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The energies for an electron in the $K , L ,$ and $M$ shells of the tungsten atom are $- 69,500 \mathrm { eV } , - 12,000 \mathrm { eV } ,$ and $- 2200 \mathrm { eV }$ , respectively. Calculate the wavelengths of the $K _ { \alpha }$ and $K _ { \beta } \mathrm { x }$ rays of tungsten.

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$K_{\alpha}-X$ -rays is $0.0216 \mathrm{nm}$the wavelength of the $K_{\beta}-x$ -rays is $[0.0184 \mathrm{nm}]$

Physics 103

Chapter 29

Atoms, Molecules, and Solids

Atomic Physics

Nuclear Physics

Cornell University

University of Washington

Simon Fraser University

University of Sheffield

Lectures

02:42

Atomic physics is the field of physics that studies atoms as an isolated system of electrons and an atomic nucleus. It is primarily concerned with the arrangement of electrons around the nucleus and the processes by which these arrangements change. The theory of quantum mechanics, a set of mathematical rules that describe the behaviour of matter and its interactions, provides a good model for the description of atomic structure and properties.

02:26

In physics, nuclear physics is the field of physics that studies the constituents and interactions of atomic nuclei. The most commonly known applications of nuclear physics are nuclear power generation and nuclear weapons technology, but the research has provided application in many fields, including those in nuclear medicine and magnetic resonance imaging, ion implantation in materials engineering, and radiocarbon dating in geology and archaeology.

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The energies for an electr…

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The energy of a K?electron…

02:57

The energy of a K-electron…

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08:35

The $\mathrm{K}$ series of…

05:31

The K-shell ionization ene…

09:14

The K series of the discre…

02:24

Use Coulomb's law to …

0:00

03:05

Use Coulomb"s law to …

05:33

09:32

Find the approximate wavel…

if a pretty fun, quick problem here for number 18. Yeah, we have. Ah, Tungsten Adam That has electrons in some given shells called the K the L in the shell. And there is an associate ID energy to each of those. Of course me. Hey, it is minus 69,000 500 electron volts. L is minus 12,000 electron volts and then em is minus 22 100 electron volts and we want to find the wavelength of the K Alfa in the gay better X rays. So to do this, then we want to find the transitions that associated with each of these bands or thes particular X rays of tungsten. So the K Alfa is going to be the l K transition, while the K better is the EMH to kay transition. So pretty much what we're going to do for both of these things is take find out first and associated energy difference for Let's table that for Alfa First, which is just e l minus e k. And that is 57,500 electron volts. And then we're going to display that in to a formal for the wavelength, which is HC over Delta E in this case, Alfa. And if you plug in it all, you get 0.216 nanometers for the wavelength. It's been an awful there the wavelength of the K Alfa Band, if you will. And I'm going to open up a new page here and there with me. You have to remember the numbers from earlier in the video. But then for the Cape Beta, I won't take a nice color for you. I have dealt E maeda, go to E m minus. Okay. And this is quite large plus 67,000 300 electron volts. So now you understand why I used the high energy blue and we do the exact same thing and we get 0.184 nanometers for the wavelength of that extra

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