IIT-JEE Questions (1982–1992): X-rays

Transcript

0:00Introduction

Hello everyone! Here is a question from IITJEE nineteen ninety two. Electrons accelerated through fifteen thousand volts strike a copper target in an X-ray tube. Find the speed of the X-rays inside the tube. Is there anything to calculate at all? Hold on to that thought. Today we solve the old IITJEE questions on X-rays, from nineteen eighty two to nineteen ninety two. Let's begin.

0:30First, the hook: how fast are the X-rays?

So, back to our first question. Many students find the speed of the electrons, root of two e V over m, as on the badge; e is the electron's charge, capital V the voltage, and m the electron's mass. Is that the speed of the X-rays? No, that is exactly the trap, so be careful. That formula gives the speed of the electrons, not of the X-rays. And X-rays are electromagnetic waves, just like light. Okay? So in the tube, which is a vacuum, their speed v equals c, the speed of light: three times ten to the power eight metres per second, the m s to the minus one on screen. So there is nothing to calculate, right? So the fifteen thousand volts only fix the highest photon energy, and so the cut-off wavelength, but never the speed. Clear?

1:23The continuous spectrum and its cut-off

Now, the left half of the figure shows the spectrum from an X-ray tube: intensity going up, and lambda, the wavelength, going along. The smooth curve is the continuous spectrum, and the sharp spikes are the K alpha and K beta lines. The right half, Moseley's plot, is not needed today. Okay? In the target, an electron loses energy in many steps, and any part can come out as a photon. The most energetic photon takes all of e V, so the curve starts at lambda min, lambda with a small min below, equal to h c over e V; h is Planck's constant and c the speed of light. In numbers, that is twelve point four over the voltage in kilovolts, in angstrom, the capital A with a small ring on top. With me? IITJEE nineteen eighty five asked this: describe the wavelengths present in the beam from an X-ray tube. What would you write? The beam has all wavelengths larger than a certain minimum. Nothing below lambda min, because no photon can carry more than e V, but everything above it. Right? Next, from nineteen eighty two: does the shortest wavelength depend on the current, the voltage, the gas in the tube, or the atomic number of the target? Yes, exactly, it depends only on the voltage. In h c over e V, the letters h, c and e are constants, so only the voltage is left. Not the current, the gas or the target, as the note under the figure says. Got it? Now nineteen eighty six: more electrons strike the anode, the target, every second. More electrons give more photons, so the intensity increases; lambda min stays put, since the voltage is unchanged. Okay? And nineteen eighty eight: the potential difference across the tube is increased. What happens to the intensity, and to the minimum wavelength? The intensity increases, and the minimum wavelength decreases, because e V sits at the bottom of h c over e V, okay? Also, a higher voltage means faster electrons, as the line says, so the second part of the nineteen eighty six question has the same answer: a smaller cut-off wavelength. Clear?

3:42Characteristic lines, treated as hydrogen-like

Now the spikes. When a fast electron knocks out an electron from the K shell, the innermost one, an electron from a higher shell falls into the gap and gives out one photon. K alpha, capital K with a small Greek alpha below, is the fall from L to K, that is, n equals two to one, where n is the principal quantum number. K beta, with a small beta below, is the fall from M to K, three to one; the arrows show the falls. Okay? So tell me, which photon carries more energy, K alpha or K beta? It is K beta, because three to one is the bigger fall. For hydrogen-like levels, the energy, and so one over lambda, is proportional to bracket one minus one over n squared; the sign like a stretched alpha means proportional to. More energy means a shorter wavelength. Right? But these lines need one condition. To knock out a K electron, e V must exceed the binding energy of the K shell, otherwise the K lines never appear. Got it?

4:48Example 1: tungsten K lines, minimum voltage and energy

Example one is from IITJEE nineteen eighty three. The binding energy of the innermost electron in tungsten is given as forty kilo electron volts, the k e V on screen. What must the accelerating voltage exceed to produce the characteristic X-rays, and what is the energy of the characteristic radiation, treating the levels as hydrogen-like? Pause the video and try it first. Done? Okay, let's check. To knock out that innermost K electron, what must the incoming energy, e V, beat? Yes, exactly, it must beat the binding energy. So e V must be greater than forty kilo electron volts, that is, the voltage must be greater than forty kilovolts, the k V on screen; the double arrow means implies. Clear? Now the radiation. The K electron was bound by forty kilo electron volts, so the photon also carries forty, right? No, this is where many students slip, so be careful. The electron that fills the gap comes from the L shell, n equals two, and it was bound too. Binding energy goes as one over n squared, so the L shell is bound by forty over two squared, that is, ten kilo electron volts. With me? So the K alpha photon gets only the difference: forty times bracket one minus one quarter, which is forty minus ten, equals thirty kilo electron volts. So the answers are forty kilovolts, and thirty kilo electron volts. If you got thirty, very good, you stayed out of the trap.

6:27Example 2: K beta wavelength from K alpha 0.32 angstrom

Now example two, from IITJEE nineteen ninety. The K alpha line of a hydrogen-like element has a wavelength of zero point three two angstrom. Find the wavelength of its K beta line. Pause the video and try it first. Done? Okay, let's check. Before calculating, should the K beta wavelength be longer or shorter than zero point three two angstrom? It must be shorter, because K beta carries more energy. Now the brackets: for K alpha, two to one, one minus one quarter is three quarters, and for K beta, three to one, one minus one ninth is eight ninths. Okay? Now the ratio lambda beta over lambda alpha; the small beta and alpha below each lambda name the lines. Many students write eight ninths over three quarters, as on the badge. Is that right? No, that is the trap again, so be careful. The brackets measure energy, and wavelength goes as one over energy, so the ratio flips. Lambda beta over lambda alpha equals three quarters over eight ninths, which is twenty seven over thirty two. Right? So lambda beta is zero point three two times twenty seven over thirty two. Zero point three two over thirty two gives zero point zero one, and times twenty seven, that is zero point two seven angstrom. And it is shorter, just as we predicted, so everything fits perfectly.

8:01Quick revision: cut-off wavelength, intensity, K lines

So, quick revision. X-rays travel at c, and lambda min, h c over e V, depends on the voltage alone. More current means more intensity; higher voltage means more intensity and a smaller lambda min. The K lines need e V above the binding energy of the K shell. And for hydrogen-like levels, the brackets are three quarters and eight ninths; flip them for wavelengths, since lambda goes as one over E, okay? Now try the practice problems on this page. All the best!