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EDS spectroscopy spectra of a zeolite sample is shown below, what is the resolution of the spectra?40000SiK35000 Counts 30000AIK2500020000I5000I000050000.0051.0152....

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EDS spectroscopy spectra of a zeolite sample is shown below, what is the resolution of the spectra?40000SiK35000 Counts 30000AIK2500020000I5000I000050000.0051.0152.0 25 Energy (keV)3.0

EDS spectroscopy spectra of a zeolite sample is shown below, what is the resolution of the spectra? 40000 SiK 35000 Counts 30000 AIK 25000 20000 I5000 I0000 5000 0.0 05 1.0 15 2.0 25 Energy (keV) 3.0



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A miniature spectrometer used for chemical analysis has a diffraction grating with 800 slits/mm set $25.0 \mathrm{mm}$ in front of the detector "screen." The detector can barely distinguish two bright lines that are $30 \mu \mathrm{m}$ apart in the first-order spectrum. What is the resolution of the spectrometer at a wavelength of $600 \mathrm{nm} ?$ That is, if two distinct wavelengths can barely be distinguished, one of them being $600.0 \mathrm{nm},$ what is the wavelength difference $\Delta \lambda$ between the two?

The location for maximum for a grating cheese de scientist Ake was. Mm. Lambda What a maximum approximating Tetteh Toby X over d We have X equals I m them the uppercase G over the workers G. It is better to put these values as, um times lambda D over de. Now you're going to find the value off Lambda D over D for both the cutters. So for red, that's use 6 90 diameter screen is two meters away and D is given by 10,000 sleeps per centimeter. So that comes out to be 10 to the minus two mater divided by 10,000 so tended to five. So that comes out to be 6 90 times potential native, nine times do times 10 to the seven meter We jeez, 6.9 times two centimeter maker. So 10,000 seats per centimeter. Yeah, it's just sick. Here we have 6 90 nanometer. Two meters is the upper Cus D and lower case T ease one centimeter or 10 to minus two. Meager developer 10 to the 5 10,000 sits started tens of minus two meter. So this comes out to be this much meter, so over here. We tried to find two meters over them, buddy, over to D. So if we try to find that, then this becomes 6.9 meter for red for blue lander D over to D. We look about to be 4.6 major. So on the screen that ready pattern with looks like like this where this is at zero. This is at 6.9 meter, the second maximize a 6.9 times too. So that's 13.8. This point is three times 6.9, 20.7 meter this maximize four times 6.9 do any third by six meter. So for Abou, these numbers will be given. As for 16 times, two is 9.2 times three 13 burnt eight, 4.6 times for 18.4. So the blue will have ah maximum over here and at four point Think they're the A minimal. So that blue we'll be like days. That is 4.6. Then at nine point do there is a maximum. So 9.2 is somewhere over here. They're the maximum 9.2 on the minimum match. Separate starting 0.8 and then at 18.4 days, a maximum and then the next maximize at 23 at 23. Think that and this will be symmetric with respect to the center of the state.

Hello students in this question we have high resolution grading of to 60 to 60. White. Okay. And 300 lines per mm. That is a small and equals to 300 lines per mm. And about at an angle theta equals two Toyota. It is equal to 75 degrees. Okay, so and next the resolving power are it is given by it is given as 10 to the power six. For wavelength lambda equals 2 500 nanometers. So we have to find its free spectral range. Okay, so a spectral range we have to determine. That is delta lambda we have to determine. Okay. No uh how do these values of R and delta lambda F S. R. That is free spectral line. Okay. A professor compared to those fabric period having one centimeter air gap. Okay so we have to compare that so we can right here that the we can right here that are it is equal to m molecular end. So which is given as 10 to the power six. And also and which is equal to 78 molecular. And to the power three. So we can calculate M. Which will be equal to 10 to the power six divided by 78 Mercury by 10 to the power three. Okay, so this lame left delta lambda, F. S. R. It will be equals two lambda divided by M. C. Lambda. Which is given as 500 nanometers. So 500 molecular orbital to the power minus nine m, develop A. M. Which is 10 to the power six divided by a 78. 10 to the power three. So we can calculate that this delta lambda F. S. R. It is equal to the 39 nano meter. Okay, so this becomes answer for this problem. Okay, now next we can calculate that are which is equal to F, marker B M. Okay, so F. and M. This can be written as to be F marker B D divided by lambda. Okay. And this is equal to 10 to the power six multiplayer beta, delta lambda, F s R. So we can solve this value. So we will obtain that this value are it is equal to the 0.125 nanometer after substituting the given values. Okay, Thank you.

And this problem will be looking at X ray. Emission lines specifically will be looking at what are called KL FA Mission lines, which are when an electron transitions to the innermost shell. So we'll be looking at three different atoms, which are emitting X rays, calcium, cobalt, cadmium with their corresponding atomic numbers listed there. So in order to do this will need a couple relationships. We need what's called Moseley's law. Moseley's law, which you can find is goes like this. It's that the observed frequency which will call new, is gonna be equal to some constant a times that atomic number Z minus b squared. And in this case, since we're talking about Kael faux mission lines, B is equal to one, so we can just right. Ah, we can write this as one when we're using it. Okay, so we'll use this and another thing will use. There's two more things will use actually will use the fact that the speed of light is equal to frequency times. Lambda wavelength. Um, and the last thing we need is the fact that the energy is gonna be equal to Plank's constant times frequency. Okay, so what we want to do now is find three things for each of these atoms and their missions. We want to find the frequency of the mission. I want to find the wavelength of that emission and the energy. Okay, so starting with calcium, um, we have Z equals 20. So if we look at this expression on B is equal, the one that we know that this term is gonna be 19 is a constant what you confined. It's equal to 2.47 times 10 to the 15th hurts. Okay, All right. So with this, we have everything we need. Start calculating values. So let's do this for the first case for calcium, we have that f is gonna be equal to a which I defined here times 20 minus one squared. So it's 19 squared. So this is gonna give you a frequency of 8.92 times 10 to the 17 hurts, okay? And we also need the wavelength so you can get that by dividing over, um, to get lambda so see, divided by f. And that's gonna give you a value of 3.36 times 10 to the negative 10 m and lastly, from this expression, we know that we can get E by taking f and multiplying by plane sconce in which is 6.63 times 10 to the negative 34. So if we go ahead and do that, you're gonna get that e for this calcium extra mission is going to be equal to 3.7 K a V, which you can get it in kv by getting which the result in jewels and then dividing by one electron fold and then converting it to, uh, ktv. Alright, so that's our first results for the next one. We have cobalt, so it's essentially the same same process. Um, I'm not sure what the I think it might be CEO, but I'm not certain. Um, so in the same way we use a this time Aziz Eagle 27. So it's gonna be 26 squared 27 minus one. And so this will give us a frequency of 1.68 times 10 to the 18 hurts. Similarly, weaken do lambda in the same fashion as we did before, and this value is gonna be 1.79 times 10 to the negative 10 m and lastly, multiplying by plank's constant multiplying frequency times Plank's constant. We're gonna get the energy, which is 6.9 K e V in this case. Okay, so there's our second result. We have one more to do. Um, last one is cadmium. So for cadmium rz value is equal to 48. So for cadmium, uh, again, I don't know what this symbol is, but F is equal to a times 48 minus one. So 47 squared, which will give us five. It's kind of ugly. 5.48 times 10 to the negative or tend to the 18 hurts. Okay, next, Lambda same ways before we get 5.47 times 10 to the negative 11 m and multiplying our frequency times plank's constant. We get our energy, which you might have predicted is gonna be higher than last time. Since we have a higher frequency corresponding to higher energy, we're going to get 22 0.7 k e v. So those are the three Adams that we looked at here and we found the frequency lambda are wavelengths and the energy of the missions for each one

For the following were asked to find the frequency the energy in terms of TV and the wavelength. So frequency is given by the mostly is law, which says that frequency F is equal to 2.48 times 10 of the 15 hurts times the atomic number Z minus one squared. Once we know frequency, we can get energy cause energy E is equal to plank's constant times the frequency. Once we know the frequency, we can also get the wavelength is wavelength is equal to the speed of light see divided by the frequency. So for Z equals 20 we simply have to plug in 20 for a value of Z into the equation for frequency. And in this case, we find that the frequency is equal to 3.7 times 10 to the third hurts, which is the same as 3.7. Excuse me, 3.7 times 10 to the third electron volts, which is the same as 3.7 k e b. Kill alike Crumbles were asked to put things in units of Cavey Sylhet. Write it in terms of Katie next calculating the energy. The energy is equal to Plank's constant, which is equal to 4.136 times 10 to the minus 15 Electron voles seconds. Um, times the frequency. Oh, I'm sorry. And I made a mistake. So I said Frequency is 3.7 k e v. That's actually the energy. I got a little ahead of myself, and you probably noticed that when you plug that in that that didn't work so again. Calculating frequency 2.48 10% to 15 hurts. Times e minus one z is 20. So that's his 20 minus. One squared gives you 8.95 times 10 to the 17 hurts. Sorry about that. So 8.95 times 10 to the 17 hertz is the frequency that we're dealing with here. So now that we know frequency unions of hurts, we can calculate the energy because energy is equal to Plank's constant times, the frequency plank's constant is 4.136 times 10 to the minus 15 electron volt seconds. So this gives us our 3.7 times 10 to the third electron volts or 3.7 ked when we can also calculate the wavelength from the frequency. Since way blank. Lando's equal to the speed of light, divided by the frequency. The speed of light is three times 10 80 meters per second, so this it gives us 3.35 times 10 to the minus 10 meters. American box set in as their solution for a the final part. Now for B, we're doing the exact same thing. But this time Z is equal to 27. So we plug in 27 to our equation for frequency for our Z value. And this gives us a value for frequency equal to 1.68 10% of the 18 hurts Now using from the frequency we can get, the energy energy is equal to plank's constant times the frequency. So this gives us a value of 6.93 times 10 to the third electron volts or 6.93 K B Linkenbach. Sudden is the second part is our solution and lastly, the wavelength wavelength is equal to speed of light, divided by the frequency three times, 10 to the eight meters per second for the speed of light. Then we get 1.79 times 10 to the minus 10 meters and lastly, when Z is equal to 48 plug ins equals 48 our equation for frequency. In this situation, we find that the frequency is equal to 5.48 times 10 to the 18 hurts. Oh, good. So now that we know are frequency weakening the energy and wavelength again, Energy is plank's constant times, the frequency so plugging in our new value for the frequency. Into that expression, we find that the energy is equal to 22.7 times 10 to the third electron volts or 22.7 ked. Lastly, the wavelength lambda is equal to the speed of light. Divide my frequency, Looking in the value for the frequency, we find that this is equal to, um 0.547 times 10 to the minus nine meters. And let's write this one in terms of animators. So this is 0.547 nanometers. We could also write the other ones in terms of nano meters if you want. Since nanometers is 10 to the minus nine meters, we wrote him in units of meters. But either way is correct, as long as your expression for units is correct,


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