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.