Production and the Continuous Spectrum

Transcript

0:00Introduction

Hello everyone! Here is a small puzzle to start. Two X-ray tubes run at the same thirty kilovolts. One has a tungsten target, and the other a copper target. Tungsten has a much bigger atomic number. So which tube's X-rays reach down to the shorter wavelength? Keep that question in mind; we will answer it today. We will see how X-rays are produced, and why their spectrum stops at a sharp limit, a topic that comes up often in JEE exams. Let's begin.

0:31Making X-rays: the Coolidge tube

So, what are X-rays? They are electromagnetic waves, just like light, but with a very short wavelength. Lambda, the Greek letter for the wavelength, is roughly zero point zero one nanometres up to ten nanometres; the wavy equals sign means about, and n m is nanometres. Röntgen discovered them in eighteen ninety five. Okay? How do we make them? In a Coolidge tube, a heated filament gives out electrons. They are accelerated through a potential difference, capital V, of tens of kilovolts, the k V on screen. Right? And all of this happens in vacuum. In a gas, the electrons would lose their energy in collisions before they ever reach the target. With me? Then they strike the target, a metal such as tungsten, molybdenum or copper. It needs a high atomic number, capital Z, because a heavy nucleus slows the electrons strongly and so gives more X-rays. It also needs a high melting point. Clear? Now let me ask you this: why such a high melting point? Where does the electrons' energy go? Yes, exactly, most of it becomes heat. About ninety nine percent of the energy turns into heat, and less than one percent comes out as X-rays. So the target gets very hot. Got it?

1:50The continuous spectrum

Now look at the figure below, at its left graph: intensity going up, against lambda, the wavelength, going across. There is a smooth curve marked continuous, with two sharp spikes on top, marked K alpha and K beta, a capital K with small Greek letters. The spikes, and the second graph, are for the next lesson. Okay? So where does the smooth part come from? Inside the target, an electron does not stop in one go. It slows down in many steps, and the radiation it gives out as it brakes is called bremsstrahlung, the long German word on screen, meaning braking radiation. Right? At each step, any part of its energy can leave as a photon, a big part or a small part. So the photons come in all sizes. Clear? Small photons mean long wavelengths, and big photons mean short ones. So all wavelengths appear, but only above lambda min, lambda with a small min written below. That is where the curve starts, on the left of the graph. With me? Now think. Which photon has the shortest wavelength of all? How much energy can it possibly carry? Yes, exactly, it is the photon that takes the electron's whole kinetic energy in one go. That energy is e V, the electron's charge, small e, times the voltage, capital V, and no photon can carry more. So no wavelength can be shorter. Okay?

3:20The cut-off wavelength: Duane–Hunt law

So, a photon's energy is h c over lambda, where h is Planck's constant and c is the speed of light. Set e V equal to h c over lambda min; the double arrow means this gives lambda min equals h c over e V, okay? This is the Duane Hunt law, and lambda min is called the cut-off wavelength. Got it? For numbers, put in the values of h, c and e. Lambda min is one thousand two hundred and forty over V in volts, in nanometres. Or, twelve point four over V in kilovolts, in angstrom, the capital A with a small ring on top. Both forms are worth remembering. Right? Now back to our puzzle. Tungsten has the bigger atomic number, so surely its X-rays reach a shorter wavelength, right? No, careful, that is the trap. Look at the formula: only V appears in it. So lambda min depends only on the voltage, not on the target, and not on the current. Both tubes have exactly the same lambda min. Clear?

4:28Intensity, quality and power

Next, how strong is the beam? The intensity is proportional to the tube current, small i; the sign like an open alpha means proportional to. More current means more electrons striking the target each second, so more X-rays. It also rises with the voltage. Okay? Now a quick question. If I raise the current, do the X-rays become more penetrating? No, they only get more intense. The quality, meaning the penetrating power, is decided by the voltage. Higher voltage gives harder X-rays, with shorter wavelengths. With me? Now two more results that come up in numericals. The number of electrons striking the target per second is i over e, the current over the electron's charge. And the power drawn is V times i. Got it? Let me test you. The voltage is doubled and the current is halved. What happens to lambda min, and to the number of electrons per second? Lambda min goes as one over V, so doubling the voltage halves it. The electrons per second go as i, so halving the current halves them too. So both are halved. Clear?

5:41Example 1: cut-off wavelength of a 30 kV X-ray tube

Now, example one, which settles our puzzle. Find the cut-off wavelength of the X-rays from a tube operated at thirty kilovolts. Pause the video and try it first. Done? Okay, let's check. Thirty kilovolts, so lambda min is one thousand two hundred and forty over thirty, about forty one nanometres, right? No, no, wait, that is the trap. This form needs V in volts, and thirty kilovolts is thirty thousand volts. Forty one nanometres is not even an X-ray. Okay? So lambda min is one thousand two hundred and forty over thirty thousand, which is about zero point zero four one three nanometres. Right? Let's check with the angstrom form: twelve point four over thirty gives about zero point four one three angstrom, the same answer. And it holds for any target, so our tungsten and copper tubes share exactly this cut-off. Well done!

6:41Example 2: electrons per second and heating at 50 kV, 20 mA

Now, example two, a little longer. An X-ray tube runs at fifty kilovolts and twenty milliamperes, the m A on screen. Find the number of electrons striking the target per second, the power drawn, and the rate of heating of the target, if one percent of the power appears as X-rays. Pause the video and try it first. Done? Okay, let's check. Twenty milliamperes is zero point zero two amperes. So i over e is zero point zero two over one point six times ten to the power minus nineteen, which gives one point two five times ten to the power seventeen electrons per second. Got it? Next, the power drawn, capital P, is V times i: fifty thousand volts times zero point zero two amperes, which gives one thousand watts, the capital W on screen. Clear? Now for the last part. So the whole one thousand watts heats the target, right? Not quite, so be careful here. One percent of the power leaves as X-rays, and one percent of one thousand watts is ten watts. With me? So the target heats up at one thousand minus ten, which is nine hundred and ninety watts. That is almost all of it, which is exactly why the target needs a high melting point. Well done!

8:03Quick revision: lambda min = 1240/V nm, voltage vs current

So, quick revision. In a Coolidge tube, electrons accelerated through V strike a target of high atomic number, and about ninety nine percent of their energy becomes heat. The continuous spectrum stops at lambda min, h c over e V, which is one thousand two hundred and forty over V nanometres, with V in volts. The voltage sets lambda min and the quality, and the current sets the intensity. Now try the practice problems on this page. You can do this. All the best!