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Discuss and explain why it is that the maximum signal in apulse-acquire experiment is seen when the flip angle of the pulseis 90 degrees. What would you expect to s...

Question

Discuss and explain why it is that the maximum signal in apulse-acquire experiment is seen when the flip angle of the pulseis 90 degrees. What would you expect to see (how will it influencethe spectrum) in such an experiment if the flip angle of the pulsewas set to i) 180 degrees or ii) 270 degrees? What about 45degrees?

Discuss and explain why it is that the maximum signal in a pulse-acquire experiment is seen when the flip angle of the pulse is 90 degrees. What would you expect to see (how will it influence the spectrum) in such an experiment if the flip angle of the pulse was set to i) 180 degrees or ii) 270 degrees? What about 45 degrees?



Answers

Explain why magnetic flux decreases as the angle $\theta$ increases from 0 to $90^{\circ}$

Here in this problem, the average intensity of the light reaching the photo sale is given by accident. A two is equal to LDS P. two multiplication causes where you to death, which is equal to asbestos, courses square. Keep the death also delight falling on the photo sale is to have the same intensity in both experiment. So I can also write the expression edge. Just look at it carefully. I can write ask even death is equal to ask a to death, solving it further. So I can write the express an edge one x 2 as desk courses square feet to is equal to S. Death multiplication causes quite a little death. On further simplification, I can write one x 2 courses squared. Theta is equal to courses square to today's. On solving it further, I can read idea expression for today. S. Is equal to cause inverse under the hood, one by two multiplication and causes quite a 60 degree on solving it further, I get the value edge 69.3 degree. Now, the additional angle made by the analyzer can be given by dairy tita, which is equal to tita, does manage to to which is equal to 69.3° -60°,, which one simplification. I finally get the value of dirty to at 9.30 degree. So this is our final answer.

In this problem, we're gonna be looking at two experiments involving Mullah risers. In the 1st 1 we have a couple of polarizer is set up. 1st 1 will call that data zero just oriented vertically and the 2nd 1 Sorry. 2nd 1 is tilted at an angled data equals 60 degrees. And we have a NPO polarized light coming through here. So lights just scattered in all directions and after it hits polarizer then is going to be polarized along that transmission axis. So it starts off with some initial intensity travels through those poll risers, and we'll have some intensity when it comes out. The other stuff. That's experiment one Experiment two set up very similar way. We also have to polarize er's this 1st 1 being data equals Syria degrees, but the 2nd 1 now is tilted an unknown number of degrees from the vertical. We're not sure what this data is yet, but we do know that the light coming in in experiment to is already polarized. Polarized along the transmission access of the first polarizer, it has the same intensity as in Experiment one, but now it just has to be polarized before it reaches that first polarizer. And so what we want to do is figure out. What will this data be in experiment to so that the intensity coming through both pull risers and this experiment will just be the same as an experiment? Let's figure out what that intensity is for Experiment one Scott intensity as not unpressurized light after it passes to the polarizer. That intensity is now going t 1/2 of what its initial intensity waas. That's true. Regardless of what data is just because unpasteurized light, always half of it, will get through the polarizer. Neither half will now, so the intensity 1/2 the initial intensity is now incident on the second polarizer here. It doesn't matter what that angle thing is because our polarized light well reduce in intensity relative to that change in the transmission angle. And that's based on malice is Law, which, you may recall, is the intensity sequel to initial intensity. Times Co. Sign Squared data with data is the relative difference in the transmission ankles. So if we go ahead and plug in data equal 60 degrees like we know, that's not Times Co sign squared 60. I know that CO sign 60 degrees. It's just 1/2 so squaring it no sign square to 60 is 1/4. Sorry I lost my 1/2 here. This is 1/2 the initial intensity because that's what was instant on the second polarize, which was half of what we started with. So now we have 1/2 the initial intensity, and we know that coastline squads 60 is 14 So the intensity leaving the poll risers and experiment one is equal to one half times 143 times the initial intensity, or just 1/8 of its initial intensity. That's how much actually gets through both poor risers. That's what we want to set up an experiment to. We need angle data to be whatever we need to be, so that 18 of the intensity gets through. Well, let's follow it along its process and see I will set that up. It comes in already polarized. So after it passes through the first polarizer, its intensity does not change because sorry, lined up exactly the way it needs to be, so all of its intensity gets through. So now incident on the he second polarizer is the full intensity s non and using malice is law. We'll see that s equals. That's not Cosmin squared Data datas are unknown, but we want this s here. Be equal to 1/8 of the original intensity like we found in government one. So let's substitute that in 1/8 of the initial intensity, and that's gonna be equal to that's no co sign square data. You see that? Yes. Knots cancel each other out. So we just get that co sign. Square data equals 18 could take the square root of both sides now cause I data equals one over square root of eight. And then finally, we want to take the co sign inverse of both sides fine in vals both sides, and you'll see then that data. It's equal to the co sign in verse of one of our spirit of eight, which is about 69 point three degrees. Now the question is asking, How many additional degrees does the second analyzer have to be rotated? Well, when we compare that to the 1st 1 60 agrees, and 60 9.3 degrees we're going from 60 degrees, 69.3 degrees, the additional degrees that needs to be rotated additional rotation of nine point three degrees. No that they didn't even increased or decreased by that additional mount, I would say increase because it's going to a larger data from 60 to the large number of 69.3.

Okay. So those questions asking if we can look at the differences in detection methods for a pretty simple physical property being bond length and see if we can explain why different detection methods might give us different answers. And this is a very important question. The more detection methods and more properties you need to look into. So it uses the example B to H seven, an ion. And we're not gonna worry about Louis structure for this because it doesn't matter and it's kind of a nightmare. So let's go ahead and just look at the fact that we have a boron hydrogen bond on a terminal end. And so if we look at X ray scattering, we get a A bond length of about 103 Pekka meters. And if we look at neutron scattering, we get a length of 118 Pekka meters Pekka meters. Okay, so both of these are incredibly accurate methods of measuring bond length, Um, to the point that X ray diffraction is used as the way that we know how whole proteins look. Onda many other molecules and neutron scattering is probably more accurate, but it's very expensive. So we need to know. We can't say that one of them is just fundamentally less certain than the other. They're both very, very good, uh, methods of detection. So we need to examine why, for this example, we see a different length for this bond. So let's go ahead and look at that. If we have an Adam, here's the nucleus. Here's the core electrons, and we're gonna say there's another one here. Here's its core electrons and they're bound with their valence electrons, right? So, nuclei, I'll do this for nucleus for a neutron are going to come in and be deflected by the nuclei, right? And so, using this angle and this distance, which is not drawn proportionately. But this is going to be the same distance is this, and using this this'll angle is going to be the same. Using the difference in distance and angles and everything of different atoms, we can get the whole shape. It's very, very accurate and very, very good. And so nuclear I will bounce off of the, uh, new neutrons will bounce off of the nuclei, but if they passed through the electron field, they'll basically just go right through because they're not charged or anything. So if we do the same thing with, um X rays do this for X ray. Right? So they're going to come in and bounce off of the core electrons. Okay, Because while x rays are here, so this is nuclear notation and X rays are also zero. They're 00 gamma like that, and they have an Elektronik wave and a magnetic wave being radiation. So some of them will also go through the magnet, the electron, the electrons in the bonding area. But some of them will also be deflected, right? And so not very many will actually be deflected for this. But some of them will. Some of them will also be deflected from the nuclei. So we have these different ways that they're deflected. And remember the new the neutrons will Onley be deflected by the nuclear. That's part of what makes it so expensive. Because nuclear neutrons are much, much harder to control and simple radiation. Oscar ahead and redraw. This didn't want that to happen. Sorry. So Okay, so let's go ahead and look at why. That leads to a difference. Mhm. If we don't have core electrons like in hydrogen. Here's a Djetou. Then the Onley things that the electron that the X rays Kenbrell ounce off of are going to be the nuclei and the meat the mean electron field. Right. So that leads to this uncertainty when you don't have core electrons. And actually, the accuracy of X ray diffraction is going to be proportional to the the nuclear charge of so the nuclei, their place on the periodic table. So hydrogen having one is going to lead to the least accurate X ray diffraction. Um, possible. Really? You can't really have an X ray diffraction worse than for H two. And if we look at what boron Boron has a z of on Lee five, which is really not that much better. But it does have the one s and two s core electrons. So at least that Boren has that going for it. But you really can't do um, X ray diffraction for hydrogen very easily. So that's where this difference comes from in these two. It's just that hydrogen is incredibly difficult to utilize the x ray diffraction method for Okay, so that's really it. Thank you.

So to start with, the have two triangles here, too. Similar. Try and goes on. Dyken, start by saying that tan Avi is X over a in boat triangles. Now you can see that the acute angle A is much larger in the second triangle done in the first triangle. You can also say that going from the first triangle to the second triangle, the value of X becomes larger. So x value goes up on DDE, the value of a goes down so a becomes shorter in the second case. So if you're denominated denominator, it keeps going down and the new manager keeps going up the scenes that the value of tan eh is going to go up as a increases as angle increases. So the relationship is angle, eh? Or actually, the town of angle, eh is directly proportional toe angle, eh? This basically means that I


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