JEE Advanced and IIT-JEE Questions (1996–2014): X-rays

Transcript

0:00Introduction

Hello everyone, here is a small puzzle for you. Double the voltage across an X-ray tube, and the shortest wavelength is halved. But the sharp K alpha line does not move at all. Why does one wavelength obey the voltage, while the other ignores it? Today we solve the JEE Advanced and IITJEE questions on X-rays, from nineteen ninety six to two thousand fourteen, and this idea comes up in several of them. So, let's begin.

0:31Two kinds of X-rays from one tube

So, the graph on the left shows intensity against wavelength, lambda. Electrons accelerated through a potential difference V strike a metal target, and lose their energy in many steps, giving a smooth, continuous spectrum. Now tell me, which photon has the shortest wavelength? Yes, exactly, the one that takes the whole kinetic energy, e times V, in one go. So lambda min, with the small word min below, equals h c over e V, where h is Planck's constant, c the speed of light, and e the charge of the electron. In numbers, it is twelve point four over V in kilovolts, in angstrom, the capital A with a little circle on top. Okay? Now look closely: there is no target and no current in this formula. So lambda min depends only on the voltage. Double the voltage, and lambda min is halved. Right? Next, the two sharp spikes, the characteristic lines. An incoming electron knocks an electron out of the innermost shell, the K shell, and an electron from the L shell falls into the gap. That gives the K alpha line, the capital K with the Greek letter alpha below; the arrow shows this fall. And K beta, with the Greek letter beta, is a fall from the M shell. Clear? Now, back to our puzzle. If we raise the voltage, does the K alpha line move? No, it stays where it is, because its energy is the gap between two energy levels of the target atom, which the voltage cannot change. So the lines are fixed by the target, not by the voltage, and our puzzle is solved! Got it? But there is a condition. The product e times V must be greater than the binding energy of the K shell, shown by the greater-than sign; otherwise no K electron is knocked out, and the K lines never appear. With me?

2:27Moseley's law for the K-alpha line

Now the graph on the right, Moseley's law. Root nu, the square root of the frequency, equals the constant a, times bracket Z minus b. Nu, the Greek letter like a v, is the frequency, capital Z is the atomic number, and the constants a and b are fixed for one series. So the graph is a straight line; the lower line is for the L alpha lines, and it cuts the Z axis further along, at b equal to seven point four. Okay? For the K alpha line, the graph cuts the Z axis at b equal to one. Why should b be exactly one here? Because one electron is still left in the K shell and screens one unit of nuclear charge, so the falling electron feels Z minus one, as in a hydrogen-like atom. One over lambda equals R times Z minus one, the whole squared, times bracket one minus one quarter, where capital R is the Rydberg constant; the bracket is the jump from level two to level one. Clear? Multiply by c, and we get the frequency. Nu equals three quarters times R times c, times Z minus one, the whole squared. Example two needs this form. Right? Now a quick question. For a heavier target, with a bigger Z, is the K alpha wavelength longer or shorter? It is shorter, because lambda goes as one over Z minus one, the whole squared. So lambda one over lambda two equals bracket Z two minus one, over Z one minus one, the whole squared; the small one and two label the two elements. Notice the flip, with Z two on top. Got it?

4:09Example 1: Cu to Mo K-alpha wavelength ratio (JEE Adv 2014)

Now, example one, from JEE Advanced two thousand fourteen. Find the ratio of the K alpha wavelength of copper, with Z equal to twenty nine, to that of molybdenum, with Z equal to forty two. Pause the video and try it first. Done? Okay, let's check. Lambda for copper over lambda for molybdenum, with their chemical symbols below, equals bracket Z of molybdenum minus one, over Z of copper minus one, the whole squared. The heavier element goes on top. Okay? Now, many students simply put forty two over twenty nine, and square it. Is that really right? No, that is exactly the trap, so be careful here. Moseley's law has Z minus one, not Z, because of the screening. So it is forty two minus one, which is forty one, over twenty nine minus one, which is twenty eight, the whole squared. Right? Forty one squared is one thousand six hundred and eighty one, and twenty eight squared is seven hundred and eighty four. Divide, and the ratio is about two point one four; the wavy equals sign means approximately. So copper, the lighter element, has the longer wavelength. Clear?

5:26Example 2: mass number and K-alpha frequency (IIT-JEE 2005)

Now a harder one, example two, from IITJEE two thousand five. X-rays are incident on a target metal whose atoms have thirty neutrons. The ratio of its nuclear radius to that of helium four is fourteen to the power one third; the small four on the top left is the mass number of helium. Find the mass number of the target atom, and the frequency of its K alpha line; capital R, the Rydberg constant, is one point one times ten to the power seven per metre. Pause the video and try it first. Done? Okay, let's check. The question starts with the size of a nucleus. So which formula links the radius of a nucleus to its mass number? Yes, exactly, the nuclear radius is R naught times A to the power one third, with capital A the mass number, and R naught, the R with a small zero below, a constant. R naught cancels, and helium has A equal to four, so A over four, to the power one third, equals fourteen to the power one third. With me? Cube both sides, and A over four equals fourteen, so A equals fifty six; the double arrow means implies. That finishes the first part. Okay? Now, for Moseley's law we need Z, the atomic number, and not the mass number. So what is Z for this atom? The mass number counts protons plus neutrons. So Z equals A minus N, where capital N is the number of neutrons. Fifty six minus thirty gives twenty six. Right? Now the second part, the frequency. Nu equals three quarters times R times c, times Z minus one, the whole squared; careful, this capital R is the Rydberg constant, not the nuclear radius. Z minus one is twenty five, and twenty five squared is six hundred and twenty five. So nu is three quarters, times one point one times ten to the power seven, times three times ten to the power eight, times six hundred and twenty five. With me? Work it out, and nu comes to about one point five five times ten to the power eighteen hertz. The usual slip is squaring twenty six in place of twenty five, so always use Z minus one. Well done!

7:48Quick revision: lambda min = hc/eV, fixed lines, use Z - 1

So, quick revision. Lambda min equals h c over e V, and it depends only on the voltage. The characteristic lines are fixed by the target, not by the voltage. And for K alpha, nu equals three quarters R c, times Z minus one, the whole squared; use Z minus one, not Z itself. Now try the other JEE questions on X-rays on this page. All the best!