 So, now that we have learned about X-rays is the time to practice few problems. So these couple of problems I have taken just to let you know what kind of questions could come. So, after you watch these videos please make sure you try at least 20 to 30 problems on your own. So, here is the first one. It says that a potential difference of 1000 volt is applied across X-ray tube. So, this is a potential difference between the cylinder and the copper. The ratio of the D Broglie wavelength of the incident electron to the shortest wavelength of the X-ray produced is what? Now, shortest wavelength of the X-ray will be produced when entire energy of the electron gets converted into a photon's energy. Because one electron can only create one photon, so if entire energy of one electron gets converted into one photon's energy, that is what the maximum energy is and if photon energy is maximum, frequency will be maximum and if frequency is maximum, wavelength will be minimum. So, for minimum wavelength we have entire kinetic energy which is E times V. Electron E is accelerated through a potential difference of V, so kinetic energy will be E times V. So, E times V will be equal to H C by lambda minimum. So, this will give us minimum wavelength equals to H C by E times V. So, this is the minimum wavelength of the photon. Now, let us try to find out what is the D Broglie wavelength of the electron. Now, D Broglie wavelength of an electron is given as what? H by momentum P. Now, momentum is not given but what is E times V? It is the energy. What kind of energy it is? It is kinetic energy. So, if kinetic energy is given, can I find out the momentum? Momentum can be written as 2 times m into kinetic energy under root of that, because kinetic energy is half mv square, so if you multiply 2 times m with half mv square, it becomes m square v square. So, take under root of that, you will get m into v which is momentum. So, D Broglie wavelength of the electron will become H by under root 2 mk. And what is K? K is E times V. So, we will get this as under root of 2 into mass of electron into charge of electron into V. So, this is D Broglie wavelength and this is minimum wavelength of the photon and the ratio of this to this is asked. So, now you can solve this your own. So, this is how you solve this particular question. So, here is the next question. In this question, what is given is K alpha wavelength and L alpha wavelengths are given. We need to find wavelength of K beta. So, let us call this as lambda 1, this as lambda 2 and this as lambda 3. What comes in your mind? That formula comes, that formula says 1 by lambda is Rieckberg constant z minus b whole square 1 minus 1 by n 1 square sorry and minus 1 by n 2 square. This comes in your mind. What is K alpha? K alpha is a radiation when electron transitions from second orbital to the first orbital. So, n 1 is 1 and n 2 is 2. So, 1 by lambda 1 is Rieckberg constant z minus b whole square 1 by 1 square minus 1 by 2 square like this. What is L alpha? L alpha is a radiation when the electron transitions from third orbital to the second orbital. Isn't it? So, 1 by lambda 2 is Rieckberg constant z minus b whole square 1 by 2 square minus 1 by 3 square. So, we need to find what is lambda 3? Essentially, I want to find out what is lambda 3 in terms of lambda 1 and lambda 2. So, let me first write down the expression for lambda 3 also. So, lambda 3 corresponds to K beta radiation. What is K beta radiation? It is a transition from third orbital to the first orbital. So, this is Rieckberg constant z minus b whole square 1 by 1 square minus 1 by 3 square. Isn't it? Now look at these three equations carefully. 1, 2 and 3. What do you think is a relation between lambda 1, lambda 2 and lambda 3? If you see closely, if you add 1 and 2, if you add 1 and 2, you will get equation number 3. Isn't it? This thing will get cancelled out when you add these two. So, you will get this equation. Right? So basically, what you get here is 1 by lambda 3 is equal to 1 by lambda 1 plus 1 by lambda 2. So, this will give me lambda 3 as lambda 1 into lambda 2 divided by lambda 1 plus lambda 2. So, this is the answer. So, in this question, we have learned a very important thing. Apart from the problem, this problem, what we have learned is that you do not need to find out, you know, do not substitute the value of Rieckberg constant and try to get right hand side. Okay? First you get what is asked in terms of variables. So, unnecessarily you will be putting value of R here and multiplying with this. Isn't it? But if you look at closely, you will get a very nice expression between lambda's. Okay? So, you can directly get lambda 3 as lambda 1, lambda 2 by lambda 1 plus lambda 2. So, you do not need to substitute value of constants and n 1 and n 2 like that. Okay? So, this is how you solve this particular question. We will take another question. Alright? Okay? So, let us solve this question. In this question, we need to find out frequency of k alpha line for z equal to 57 if frequency of k alpha line of z equal to 21 is given. Okay? So, what is given is frequency of atomic number 29 for k alpha and we need to find frequency of z equal to 57 for k alpha. Fine? Now, again, that we will start with the basics. 1 by lambda is Rittberg constant z minus 1, 1 by n 1 square minus 1 by n 2 square, right? For k alpha, n 1 is 1 and n 2 is 2, right? So, we will get 1 by lambda is 3R by 4, sorry, this is z minus 1 square. Screening constant has a value of 1 for k. Okay? So, 3R by 4 z minus 1 whole square. Fine? So, it is frequency that is given. So, I will multiply c both sides. So, c by lambda is 3R c by 4 z minus 1 whole square, right? This is the frequency. Okay? So, this will give me frequency for z equal to 29 is equal to 3R c by 4 29 minus 1 whole square. Fine? Similarly, when you put z equal to 57, you will get frequency for z equal to 57 equals to 3R c by 4 57 minus 1 whole square, okay? And when you divide left-hand side and right-hand side, this constant will go away. So, you will get frequency of z equal to 29 divided by frequency of z equal to 57 is equal to 28 square by 56 square. Okay? So, if you know this, you will get the value of frequency for z equal to 57 for k alpha radiation. Okay? Fine. I want to take a pause here and, you know, also touch base on few things. For example, here one more thing which you could have used is Mousselie's law, which says that root of frequency is some constant c into z minus b. Okay? We have derived this Mousselie's law for k radiation. Okay? What about for other type of radiation, l alpha, l beta and things like those? So, if you see that 1 by lambda is Rigberg constant z minus b whole square 1 by n1 square minus 1 by n2 square. If I multiply c both sides, I will get frequency equal to Rigberg constant c into z minus b whole square 1 by n1 square minus 1 by n2 square. Okay? Now, here Rigberg constant c and 1 by n1 and 1 by n2. Suppose n1 and n2 are fixed. We are talking about transition between two fixed shells. Right? So, again you will get from this, you will get z minus b is equal to some constant into root of frequency. Right? Or you can write it like that also. If you take c left hand side, you will get root of frequency is this. Okay? So, constant remains constant, right? 1 by c is, we can say, c1. So, this Mousli's law is true for any kind of characteristic wavelength for the x-ray. Fine? It is not just for the k. So, you could have used Mousli's law here as well. Root of frequency of first k alpha, sorry, first atomic number that is 29, root of that divided by root of frequency of this atomic number's frequency is equal to z minus 1, which is 28 by 57. So, when you square it, you will get the same equation. This by that is equal to 28 square by 56 square, that's the only thing, that's the same thing which you will get. Okay? So, this is all about the x-rays. Hope you like the video and make sure once you watch this video, you have to practice at least 20 to 30 questions to get comfortable with it. Okay? Thank you.