Characteristic X-rays and Moseley's Law

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle. An X-ray tube gives a smooth spread of wavelengths, plus a few sharp spikes. Raise the voltage, and the smooth part shifts, but the spikes, once they appear, stay put. Change the target metal, and they jump. So if the tallest spike sits at zero point one nine four nanometres, which metal is in the tube? Today we learn characteristic X-rays and Moseley's law, which come up often in JEE exams. Let's begin.

0:32Where the sharp spikes come from

First, the figure: an X-ray spectrum, intensity against lambda, the wavelength. The smooth continuous spectrum starts at lambda min, with min written small, equal to h c over e V: Planck's constant times the speed of light, over the electron's charge times the voltage. On top stand two sharp spikes, labelled K alpha and K beta: a capital K with a small Greek alpha or beta below. The second graph comes later. Okay? Where do these spikes come from? Inside the target, a fast electron knocks an electron out of an inner shell, say the innermost K shell, leaving an empty place, a vacancy. Right? Let me ask you this: what will the atom do with that empty place? Yes, exactly, an electron from a higher shell falls into the vacancy, and gives out the energy difference as one photon. Got it? The shells of an element sit at fixed energies, so the photon has a definite energy, which gives a sharp line. We call these lines characteristic, because they belong to the target element. Clear?

1:39Naming the lines: K alpha, K beta, L alpha

Now, the names. The capital letter is the shell with the vacancy. K alpha, capital K with a small Greek alpha below, is L to K, as the arrow shows. K beta, with a small Greek beta, is M to K, and L alpha is M to L, okay? The K alpha energy is E K minus E L, capital E with a small K or L below. These are energies of the whole atom, with a vacancy in the K or the L shell. With me? Now a quick question. Beta comes after alpha, so the K beta line should have the longer wavelength, right? No, that is exactly the trap. M to K is a bigger fall, so K beta has more energy and a shorter wavelength. As the less-than signs show, lambda K beta is less than lambda K alpha, which is less than lambda L alpha, and K alpha is the most intense. Okay? Here is a neat link. M to K is M to L, plus L to K, so the energies add. Energy goes as one over lambda, so one over lambda K beta equals one over lambda K alpha, plus one over lambda L alpha. Clear?

2:50The lines ignore the voltage

Now back to our puzzle. Suppose I double the voltage of the tube. Does the K alpha spike move? No, the K alpha spike stays exactly where it is. The line comes from the shells of the target atoms, which the voltage cannot change. So the lines are fixed by the target, not by the voltage, as the note under the figure says. Okay? But there is one condition. The electron's energy, e V, its charge times the voltage, must exceed the binding energy of the inner shell, or those lines do not appear. Right? So what does a higher voltage change? Lambda min falls while the lines stay, so the gap, lambda K, a K line's wavelength, minus lambda min, increases. Got it?

3:36Moseley's law (1913)

Now Moseley's law of nineteen thirteen. Root nu equals a, times Z minus b in brackets. Nu, the Greek letter like a v, is the line's frequency, and capital Z is the atomic number; the letters a and b are constants for a given line. The second graph, root nu against Z, gives straight lines for K alpha and L alpha. Okay? For K alpha, the line meets the Z axis at b equal to one, and for L alpha, at seven point four. Why exactly one? Think about who is still in the K shell. Yes, exactly, it is the electron still left in the K shell. It screens one unit of the nuclear charge, so the falling electron feels an effective charge of Z minus one. Clear? So K alpha is a hydrogen-like jump from two to one, with Z minus one for the atomic number. One over lambda equals capital R, the Rydberg constant, times Z minus one, all squared, times one minus one quarter in brackets. With me?

4:44Topper's trick and the ratio rule

Now a topper's trick. Hydrogen's first Lyman line, two to one, carries ten point two electron volts, the e V on screen. K alpha is the same jump with Z minus one, so its energy is ten point two times Z minus one, all squared, in electron volts. That is easy to remember, right? Now compare two elements. Does the element with the bigger atomic number give a longer or a shorter K alpha wavelength? It is shorter, because the energy is bigger, so the ratio flips. Lambda one over lambda two equals Z two minus one, over Z one minus one, all squared; the small one and two name the elements. Watch the order here. Clear? Finally, this law showed that the atomic number Z, not the atomic mass, fixes an element's place in the periodic table. Okay?

5:38Example 1: K alpha energy and wavelength of copper

Now, example one. Use Moseley's law to estimate the energy and wavelength of the K alpha line of copper, with Z equal to twenty nine. Pause the video and try it first. Done? Okay, let's check. Copper has Z equal to twenty nine, so the energy is ten point two times twenty nine squared, right? No, that is the trap. Moseley's law uses Z minus one, because of the screening, so for copper it is twenty nine minus one, which is twenty eight. Clear? Twenty eight squared is seven hundred and eighty four, and ten point two times that gives about seven thousand nine hundred and ninety seven electron volts; the wavy equals sign means about. That is eight point zero kilo electron volts, the k e V on screen. Okay? Now the wavelength: lambda in nanometres, the n m, is one thousand two hundred and forty over the energy in electron volts. For seven thousand nine hundred and ninety seven, that gives about zero point one five five nanometres. Well done!

6:46Example 2: identify the metal from a 0.194 nm K alpha line

Now, remember our puzzle? The tallest spike is the K alpha line, so here is example two. The K alpha line of an element has wavelength zero point one nine four nanometres. Identify the element. Pause the video and try it first. Done? Okay, let's check. The energy is one thousand two hundred and forty over zero point one nine four, which is about six thousand three hundred and ninety two electron volts. Right? Now use the topper's trick backwards. Z minus one, all squared, is six thousand three hundred and ninety two over ten point two, which is about six hundred and twenty six point seven. Got it? Twenty five squared is six hundred and twenty five, very close, so Z minus one is twenty five. So the element has Z equal to twenty five, right? Careful here, because that is the last trap. We found Z minus one, not Z, so add the one back, and Z is twenty six, which is iron. Puzzle solved: the metal in our tube is iron. Well done!

7:56Quick revision: K and L lines, voltage rule, Moseley's law

So, quick revision. An inner vacancy is filled from above: K alpha is L to K, K beta is M to K, and M to L gives L alpha. The lines depend on the target, not the voltage, once e V beats the binding energy. And Moseley's law, root nu equals a, times Z minus b, gives K alpha ten point two times Z minus one, all squared, in electron volts. Now try the practice problems on this page. All the best!