Why Are We Bombarded by Muons? Muons are unstable subatomic particles that decay to electrons with a mean lifetime of 2.2 ms. They are produced when cosmic rays bombard the upper atmosphere about 10 km above the earth’s surface, and they travel very close to the speed of light. The problem we want to address is why we see any of them at the earth’s surface. (a) What is the greatest distance a muon could travel during its 2.2@ms lifetime? (b) According to your answer in part (a), it would seem that muons could never make it to the ground. But the 2.2@ms lifetime is measured in the frame of the muon, and muons are moving very fast. At a speed of 0.999c, what is the mean lifetime of a muon as measured by an observer at rest on the earth? How far would the muon travel in this time? Does this result explain why we find muons in cosmic rays? (c) From the point of view of the muon, it still lives for only 2.2 ms, so how does it make it to the ground? What is the thickness of the 10 km of atmosphere through which the muon must travel, as measured by the muon? Is it now clear how the muon is able to reach the ground?

Video transcript

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let's read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A short lived particle created at an altitude of 38.0 kg meters has an average life span of 2.20 microseconds. The particle moves at a speed of 0.996 C I compute the maximum distance covered by the particle during its life span. I I, according to the frame of the particle, its life span is 2.20 microseconds determine the altitude. It has to measured relative to its frame. II I interpret what the result in part I and I I imply regarding the possibility of the particle covering the entire altitude of the earth's surface. OK. So we have three goals that we're trying to accomplish for this problem. The first one, we're trying to find the maximum distance covered by the particle during its lifespan. And then the second goal is to determine the altitude it has to descend, measured relative to its frame. And then the third goal is to, to take what we found in part one or part I and I, I imply regarding the possibility of the particle covering the entire altitude in earth space. So we need to determine if it's possible or not possible. OK. So we're given some multiple choice answers here. Let's read them off to see what our final answer set might be. A is I 657 m I I 3.39 kilometers II I, it is possible for the particle to hit the surface. B is I 657 m I I 3.39 kilometers II I, it is highly not possible for the particle to hit the surface. C is I 7.36 multiplied by 10 to the power of negative 15 m I I 7. kilometers II I, it is possible or it is highly not possible for the particle to hit the surface. D is I 7.36 multiplied by 10 to the power of negative m I I 7.35 kilometers II I, it is possible for the particle to hit the surface. OK. So let us start for solving for part I first. So let us recall and use the equation to determine the distance. And let's remember that we can find distance by saying that distance is equal to the velocity multiplied by the change in time. Delta T so let's plug in our known variables. So D is equal to the velocity which was 0.996 multiplied by C which C is the speed of light. And the numerical value for that is 3.0 multiplied by 10 to the power of 8 m per second multiplied by delta T which is the life span, which was 2. microseconds when we need to convert microseconds to seconds. So we need to multiply 2.20 by 10 of the power negative six. So when we plug it into a calculator, we should get 657 m. Which is our answer for part I, OK. So let's make a quick note here that our answer for I in kilometers would be 0. kilometers. OK. So to solve for part you, I'll make a note here. So this is for part I, so for part I I to solve for this, we need to use the equation for the contracted length. So the contracted lengths equation states that I is equal to I zero multiplied by the square root of one minus the velocity squared divided by the speed of light squared. OK? And I zero is 0.0 kilometers because this is the altitude. So I zero represents the altitude. So let's plug in our known values to solve for I. So I zero, like we said was 38 kilometers, 38.0 kilometers multiplied by the square root of one minus the velocity, which was 0.996 C squared divided by C squared note that the C squares cancel out. So the speed of light cancels out. So when we plug this into a calculator, we should get 3.39 kilometers. So that is our answer for part I I, OK. So now to solve for II I note that the particle has the ability to hit the surface if the contracted length is less than, than the result found in part I. So part I is the length moved during the entire lifespan of the particles. So 3.39 kilometers is greater than 0. kilometers. Thus it is highly not possible or the particle to hit the surface. A so we found her on the answer or part I OK. So going back up to look at our multiple choice answers, the correct answer has to be the letter B I is 657 m I I is 3. kilometers and II I, it is highly not possible for the particle to hit the surface. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.