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Within three years after it begins operation, the proton beam at the Large Hadron Collider at CERN is expected to reach a luminosity of $10^{34} \mathrm{~cm}^{-2} \mathrm{~s}^{-1}$ (this means that in a $1-\mathrm{cm}^{2}$ area, $10^{34}$ protons encounter each other every second). The cross section for collisions, which could lead to direct evidence of the Higgs boson, is approximately $1 \mathrm{pb}$ (picobarn). [These numbers were obtained from "Introduction to LHC physics," by G. Polesello, Journal of Physics: Conference Series $53(2006), 107-116 .]$ If the accelerator runs without interruption, approximately how many of these Higgs events can one expect in one year at the LHC?

Hi there suffer this problem. We have a 70 m tera electron volts proton bunches that circulate in opposite directions as they've shown in this figure. And it has um As a radio of seven of 27 km. So the radios of this is seven At 27 km. And we are also given that and these large hadron collider as much together at bean crozes every 25 nanoseconds. So that's also an information for this problem. Nanoseconds is 10 to the -9 seconds. So the first question for this problem is um, how many bunches met every second? So to solve this problem, um we know that with one bunch For 25/9 seconds, the number of pounds she's per second is going to be the inverse of that. So I want to be One over 25/9 seconds. This can also be reading us 25 times 10 to the -9 um seconds. So, plugging this into the calculator, we obtain a value of four Times 10 to the minus Out to the 7th um bunches per second. Now for part B of this problem we are asked. So we are told that each bunch has 150 million, Protestants billion products that we have billion bra items. Okay. And Of this uh the department also tell us that typically 20 collide during each crossing. So we need to estimate the fraction of proteins that collided per crossing. So to do that description of proteins that collide are going to give it by 20 over the number of problems that Um the total number of protons which is 115 billions. So when we put this into the calculator, we obtain that The fraction of proteins acolyte is 1.7 Times 10 to the -10 products. So this is the fraction of Britain's direct collide now um for part C of this problem, we need to estimate how many collisions take place each second. So to solve that, um we want the number of collisions that occur age second. So using the information in the result that we obtain in part eight that give us for Times 10 to the seven bunches per second and We know now that they are 20 collisions per bunch. So when we multiply this, we obtain that they are approximately 800 million of collisions. So that's the solution for parsi of this problem. Now, party of this problem tell us that the bunches are 30 cm long and artist squeeze to a diameter of 20. So Party of this tell us that there is 30s and temperatures. Um lawn of the bunches and artist squeezed with a diameter of 20 micro meters. This is a die amateur. So we need to estimate the density of protons in a punch in units of proteins per millimeter square. So we want the protons density in a bunch with n the number of protons. So we're going to say that end is the number of problems in a slender yeah. And of length are and radios length. Our and radios art. So with this said, we know that the equation for the density is the number of protons over the pie are square Times L'd that for this lingers. So we will have 150 million of problems that we know are in each bunch. And we introduced the value for that. So we know that the diameter is 20 micrometers. So From this we can obtain the radios, which is the half of this value, which is 10 μm. So we introduced that in here 10 me crow meters squared and the length of this is given also for this problem areas 30 centimeters. We can convert this into meters and it's going to be 0.3 m. So from the density of this, we obtain going to put that in here. So it is 1.2 times 10 to the 12 protons per mil immature cube. So this is a solution for part uh the of this problem. Now, the final part of this problem part E tell us to estimate the density of hydrants in ordinary matter. So to do that. Um we follow the hint given for this problem. Yeah. Mhm. And we know that the mast Is 75 kg. Um We are assuming that for our body, the body is a salinger. If we assume that the body is a slumber of 30 cm in damage in diameter. So we are telling that a diameter in this case is 30 cm. Mhm. And we are assuming also that the length of the body is or the head of the body is going to be Going to call that ELT and it's going to be 1.75 majors. Um So the first thing that we can obtain is the volume of this body and we know that the volume of a cylinder is pi artist where times. Eld. So this will be pine times the radio. So we know that the radios can be obtained from here. So the radio is half this value of the diameter. So it is faith things and dimitar's. So we will have in here 50 cm that can be written in meters and it is 0.15 m square. And the length of the body is 1.75 m. So when we plot this into the calculator, we obtain that the volume of the body is 0.124 meters cube. Now the numbers of hard drums is going to be the mask over the the weight of the mass of the price of a product of one product. So we can obtain how many protons there is in a body. So we will have um 75 kg for the party. And we know that the Mass of a proton is 1.65, 7 times 10 two minus 27 kg. So when we plot this into the calculator, we obtain a value of 40 four points five times 10 to the 28 a drum. So that's the number of bedrooms in a body. So we know that the density can be obtain of the number of had drums over the volume. So we just do that number of hundreds over the volume. So we know all. And these two quantities, we know that the poly The number is 4.75 times 10 to the 80 28 Odd rooms. And this over the volume which is zero coin, 124 um meters cube. So this will give us for the density about you off 3 3.6 times 10 To the 90 to them. 29 um hum drones, perimeter Q. We can also convert this into the limiters cube, but just multiplying dad expression. So we will obtain that. This is going to be approximately four times 10 to the 20 hadron 30 millimeter. Give so that is the solution for part A of this problem. So this is it. Thank you.

Hey, the scattering off a crime and the pro TEM goingto arrested once and then going back to by bus, brutal and for But they want to find the lifetime of these rationales. There is doubt, a plus plus, knowing that the approximate with is 100. Maybe so we have felt I belt plus plus is 103 and we can find the the lifetime using the uncertainty principle. So we have four part, eh? Don't that t Delta plus plus is age bar 80 value by 45 They're dirty, not the plus plus. And then there's is just 4.135 times into the mine. 15 Blanton boats second, divided by four pine and 100 maybe times and ballistics troubles. And this gives us 3.3 time Stand to the manor, Stoney for seconds for part B one toe, right, The clerical position right in the diagram. So we have to be We want toe first, verify that the composition of the particle highly I amulets and then re creates a the bar. So we have the car composition for being bus must be going to delta was was way have you the bar. Bless you. Be going to you. You you so dearly but annihilate. And then this goes back to you. The bar. Thus you. You The drawing will be something like this. We have you. You be the bar you we have. It's less here that broken here. And then this goes like this. So you, you you. So this is Delta Plus. And when we draw something like this is because there's annihilate them It the omitted part will get reabsorbed by one of this Barth cruise. And then here we have you. You you going to you the bar? You you Then again have this. And then when we draw something like this is because only this part was omitted overto particle. And then this Create a pair off the the bar in the final state. So this will be the answer for a C.

This question asks us to write all the possible combinations that would result in a barrier on made up of ups downs or strange quirks. And the key idea here is that barry on is made up of three works and we're ignoring the anti Deloreans for now. So we can have an up Uh huh up pork. You could have a down down down and it's strange. Strange. Strange. You could have an up up down and then up up. Strange. Yeah. Okay. We could have an up down, down and then up. Strange. Strange. Okay, then we can have a down down strange and down. Strange. Strange. Yeah. And lastly, we can have Strange up down, sadly. This is 123456789, 10 possible combinations.

There are six Hadrian's with quantum numbers or we're going to charge up strange charm. Bottom for this be so let's label them A B C D E and F. So for the first one A it has a charge of two and up nous of one. A strange of zero uh charm of one and a bottom of zero. So no bottom courts, no strange courts and needs to have a charge of two. So one charm cork and two up courts report be a charge of zero and up nous of one. A strangeness of negative two. So what the strangest of negative two we know this is at least two strange ports. A charm of one. So that the only way to get a charm of one is to put a charm work in their part. C. So we have a strangest of one and a bottom nous of negative one. So a strangest of one means it's an anti strange work and a bottom nous of negative one means this is a bottom work for the next one we have up nous of negative one. Strangeness of 10 everywhere else in the charge must be zero. So we have an anti strange work and an anti up quark. The only way to make this be charge of zero is with an anti down pork. The next one charge of zero up nous of one. Strangeness of negative one. So we need a strange work in their charm of ones. We need a charm work in there and the bottom is zero. The only way to make this charge zero is with a down work. Last one you have a strangeness of negative three. The only way to get a strangeness of negative three is strange. Strange, strange looking at everything else and we still get an up nous of one and a charge of negative one.


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