself. Z is three, so the ion is L i two plus. Oh, and this is our number from the puzzle: one twenty two point four, nine times hydrogen! Now the two to one line. Delta, the Greek triangle, means change. Delta E equals one twenty two point four, times, in brackets, one over one squared, minus one over two squared. That is one twenty two point four times three by four: ninety one point eight electron volts. And the wavelength, lambda, is twelve forty over ninety one point eight, in nanometres: approximately thirteen point five nanometres. Remember, twelve forty over the energy in electron volts gives lambda in nanometres, from the photon chapter. Clear?

7:45Quick revision: find Z, then scale r, v, E and T

So, quick revision. Step one: find Z, the nuclear charge, not the charge of the ion. L i two plus has Z three. Then the powers: r as n squared over Z, v as Z over n, E as Z squared over n squared, and T as n cubed over Z squared. And the ionisation energy from level n is thirteen point six Z squared over n squared, electron volt. Now go and try the practice problems on this page. Drill them well, and these become sure marks in your JEE paper. All the best!